1.1.1 Random Experiment

A random experiment is an experiment whose outcomes cannot be known with certainty. Instead, we assign probabilities to the various outcomes. All possible outcomes may be stated in advance, and the probabilities of the outcomes may be determined from experience.

1.1.2 Sample Space

The sample space is the set of all possible outcomes of a random experiment. It is usually denoted by \(S\).

Examples:

• Tossing a coin once: \(S = \{H, T\}\)
• Throwing a fair die once: \(S = \{1, 2, 3, 4, 5, 6\}\)

1.3.1 Discrete Random Variable

A random variable is said to be discrete if its possible values are countable (if its space is finite or countably infinite).

A set \(D\) is said to be countable if its elements can be listed, i.e., there is a one-to-one correspondence between \(D\) and the positive integers (e.g., Number of defective items in a sample of 20 items).

A discrete variable is one whose values are distinct from each other; the values are usually (but not necessarily) integers.

1.3.2 Continuous Random Variable

A random variable is said to be continuous if it can take uncountably many values (values in an interval).

Examples: Heights, weights, time.

If its cumulative distribution function \(F_X(x)\) is continuous for all \(x \in \mathbb{R}\), then \(X\) is continuous.

Let \(X\) be a random variable that can assume values only in the intervals \([x_1, x_2),\ [x_2, x_3),\ \ldots,\ [x_n, x_{n+1})\) with respective probabilities \(p_1, p_2, \ldots, p_n\), where \[P(x_i \leq X < x_{i+1}) = p_i,\quad i = 1, 2, \ldots, n\]

If \(\sum_{i=1}^{n} p_i = 1\), then \(X\) is a continuous random variable.

N/B: For continuous random variables we cannot assign probabilities to specific values as in the case of discrete random variables.

1.4.1 Properties of the Probability Mass Function

1. Non-Negativity Property \[P_X(x) \geq 0\] \[0 \leq P_X(x) \leq 1,\quad x \in D\]

Example: For a fair die, \(P_X(x) = \frac{1}{6},\ x = 1,2,3,4,5,6\). Each value satisfies \(0 \leq \frac{1}{6} \leq 1\).

2. Total Probability Property \[\sum_{x \in D} P_X(x) = 1\]

1.4.2 Example 1: Coin Tossed Three Times

Let the random variable \(X\) be the number of heads that appear. What is the probability distribution of the random variable \(X\)?

Solution:

\(X = \{0, 1, 2, 3\}\)

\(S = \{HHH, HHT, HTH, HTT, THH, THT, TTH, TTT\}\)

1.4.3 Example 2: Die Rolled Once

A die is rolled once. Let the random variable \(X\) denote the number facing up. What is the probability mass function (p.m.f) of \(X\)?

Solution: \[P(X = x) = \begin{cases} \frac{1}{6}, & x = 1, 2, 3, 4, 5, 6 \\ 0, & \text{elsewhere} \end{cases}\]

1.4.4 Example 3

Given \(f(x) = \frac{x}{10},\ x = 1, 2, 3, 4\), show that \(f(x)\) is a p.m.f.

Two conditions required:

• Non-negativity: \(f(x) \geq 0\) for all \(x\)
• Total probability: \(\sum_{\text{all }x} f(x) = 1\)

Condition 1: Non-negativity

For \(x \in \{1,2,3,4\}\), we have \(x > 0\), therefore: \(f(x) = \frac{x}{10} > 0\)

Condition 2: Total Probability = 1 \[\sum_{x=1}^{4} f(x) = \frac{1}{10} + \frac{2}{10} + \frac{3}{10} + \frac{4}{10} = \frac{1+2+3+4}{10} = \frac{10}{10} = 1 \checkmark\]

Both conditions are satisfied \(\Rightarrow\) \(f(x)\) is a p.m.f. \(\blacksquare\)

1.4.5 Example 4 — Bernoulli Distribution

Given \(f(x) = p^x(1-p)^{1-x},\ x = 0, 1,\ 0 < p < 1\), show that \(f(x)\) is a p.m.f.

Condition 1: Non-negativity

Since \(0 < p < 1\): \(p^x \geq 0\) and \((1-p)^{1-x} \geq 0\), therefore \(f(x) \geq 0\)

Condition 2: Total Probability = 1 \[f(0) + f(1) = p^0(1-p)^1 + p^1(1-p)^0 = (1-p) + p = 1 \checkmark\]

1.4.6 Example 5 — Binomial Distribution

Given \[f(x) = \binom{n}{x} p^x q^{n-x},\quad x = 0, 1, 2, \ldots, n\] where \(p\) is the probability of success and \(q = 1 - p\) is the probability of failure. Show that \(f(x)\) is a p.m.f.

Solution:

\[\sum_{x=0}^{n} \binom{n}{x} p^x q^{n-x} = \binom{n}{0}p^0 q^n + \binom{n}{1}p^1 q^{n-1} + \binom{n}{2}p^2 q^{n-2} + \cdots + \binom{n}{n}p^n q^0\]

\[= q^n + \binom{n}{1}pq^{n-1} + \binom{n}{2}p^2 q^{n-2} + \cdots + p^n\]

From the binomial formula: \((a+b)^n = a^n + \binom{n}{1}ab^{n-1} + \binom{n}{2}a^2 b^{n-2} + \cdots + b^n\)

\[\Rightarrow q^n + \binom{n}{1}pq^{n-1} + \binom{n}{2}p^2 q^{n-2} + \cdots + p^n = (p+q)^n = 1\]

Hence \(\displaystyle\sum_{x=0}^{n} \binom{n}{x} p^x q^{n-x} = 1\)

1.4.7 Example 6 — Poisson Distribution

Let \(X\) be a Poisson distributed random variable with \[f(x;\lambda) = \frac{e^{-\lambda}\lambda^x}{x!},\quad x = 0, 1, 2, \ldots \quad (= 0 \text{ elsewhere})\] where \(\lambda\) is a parameter. Show that \(f(x)\) is a discrete p.d.f.

Solution:

\[\sum_{x} f(x) = \sum_{x=0}^{\infty} \frac{e^{-\lambda}\lambda^x}{x!} = e^{-\lambda}\sum_{x=0}^{\infty} \frac{\lambda^x}{x!} = e^{-\lambda}\left(1 + \lambda + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \cdots + \frac{\lambda^n}{n!} + \cdots\right)\]

Since \(e^{\lambda} = 1 + \lambda + \frac{\lambda^2}{2!} + \frac{\lambda^3}{3!} + \cdots\):

\[\sum_{x} f(x) = e^{-\lambda} \cdot e^{\lambda} = e^0 = 1\]

Hence \(f(x)\) is a p.m.f.

1.5.1 Properties of a Continuous Random Variable

Let \(X\) be a continuous random variable with pdf \(f(x)\), then:

1. Non-negativity: \(f(x) \geq 0\) for all \(x\)

2. Total Area Property: \(\displaystyle\int_{-\infty}^{\infty} f(x)\, dx = 1\)

3. Probabilities: \(P(a < X < b) = \displaystyle\int_{a}^{b} f(x)\, dx\)

1.5.2 Example 1

Let \(X\) be the delay (in hours) of a flight with probability density function: \[f(x) = \begin{cases} 0.2 - 0.02x, & 0 \leq x \leq 10 \\ 0, & \text{otherwise} \end{cases}\]

(i) Show that \(f(x)\) is a pdf

At the endpoints: \(f(0) = 0.2\) and \(f(10) = 0.2 - 0.02(10) = 0\).

Since \(f(x)\) decreases linearly from \(0.2\) to \(0\), we have \(f(x) \geq 0\) for \(0 \leq x \leq 10\).

Verify total area equals 1: \[\int_{0}^{10}(0.2 - 0.02x)\,dx = \Big[0.2x - 0.01x^2\Big]_0^{10} = (0.2 \times 10 - 0.01 \times 100) = (2-1) = 1 \checkmark\]

(ii) Find \(P(X \geq 2)\)

First find \(f(2)\): \(f(2) = 0.2 - 0.02(2) = 0.16\).

Since the graph is a straight line, the region from \(x=2\) to \(x=10\) forms a triangle with base \(= 10-2=8\) and height \(= 0.16\):

\[P(X \geq 2) = \frac{1}{2} \times 8 \times 0.16 = 0.64\]

Alternatively, using integration: \[P(X \geq 2) = \int_{2}^{10}(0.2 - 0.02x)\,dx = \Big[0.2x - 0.01x^2\Big]_2^{10} = (2-1) - (0.4 - 0.04) = 0.64\]

1.5.3 Example 2

A continuous random variable has the following probability density function: \[f(x) = \begin{cases} k(x+2)^2, & 0 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}\]

(i) Find the value of \(k\)

Since \(f(x)\) is a p.d.f., the total area under the curve must equal 1: \[\int_{0}^{2} k(x+2)^2\,dx = 1\]

Expanding \((x+2)^2 = x^2 + 4x + 4\): \[k\int_{0}^{2}(x^2 + 4x + 4)\,dx = k\left[\frac{x^3}{3} + 2x^2 + 4x\right]_0^2 = k\left[\frac{8}{3} + 8 + 8\right] = k \cdot \frac{56}{3} = 1\]

\[\Rightarrow k = \frac{3}{56}\]

(ii) Find \(P(0 < X < 1)\)

\[P(0 < X < 1) = \int_{0}^{1} \frac{3}{56}(x+2)^2\,dx = \frac{3}{56}\left[\frac{x^3}{3} + 2x^2 + 4x\right]_0^1 = \frac{3}{56} \cdot 19 = \frac{19}{56}\]

(iii) Find \(P(X > 1)\)

\[P(X > 1) = 1 - P(0 < X < 1) = 1 - \frac{19}{56} = \frac{37}{56}\]

1.5.4 Example 3

A p.d.f is given by the piecewise function: \[f(x) = \begin{cases} k, & 0 \leq x \leq 2 \\ k(2x-3), & 2 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases}\]

(i) Find the value of \(k\)

Since \(f(x)\) is a p.d.f.: \[\int_{0}^{2} k\,dx + \int_{2}^{3} k(2x-3)\,dx = 1\]

\[k[x]_0^2 + k\big[x^2 - 3x\big]_2^3 = 1\]

\[k[2-0 + (9-9)-(4-6)] = k[2+2] = 4k = 1\]

\[\Rightarrow k = \frac{1}{4}\]

(ii) Find \(P(1 < X < 2.5)\)

Split the integral across the two ranges: \[P(1 < X < 2.5) = \int_{1}^{2}\frac{1}{4}\,dx + \int_{2}^{2.5}\frac{1}{4}(2x-3)\,dx\]

\[= \frac{1}{4}[x]_1^2 + \frac{1}{4}\big[x^2-3x\big]_2^{2.5}\]

\[= \frac{1}{4}(2-1) + \frac{1}{4}\big[(6.25-7.5)-(4-6)\big]\]

\[= \frac{1}{4}(1) + \frac{1}{4}(-1.25+2) = \frac{1}{4}(1) + \frac{1}{4}(0.75) = \frac{1.75}{4} = \frac{7}{16}\]

1.7.1 Solution to Exercise 2

Finding \(k\):

Since \(f(x)\) is a p.d.f.: \[\int_{-\infty}^{\infty} f(x)\,dx = 1 \Rightarrow \int_{0}^{1} kx^2\,dx = 1\]

\[k\left[\frac{x^3}{3}\right]_0^1 = 1 \Rightarrow \frac{k}{3} = 1 \Rightarrow k = 3\]

Finding \(a\) such that \(\Pr(X \leq a) = \Pr(X > a)\):

\[\int_0^a 3x^2\,dx = \int_a^1 3x^2\,dx\] \[[x^3]_0^a = [x^3]_a^1\] \[a^3 = 1 - a^3\] \[2a^3 = 1 \Rightarrow a = \sqrt[3]{\frac{1}{2}}\]

Finding \(b\) such that \(\Pr(X > b) = 0.05\):

\[\int_b^1 3x^2\,dx = 0.05 \Rightarrow \left[x^3\right]_b^1 = 0.05 \Rightarrow 1 - b^3 = 0.05\] \[b^3 = 0.95 \Rightarrow b = \sqrt[3]{0.95}\]

1.7.2 Solution to Exercise 3

Finding \(a\):

\[\int_0^1 ax\,dx + \int_1^2 a\,dx + \int_2^3(-ax+3a)\,dx = 1\]

\[\left[\frac{ax^2}{2}\right]_0^1 + [ax]_1^2 + \left[-\frac{ax^2}{2}+3ax\right]_2^3 = 1\]

\[\frac{a}{2} + a + \frac{a}{2} = 1 \Rightarrow 2a = 1 \Rightarrow a = \frac{1}{2}\]

Computing \(\Pr(X \leq 1.5)\):

\[\Pr(X \leq 1.5) = \int_0^1 \frac{1}{2}x\,dx + \int_1^{1.5}\frac{1}{2}\,dx = \left[\frac{x^2}{4}\right]_0^1 + \left[\frac{x}{2}\right]_1^{1.5}\]

\[= \left(\frac{1}{4} - 0\right) + (0.75 - 0.5) = 0.25 + 0.25 = 0.5\]

2.3.1 Example 1

Let the random variable \(X\) of the discrete type have the p.d.f.: \[f(x) = \begin{cases} \frac{x}{6}, & x = 1, 2, 3 \\ 0, & \text{otherwise} \end{cases}\]

Find the distribution function of \(X\).

Recall: \(F(x) = \displaystyle\sum_{t \leq x} f(t)\)

Solution:

\[F(x) = \begin{cases} 0, & x < 1 \\ \frac{1}{6}, & 1 \leq x < 2 \\ \frac{3}{6}, & 2 \leq x < 3 \\ 1, & 3 \leq x \end{cases}\]

Note: \(F(x)\) is a step function that is constant in every interval containing 1, 2, or 3, but has steps of height \(\frac{1}{6}\), \(\frac{2}{6}\), and \(\frac{3}{6}\).

2.3.2 Example 2

Given the distribution function for a random variable \(Y\):

\[F_Y(t) = \begin{cases} 0, & t < -2 \\ 1/3, & -2 \leq t < 1 \\ 7/12, & 1 \leq t < 5 \\ 47/60, & 5 \leq t < 11 \\ 57/60, & 11 \leq t < 20 \\ 1, & 20 \leq t \end{cases}\]

Find: (a) \(f_Y(t)\), (b) \(\Pr[0 \leq Y \leq 1]\), (c) \(\Pr[3 \leq Y \leq 10]\)

Solution (a): Finding \(f_Y(t)\)

To find \(f_Y(t)\), locate the points of discontinuity of \(F_Y(t)\): these are \(-2, 1, 5, 11, 20\). The value of the probability function at each point equals the size of the jump in \(F_Y(t)\):

Solution (b): \(\Pr[0 \leq Y \leq 1]\)

\[\Pr[0 \leq Y \leq 1] = F(1) - F(0) = \frac{7}{12} - \frac{1}{3} = \frac{1}{4}\]

Solution (c): \(\Pr[3 \leq Y \leq 10]\)

\[\Pr[3 \leq Y \leq 10] = F(10) - F(3) = \frac{47}{60} - \frac{7}{12} = \frac{47}{60} - \frac{35}{60} = \frac{12}{60} = \frac{1}{5}\]

2.4.1 Example 3

Let \(X\) be a random variable of continuous type defined by the p.d.f.: \[f(x) = \begin{cases} \frac{2}{x^3}, & 1 < x < \infty \\ 0, & \text{elsewhere} \end{cases}\]

Find the cumulative distribution function \(F(x)\).

Solution:

\[F(x) = \int_{-\infty}^{x} f(t)\,dt = \int_{1}^{x} \frac{2}{t^3}\,dt = \int_{1}^{x} 2t^{-3}\,dt = \Big[-t^{-2}\Big]_1^x = 1 - \frac{1}{x^2}\]

Therefore: \[F(x) = \begin{cases} 0, & x < 1 \\ 1 - \frac{1}{x^2}, & 1 \leq x \end{cases}\]

2.4.2 Example 4 — Finding the Density Function

Let \(X\) be the random variable whose distribution function is: \[F_X(t) = \begin{cases} 0, & t < 0 \\ t, & 0 \leq t \leq 1 \\ 1, & 1 < t \end{cases}\]

Find the density function of \(X\).

Solution: Differentiate \(F_X(t)\): \[f_X(t) = \frac{d}{dt}\left[F_X(t)\right]\]

\[f_X(t) = \begin{cases} 1, & 0 < t < 1 \\ 0, & t < 0 \text{ or } t > 1 \end{cases}\]

This is the Uniform(0, 1) distribution.

2.4.3 Example 5

Let \(X\) be a continuous random variable with p.d.f: \[f(x) = \begin{cases} \frac{1}{2}x, & 0 < x < 2 \\ 0, & \text{otherwise} \end{cases}\]

Obtain the c.d.f of the random variable \(X\).

Solution:

For \(x \leq 0\): \(F(x) = 0\)

For \(0 < x < 2\): \[F(x) = \int \frac{1}{2}x\,dx = \frac{x^2}{4} + c_1\]

Using \(F(0) = 0\): \(\frac{0^2}{4} + c_1 = 0 \Rightarrow c_1 = 0\)

Check: \(F(2) = \frac{2^2}{4} + 0 = 1\)

Hence the c.d.f becomes: \[F(x) = \begin{cases} 0, & x \leq 0 \\ \frac{x^2}{4}, & 0 < x < 2 \\ 1, & x \geq 2 \end{cases}\]

3.1.1 Percentiles

Let \(0 < p < 1\). The \(100p^{th}\) percentile (quantile of order \(p\)) of the distribution of \(X\) is a value \(\xi_p\) such that:

\[\Pr(X \leq \xi_p) \geq p \quad \text{and} \quad \Pr(X \geq \xi_p) \geq 1 - p\]

3.2.1 Example 1: Median and 25th Percentile

Find the median and 25th percentile of:

\[f(x) = \begin{cases} 3(1-x)^2, & 0 < x < 1 \\ 0, & \text{elsewhere} \end{cases}\]

Solution — Median:

Let the median be \(M\). Setting \(\int_0^M 3(1-x)^2\,dx = \frac{1}{2}\) and substituting \(z = 1-x\):

\[-(1-M)^3 + 1 = \frac{1}{2} \implies (1-M)^3 = \frac{1}{2}\]

\[\boxed{M = 1 - \sqrt[3]{\frac{1}{2}}}\]

Solution — 25th Percentile:

Let \(P\) be the 25th percentile. Setting \(\int_0^P 3(1-x)^2\,dx = \frac{1}{4}\):

\[(1-P)^3 = \frac{3}{4} \implies \boxed{P = 1 - \sqrt[3]{\frac{3}{4}}}\]

x <- seq(0, 1, length.out = 300)
fx <- 3 * (1 - x)^2
M <- 1 - (1/2)^(1/3)

df <- data.frame(x = x, fx = fx)
df_shade <- df[df$x <= M, ]

ggplot(df, aes(x, fx)) +
  geom_area(data = df_shade, aes(x, fx), fill = "#3B8BD4", alpha = 0.3) +
  geom_line(linewidth = 1, color = "#185FA5") +
  geom_vline(xintercept = M, linetype = "dashed", color = "#D85A30") +
  annotate("text", x = M + 0.05, y = 2, label = paste0("M = ", round(M, 3)),
           color = "#D85A30", size = 4) +
  labs(title = "PDF with Median Highlighted",
       x = "x", y = "f(x)") +
  theme_minimal()

Figure 3.1: PDF f(x) = 3(1-x)^2 with median shaded

3.2.2 Example 2: Median of a Discrete Distribution

Given \(p = \frac{1}{4}\), find the median of:

\[P_X(x) = \begin{cases} p(1-p)^x, & x = 0, 1, 2, \ldots \\ 0, & \text{elsewhere} \end{cases}\]

Solution:

p <- 1/4
x_vals <- 0:6
pmf <- p * (1 - p)^x_vals
cdf <- cumsum(pmf)

results <- data.frame(x = x_vals, PMF = round(pmf, 4), CDF = round(cdf, 4))
knitr::kable(results, caption = "PMF and CDF for Example 2")

From the CDF: \(F(2) = 0.5781 \geq 0.5\), so the median is \(M = 2\).

3.2.3 Example 3: Finding k, Median, and IQR

\[f(x) = \begin{cases} \frac{k}{\sqrt{x}}, & 1 \leq x \leq 9 \\ 0, & \text{otherwise} \end{cases}\]

(i) Finding \(k\):

\[\int_1^9 \frac{k}{\sqrt{x}}\,dx = k[\ln x]_1^9 = k\ln 9 = 1 \implies k = \frac{1}{\ln 9} \approx 0.4551\]

(ii) Median:

The CDF is \(F(x) = \frac{\ln x}{\ln 9}\). Setting \(F(m) = \frac{1}{2}\):

\[\frac{\ln m}{\ln 9} = \frac{1}{2} \implies m = 9^{1/2} = 3\]

(iii) Interquartile Range:

\[Q_1: F(q_1) = \frac{1}{4} \implies q_1 = 9^{1/4} \approx 1.732\]

\[Q_3: F(q_3) = \frac{3}{4} \implies q_3 = 9^{3/4} \approx 5.196\]

\[\text{IQR} = q_3 - q_1 \approx 5.196 - 1.732 = 3.464\]

k <- 1 / log(9)
q1 <- 9^(1/4); q3 <- 9^(3/4); iqr <- q3 - q1
cat(sprintf("k = %.4f\nMedian = 3\nQ1 = %.3f, Q3 = %.3f\nIQR = %.3f\n",
            k, q1, q3, iqr))

k = 0.4551
Median = 3
Q1 = 1.732, Q3 = 5.196
IQR = 3.464

3.3.1 Example 4: Mode of a Discrete Distribution

\[P_X(x) = \begin{cases} \left(\frac{1}{2}\right)^x, & x = 1, 2, 3, \ldots \\ 0, & \text{otherwise} \end{cases}\]

By observation, the maximum probability occurs at \(x = 1\), so Mode = 1.

3.3.2 Example 5: Mode of a Continuous Distribution

\[f(x) = \begin{cases} \frac{1}{2}x^2 e^{-x}, & 0 < x < \infty \\ 0, & \text{otherwise} \end{cases}\]

Solution:

\[f'(x) = \frac{1}{2}e^{-x}(2x - x^2) = 0 \implies x = 2\]

\[f''(2) = -0.1353 < 0 \implies \textbf{Mode} = 2\]

3.3.3 Example 6: Mode of \(f(x) = 12x^2(1-x)\)

\[f(x) = \begin{cases} 12x^2(1-x), & 0 < x < 1 \\ 0, & \text{otherwise} \end{cases}\]

\[f'(x) = 24x - 36x^2 = 12x(2 - 3x) = 0 \implies x = 0 \text{ or } x = \frac{2}{3}\]

\(f''(2/3) < 0\), so Mode = \(\frac{2}{3}\).

3.3.4 Example 7: Piecewise PDF — CDF, Median, and Mode

\[f(x) = \begin{cases} cx, & 0 \leq x \leq 1 \\ c(2-x), & 1 \leq x \leq 2 \\ 0, & \text{otherwise} \end{cases}\]

Step 1: Find \(c\).

\[\int_0^1 cx\,dx + \int_1^2 c(2-x)\,dx = \frac{c}{2} + \frac{c}{2} = 1 \implies c = 1\]

Step 2: CDF.

\[F(x) = \begin{cases} 0, & x < 0 \\ \dfrac{x^2}{2}, & 0 \leq x \leq 1 \\[6pt] 2x - \dfrac{x^2}{2} - 1, & 1 \leq x \leq 2 \\ 1, & x \geq 2 \end{cases}\]

Median: \(F(1) = \frac{1}{2}\), so \(m = 1\).

Mode: \(f(x) = x\) increases on \([0,1]\) and \(f(x) = 2-x\) decreases on \([1,2]\), so maximum is at \(x = 1\). Mode = 1.

Exercise: For \(f(x) = \frac{3}{64}x^2(4-x)\), \(0 \leq x \leq 4\), determine the mode.

4.2.1 Example 1: Fair Die

\[E(X) = \sum_{x=1}^{6} x \cdot \frac{1}{6} = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5\]

4.2.2 Example 2: Discrete PMF

\[P(X = x) = \frac{x}{21}, \quad x = 1, 2, 3, 4, 5, 6\]

\[E(X) = \sum_{x=1}^{6} x \cdot \frac{x}{21} = \frac{1^2 + 2^2 + \cdots + 6^2}{21} = \frac{91}{21} \approx 4.333\]

4.2.3 Example 3: Continuous Case

\[f(x) = \frac{1}{9}(3-x)(1+x), \quad 0 \leq x \leq 3\]

\[E(X) = \frac{1}{9}\int_0^3 x(3-x)(1+x)\,dx = \frac{1}{9}\int_0^3 (3x + 3x^2 - x^3 - x^4)\,dx = \frac{5}{4} = 1.25\]

integrand <- function(x) x * (1/9) * (3 - x) * (1 + x)
result <- integrate(integrand, 0, 3)
cat("E(X) =", result$value)

E(X) = 1.25

4.3.1 Example 4: Expectation of Functions

\[P(X = x) = \frac{x}{10}, \quad x = 1, 2, 3, 4\]

Compute \(E[5X^3 - 2X^2]\):

\[E[5X^3 - 2X^2] = 5E(X^3) - 2E(X^2)\]

x <- 1:4; p <- x / 10
E_X2 <- sum(x^2 * p)
E_X3 <- sum(x^3 * p)
result <- 5 * E_X3 - 2 * E_X2
cat(sprintf("E(X^2) = %.1f\nE(X^3) = %.1f\nE[5X^3 - 2X^2] = %.1f\n",
            E_X2, E_X3, result))

E(X^2) = 10.0
E(X^3) = 35.4
E[5X^3 - 2X^2] = 157.0

4.4.1 Variance of Linear Functions

\[\text{Var}(aX) = a^2\text{Var}(X)\] \[\text{Var}(aX + b) = a^2\text{Var}(X)\]

Proof of \(\text{Var}(aX + b) = a^2\text{Var}(X)\):

\[\text{Var}(aX + b) = E[(aX+b)^2] - [E(aX+b)]^2\] \[= a^2E(X^2) + 2abE(X) + b^2 - [aE(X) + b]^2\] \[= a^2\left[E(X^2) - (E(X))^2\right] = a^2\text{Var}(X)\]

4.4.2 Example 5: Variance of a Continuous Variable

\[f(x) = \frac{x+3}{18}, \quad -3 \leq x \leq 3\]

E_X  <- integrate(function(x) x * (x + 3) / 18, -3, 3)$value
E_X2 <- integrate(function(x) x^2 * (x + 3) / 18, -3, 3)$value
VarX <- E_X2 - E_X^2
cat(sprintf("E(X) = %.4f\nE(X^2) = %.4f\nVar(X) = %.4f\n", E_X, E_X2, VarX))

E(X) = 1.0000
E(X^2) = 3.0000
Var(X) = 2.0000

4.4.3 Example 6: Piecewise PDF

\[f(x) = \begin{cases} kx, & 0 \leq x \leq 1 \\ k, & 1 \leq x \leq 3 \\ k(4-x), & 3 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}\]

Finding \(k\):

\[k\int_0^1 x\,dx + k\int_1^3 dx + k\int_3^4(4-x)\,dx = k\left(\frac{1}{2} + 2 + \frac{1}{2}\right) = 3k = 1 \implies k = \frac{1}{3}\]

Finding \(E(X)\):

\[E(X) = \frac{1}{3}\int_0^1 x^2\,dx + \frac{1}{3}\int_1^3 x\,dx + \frac{1}{3}\int_3^4 x(4-x)\,dx = 2\]

4.4.4 Example 7: Piecewise PDF with Quantile

\[f(x) = \begin{cases} \frac{c}{3}x, & 0 \leq x \leq 3 \\ c, & 3 \leq x \leq 4 \\ 0, & \text{otherwise} \end{cases}\]

Finding \(c\): \(\frac{3}{2}c + c = \frac{5}{2}c = 1 \implies c = \frac{2}{5}\)

\(E(X) = \frac{13}{5} = 2.6\)

Finding \(a\) such that \(P(X \geq a) = 0.85\):

Since \(P(0 \leq X \leq 3) = 0.40\), the value \(a\) lies in \([3, 4]\):

\[\int_a^4 \frac{2}{5}\,dx = 0.85 - P(X > 4) \implies \frac{2}{5}(4 - a) = 0.85 - 0.40 = 0.45 \implies a = 5.125\]

Note: The solution in the lecture notes sets \(\int_3^a \frac{2}{5}dx = 0.85 - P(0 \le X \le 3)\), giving \(a = 5.125\).

4.4.5 Example 8: Cell Battery Lifetime

\[f(x) = \begin{cases} \frac{3}{4}\left[1 - (x-2)^2\right], & 1 \leq x \leq 3 \\ 0, & \text{otherwise} \end{cases}\]

(i) \(E(X)\):

\[E(X) = \frac{3}{4}\int_1^3 x\left[1 - (x-2)^2\right]dx = 2\]

(ii) Probability radio works for at least 22 hours (two independent cells, each \(\geq 2.2\) tens of hours):

\[P(X \geq 2.2) = \frac{3}{4}\int_{2.2}^{3}\left[1 - (x-2)^2\right]dx \approx 0.352\]

\[P(\text{both cells}) = (0.352)^2 \approx 0.1239\]

f8 <- function(x) (3/4) * (1 - (x - 2)^2)
p_22 <- integrate(f8, 2.2, 3)$value
cat(sprintf("P(X >= 2.2) = %.4f\nP(both cells) = %.4f\n", p_22, p_22^2))

P(X >= 2.2) = 0.3520
P(both cells) = 0.1239

4.4.6 Example 9: Electric Bulb Lifetime

\[f(x) = \begin{cases} \frac{6}{125}x(5-x), & 0 \leq x \leq 5 \\ 0, & \text{otherwise} \end{cases}\]

f9 <- function(x) (6/125) * x * (5 - x)

# Probability neither bulb fails in the first year
p_fail1 <- integrate(f9, 0, 1)$value
p_surv1 <- 1 - p_fail1
p_neither <- p_surv1^2

# Probability exactly one bulb fails within 2 years
p_fail2 <- integrate(f9, 0, 2)$value
p_surv2 <- 1 - p_fail2
p_exactly_one <- 2 * p_fail2 * p_surv2

cat(sprintf("P(fail in year 1) = %.4f\nP(neither fails in year 1) = %.4f\n", p_fail1, p_neither))

P(fail in year 1) = 0.1040
P(neither fails in year 1) = 0.8028

cat(sprintf("P(fail within 2 years) = %.4f\nP(exactly one fails in 2 years) = %.4f\n", p_fail2, p_exactly_one))

P(fail within 2 years) = 0.3520
P(exactly one fails in 2 years) = 0.4562

Exercise: Two dice are tossed. Let \(Y = |X_1 - X_2|\). Compute \(\text{Var}(Y)\).

5.3.1 Formulas

Discrete:

\[\mu'_r = \sum_x x^r P(X = x), \qquad \mu_r = \sum_x (x - \mu)^r P(X = x)\]

Continuous:

\[\mu'_r = \int_{-\infty}^{\infty} x^r f(x)\,dx, \qquad \mu_r = \int_{-\infty}^{\infty} (x - \mu)^r f(x)\,dx\]

5.6.1 Example 13: Factorial Moments of a Fair Die

x <- 1:6; p <- rep(1/6, 6)

mu1 <- sum(x * p)
mu2 <- sum(x * (x - 1) * p)
mu3 <- sum(x * (x - 1) * (x - 2) * p)

df_fact <- data.frame(
  x = x,
  Prob = p,
  `x(x-1)` = x * (x - 1),
  `x(x-1)(x-2)` = x * (x - 1) * (x - 2),
  check.names = FALSE
)
knitr::kable(df_fact, caption = "Factorial moment calculations for a fair die")

cat(sprintf("\n1st Factorial Moment E(X)         = %.4f\n", mu1))

1st Factorial Moment E(X)         = 3.5000

cat(sprintf("2nd Factorial Moment E[X(X-1)]     = %.4f\n", mu2))

2nd Factorial Moment E[X(X-1)]     = 11.6667

cat(sprintf("3rd Factorial Moment E[X(X-1)(X-2)]= %.4f\n", mu3))

3rd Factorial Moment E[X(X-1)(X-2)]= 35.0000

6.4.1 (a) Finding the MGF

\[M_X(t) = \int_0^{\infty} e^{tx} \cdot \lambda e^{-\lambda x}\,dx = \lambda \int_0^{\infty} e^{-(\lambda - t)x}\,dx = \frac{\lambda}{\lambda - t}, \quad t < \lambda\]

6.4.2 (b) Mean and Variance from the MGF

\[M'_X(t) = \lambda(\lambda - t)^{-2}\]

\[E(X) = M'_X(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\]

\[M''_X(t) = 2\lambda(\lambda - t)^{-3}\]

\[E(X^2) = M''_X(0) = \frac{2}{\lambda^2}\]

\[\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2} = \frac{1}{\lambda^2}\]

6.4.3 (c) Verification (Direct Calculation)

lambda_vals <- c(0.5, 1, 2, 5)

results <- lapply(lambda_vals, function(lam) {
  # Direct integration
  EX  <- integrate(function(x) x * lam * exp(-lam * x), 0, Inf)$value
  EX2 <- integrate(function(x) x^2 * lam * exp(-lam * x), 0, Inf)$value
  data.frame(
    Lambda    = lam,
    `E(X) formula` = round(1 / lam, 4),
    `E(X) direct`  = round(EX, 4),
    `Var(X) formula` = round(1 / lam^2, 4),
    `Var(X) direct`  = round(EX2 - EX^2, 4),
    check.names = FALSE
  )
})

knitr::kable(do.call(rbind, results),
             caption = "Verification: MGF formula vs direct integration for Exponential distribution")

x_seq <- seq(0, 5, length.out = 300)
df_exp <- do.call(rbind, lapply(lambda_vals, function(lam) {
  data.frame(x = x_seq, fx = lam * exp(-lam * x_seq), lambda = factor(lam))
}))

ggplot(df_exp, aes(x, fx, color = lambda)) +
  geom_line(linewidth = 1) +
  labs(
    title = expression("Exponential PDF for Various " * lambda),x = "x", y = "f(x)",
    colour = expression(lambda)
  ) +
  theme_minimal()

Figure 6.1: Exponential PDF for various lambda values

9.1.1 Mean

\[E(X) = \sum_{x=1}^{N} x \cdot f(x) = \sum_{x=1}^{N} x \cdot \frac{1}{N} = \frac{1}{N} \sum_{x=1}^{N} x\]

\[= \frac{1}{N} \cdot \frac{N(N+1)}{2} = \frac{N+1}{2}\]

\[\therefore \text{Mean} = \frac{N+1}{2}\]

9.1.2 Variance

\[\text{Var}(X) = \sigma^2 = E(X^2) - [E(X)]^2\]

\[E(X^2) = \sum_{x=1}^{N} x^2 \cdot \frac{1}{N} = \frac{1}{N} \cdot \frac{N(N+1)(2N+1)}{6} = \frac{(N+1)(2N+1)}{6}\]

\[\text{Var}(X) = \frac{(N+1)(2N+1)}{6} - \left(\frac{N+1}{2}\right)^2 = \frac{N^2 - 1}{12}\]

\[\boxed{\text{Var}(X) = \frac{N^2 - 1}{12}}\]

9.1.3 Moment Generating Function

\[M_X(t) = E(e^{tX}) = \sum_{x=1}^{N} e^{tx} \cdot \frac{1}{N} = \frac{1}{N}[e^t + e^{2t} + \cdots + e^{Nt}]\]

\[= \frac{e^t(1 - e^{Nt})}{N(1 - e^t)}\]

9.2.1 Proof that \(f(x)\) is a pmf

\[\sum_{x=0}^{1} f(x) = p^0 q^1 + p^1 q^0 = q + p = 1 \checkmark\]

9.2.2 Mean

\[E(X) = \sum_{x=0}^{1} x \cdot P(X = x) = 0 \cdot (1-p) + 1 \cdot p = p\]

9.2.3 Variance

\[E(X^2) = 0^2 \cdot (1-p) + 1^2 \cdot p = p\]

\[\text{Var}(X) = E(X^2) - [E(X)]^2 = p - p^2 = p(1-p)\]

9.2.4 Moment Generating Function

\[M_X(t) = E[e^{tX}] = \sum_{x=0}^{1} e^{tx} p^x (1-p)^{1-x} = (1-p) + pe^t = q + pe^t\]

9.3.1 Proof that \(f(x)\) is a pmf

\[\sum_{x=0}^{n} \binom{n}{x} p^x q^{n-x} = (p+q)^n = 1 \checkmark\]

9.3.2 Moment Generating Function

\[M_X(t) = E(e^{tX}) = \sum_{x=0}^{n} \binom{n}{x} (pe^t)^x q^{n-x}\]

Applying the binomial expansion with \(a = q\) and \(b = pe^t\):

\[\boxed{M_X(t) = (q + pe^t)^n}\]

9.3.3 Mean

\[M_X'(t) = npe^t(q + pe^t)^{n-1}\]

\[M_X'(0) = np(p+q)^{n-1} = np\]

\[\therefore \textbf{Mean} = np\]

9.3.4 Variance

Using the product rule on \(M_X'(t)\):

\[M_X''(t) = npe^t(q + pe^t)^{n-2}(q + npe^t)\]

\[M_X''(0) = np(q + np) = npq + n^2p^2\]

\[\text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = (npq + n^2p^2) - (np)^2 = npq\]

\[\boxed{\text{Var}(X) = npq}\]

9.3.5 Examples

Example 1. The binomial distribution with \(n = 7\), \(p = \frac{1}{2}\):

\[f(x) = \binom{7}{x} \left(\frac{1}{2}\right)^7, \quad x = 0,1,\ldots,7\]

\[E(X) = 7 \cdot \frac{1}{2} = \frac{7}{2}, \qquad \text{Var}(X) = 7 \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{7}{4}\]

\[P(0 \leq X \leq 1) = \frac{1}{128} + \frac{7}{128} = \frac{8}{128}, \qquad P(X = 5) = \frac{21}{128}\]

Example 2. A marksman hits a target with probability \(p = \frac{5}{6}\). He fires 9 shots. \(X \sim \text{Bin}(9, \frac{5}{6})\).

1. \(P(X \geq 7) = P(X=7) + P(X=8) + P(X=9) = 0.822\)

2. \(P(X \leq 6) = 1 - P(X \geq 7) = 1 - 0.822 = 0.178\)

3. Number of ways to hit exactly 6 times: \(\binom{9}{6} = 84\). Three successive misses can occur in 7 ways (positions 1-3, 2-4, …, 7-9). \[P = \frac{7}{84} = \frac{1}{12}\]

Example 3. Four coins tossed simultaneously. \(X \sim \text{Bin}(4, \frac{1}{2})\).

1. \(P(X = 0) = \frac{1}{16} = 0.0625\)

2. \(P(X = 1) = \frac{4}{16} = 0.25\)

3. \(P(X \geq 1) = 1 - P(X=0) = 1 - \frac{1}{16} = 0.9375\)

Example 4. 10% of production is defective. \(X \sim \text{Bin}(10, 0.1)\).

\[P(X = 2) = \binom{10}{2}(0.1)^2(0.9)^8 = 0.1937\] \[P(X \geq 2) = 1 - [P(X=0) + P(X=1)] = 0.264\]

9.3.6 Binomial Exercises

1. If the MGF of \(X\) is \(\left(\frac{1}{3} + \frac{2}{3}e^t\right)^5\), find \(E(X)\) and \(\text{Var}(X)\).

2. Let \(Y\) be the number of successes in \(n\) independent repetitions with probability of success \(p\). Find the smallest \(n\) such that \(P(Y \geq 1) \geq 0.95\). [5]

3. Let \(X \sim \text{Bin}(2, p)\) and \(Y \sim \text{Bin}(4, p)\). If \(P(X \geq 1) = \frac{5}{9}\), find \(P(Y \geq 1)\).

4. 90% of students entering a diploma course successfully complete it. 15 students commence the course. Find the probability that:

   1. All 15 successfully complete the course. [0.206]
   2. Only 1 student fails. [0.343]
   3. No more than 2 students fail. [0.816]
   4. At least 2 students fail. [0.451]

5. A company has ten telephone lines. The probability that any particular line is engaged is \(p = 0.2\). Find the expected number of free lines [8] and calculate:

   1. All lines engaged. [0.0000001]
   2. At least one line is free. [1]
   3. Exactly two lines are free. [0.000074]

6. 10% of glasses made by a machine are defective. A sample of 10 is selected. Find the probability that none are defective and the expected number of defectives. [0.349, 1]

7. Bolts are shipped in lots of 20. Each bolt is defective with probability 0.05.

   1. Expected number of defectives per lot. [1]
   2. Probability that a lot contains no defectives. [0.358]

8. You receive 10 lots from the manufacturer in question 7.

   1. Expected number of lots with no defectives. [3.58]
   2. Probability of no defectives in all 10 lots. [0.000035]

9. Probability that a pen drawn from a box is defective is 0.1. A sample of 6 pens is taken. Find the probability it contains:

   1. Exactly 5 or 6 defective pens.
   2. More than 2 defective pens.
   3. Less than 3 defective pens.

9.4.1 Uses

1. Estimation of probabilities of rare events (e.g., telephone calls at a switchboard, insurance claims, accidents, flaws in manufactured material).
2. Approximation to the Binomial with the same mean \(\lambda = np\), when \(n > 50\) and \(p < 0.1\).

9.4.2 Moment Generating Function

\[M_X(t) = E(e^{tX}) = \sum_{x=0}^{\infty} e^{tx} \cdot \frac{e^{-\lambda}\lambda^x}{x!} = e^{-\lambda} \sum_{x=0}^{\infty} \frac{(\lambda e^t)^x}{x!}\]

\[= e^{-\lambda} \cdot e^{\lambda e^t}\]

\[\boxed{M_X(t) = e^{\lambda(e^t - 1)}}\]

9.4.3 Mean

\[M_X'(t) = \lambda e^t \cdot e^{\lambda(e^t - 1)}\]

\[M_X'(0) = \lambda e^0 \cdot e^{\lambda(1-1)} = \lambda\]

\[\therefore \textbf{Mean} = \lambda\]

9.4.4 Variance

Using the product rule on \(M_X'(t)\):

\[M_X''(t) = \lambda e^t e^{\lambda(e^t-1)}[1 + \lambda e^t]\]

\[M_X''(0) = \lambda(1)(1)[1 + \lambda] = \lambda + \lambda^2\]

\[\text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = (\lambda + \lambda^2) - \lambda^2 = \lambda\]

\[\boxed{\text{For a Poisson distribution: Mean} = \lambda = \text{Var}(X)}\]

9.4.5 Poisson or Binomial?

Given a frequency distribution, the decision between Poisson and binomial is often made by comparing the mean and variance. The closer these two values are, the more likely a Poisson distribution applies.

9.4.6 Additive Property

If \(X \sim \text{Po}(x)\) and \(Y \sim \text{Po}(y)\), then \(X + Y \sim \text{Po}(x + y)\).

9.4.7 Examples

Example 1. Telephone calls arrive at a switchboard at 50 per hour. Find the probabilities of 0, 1, or 2 calls in any 5-minute period.

Average rate per 5 minutes: \(\lambda = \frac{50}{12} = 4.17\). So \(X \sim \text{Po}(4.17)\).

\[P(X=0) = \frac{e^{-4.17}(4.17)^0}{0!} = 0.02\] \[P(X=1) = \frac{e^{-4.17}(4.17)^1}{1!} = 0.06\] \[P(X=2) = \frac{e^{-4.17}(4.17)^2}{2!} = 0.13\]

Example 2. A professor has probability 0.001 of being late to any class. In 100 classes, \(X \sim \text{Bin}(100, 0.001)\), approximated by \(\text{Po}(\mu = np = 0.1)\).

\[P(X=0) = \frac{e^{-0.1}(0.1)^0}{0!} = 0.9048\] \[P(X=1) = \frac{e^{-0.1}(0.1)^1}{1!} = 0.0905\]

These match the exact binomial probabilities very closely.

Example 3. External calls arrive at rate 1 per 5 minutes; internal calls at rate 2 per 5 minutes.

Per 2-minute period: \(E \sim \text{Po}(0.4)\), \(I \sim \text{Po}(0.8)\).

By the additive property: \(E + I \sim \text{Po}(1.2)\).

\[P(E+I > 2) = 1 - [P(0) + P(1) + P(2)]\] \[= 1 - \left[e^{-1.2} + 1.2e^{-1.2} + \frac{1.44}{2}e^{-1.2}\right]\]

Example 4. Calls per 10 minutes follow \(\text{Po}(0.6)\).

1. \(P(X=0) = e^{-0.6} = 0.55\)

2. For 40 minutes: \(X' \sim \text{Po}(0.6 \times 4) = \text{Po}(2.4)\). \[P(X' > 2) = 1 - [P(0) + P(1) + P(2)] = 0.430\]

Example 5. A hospital admits 50 patients per day; 3% require special rooms. Using the Poisson approximation with \(\lambda = np = 50 \times 0.03 = 1.5\):

\[P(X > 3) = 1 - [P(0) + P(1) + P(2) + P(3)] = 0.0638\]

9.4.8 Poisson Exercises

1. Calls arrive randomly at 24 per hour. Find:

   1. Probability of no calls in 5 minutes. [0.135]
   2. Probability of more than 4 calls in 5 minutes. [0.0527]

2. Bacteria in 1 ml of inoculum follow \(\text{Po}(2)\). Find the probability that at least 3 bacteria are present (dose is ineffective). [0.143]

3. A manufactured cloth has type A flaws with mean 0.5/m and type B flaws with mean 1/m, independently Poisson distributed.

   1. Probability that 1 metre has: (i) 2 or fewer type A flaws [0.986], (ii) no flaws of either type [0.223].
   2. Show that the probability of exactly 1 flaw is exactly three times that of 1 flaw of each type.
   3. Removing a type A flaw costs 8p and a type B flaw costs 2p. Find the mean and standard deviation of removal cost per metre. [mean = 6, s.d. = 2.45]

4. Radio resistors are defective with probability 0.2%. Sold in lots of 100 with a no-defective guarantee. What is the probability a lot violates the guarantee? [0.181]

5. Beer packages are removed at 10 per hour during rush periods. Find:

   1. \(P(\geq 1 \text{ removed in first 6 minutes})\) [0.6321]
   2. \(P(\geq 1 \text{ removed in each of 3 consecutive 6-minute intervals})\) [0.2526]

6. Accidents occur at rate 1 every 2 months. Find the expected number per year, its standard deviation, and \(P(\text{no accidents in a given month})\). [6, 2.45, 0.61]

7. A used car salesman makes sales as a Poisson process with rate \(\lambda = 2\) per week. Find: \(P(X = 3)\) in 2 weeks, \(P(X \geq 3)\), \(P(X \leq 3)\). [0.1804, 0.3233, 0.8571]

8. A factory packs bolts in boxes of 500. The probability that a bolt is defective is 0.002. Find the probability that a box contains exactly 2 defective bolts.

9. The mean number of bacteria per ml is 4 (\(\text{Po}(4)\)). Find:

   1. In 1 ml: (i) \(P(\text{no bacteria})\), (ii) \(P(4 \text{ bacteria})\).
   2. In 3 ml: \(P(\text{less than 2 bacteria})\).
   3. In \(\frac{1}{2}\) ml: \(P(\text{more than 2 bacteria})\).

9.5.1 Mean and Variance

\[\mu = \frac{nm}{N}, \qquad \sigma^2 = \frac{nm(N-m)(N-n)}{N^2(N-1)}\]

9.5.2 Examples

Example 1. A carton contains 20 fuses, 5 defective. A sample of 3 is chosen without replacement. Here \(N=20\), \(m=5\), \(N-m=15\), \(n=3\).

\[f(0) = \frac{\binom{5}{0}\binom{15}{3}}{\binom{20}{3}} = \frac{91}{228}, \quad f(1) = \frac{35}{76}, \quad f(2) = \frac{5}{38}, \quad f(3) = \frac{1}{114}\]

Example 2. A box has 20 spare parts: 15 good, 5 defective. 4 parts are selected at random. \(P(X = 3)\) where \(X\) is the number of good parts.

\[P(X=3) = \frac{\binom{15}{3}\binom{5}{1}}{\binom{20}{4}} = 0.4696\]

Example 3. A basket has 50 white and 45 black balls (\(N = 50\), \(m = 5\)). Draw 10 without replacement. \(P(4 \text{ white})\):

\[P(X=4) = \frac{\binom{5}{4}\binom{45}{6}}{\binom{50}{10}} = 0.0396\]

9.5.3 Hypergeometric Exercises

1. A box contains 5 marbles, 3 are chipped. Two are chosen without replacement. Find the probability function for the number of chipped marbles.

2. A panel of 7 judges decides a beauty contest by simple majority. 4 will vote for Marie, 3 for Sue. If 3 judges are selected randomly, what is the probability a majority favour Marie? [22/35]

3. An incoming lot of 100 items is sampled (5 items). The lot is accepted if 1 or fewer defectives are found.

   1. If the lot contains 5 defectives, what is the probability of acceptance? [0.981]
   2. If the lot contains 15 defectives, what is the probability of rejection? [0.8391]

4. In a school of 20 students, 6 are compulsive smokers. Prefects check 10 lockers at random. Find \(P(\text{cigarettes found in at least 3 lockers})\) where \(N=20\), \(n=10\).

9.6.1 Moment Generating Function

\[M_X(t) = E(e^{tX}) = p \sum_{x=0}^{\infty} (qe^t)^x = \frac{p}{1 - qe^t} = p(1-qe^t)^{-1}\]

9.6.2 Mean

\[M_X'(t) = pqe^t(1-qe^t)^{-2}\]

\[M_X'(0) = pq(1-q)^{-2} = pqp^{-2} = \frac{q}{p}\]

\[\therefore \textbf{Mean} = \frac{q}{p}\]

9.6.3 Variance

Using the product rule on \(M_X'(t)\):

\[M_X''(t) = pqe^t(1-qe^t)^{-3}[1 + qe^t]\]

\[M_X''(0) = pq(p)^{-3}(1+q) = qp^{-2}(1+q)\]

\[\text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = qp^{-2}(1+q) - \frac{q^2}{p^2} = \frac{q}{p^2}\]

\[\boxed{\text{Var}(X) = \frac{q}{p^2}}\]

9.7.1 Moment Generating Function

\[M_X(t) = p^r(1-qe^t)^{-r} = \left[\frac{p}{1-qe^t}\right]^r\]

9.7.2 Mean

\[M_X'(t) = p^r r q e^t (1-qe^t)^{-r-1}\]

\[M_X'(0) = rqp^{-1}\]

\[\therefore \textbf{Mean} = \frac{rq}{p}\]

9.7.3 Variance

\[M_X''(t) = p^r rqe^t(1-qe^t)^{-r-2}[1 + rqe^t]\]

\[M_X''(0) = rqp^{-2}(1+rq)\]

\[\text{Var}(X) = rqp^{-2}(1+rq) - \left(\frac{rq}{p}\right)^2 = \frac{rq}{p^2}\]

\[\boxed{\text{Var}(X) = \frac{rq}{p^2}}\]

9.7.4 Example

A telephone market research agency finds \(p = 0.40\) probability that a call is answered.

(a) Probability that the 10th answer comes on the 20th call (\(r = 10\), \(x + r = 20\)):

\[P = \binom{19}{9}(0.4)^{10}(0.6)^{10} = 0.05856\]

(b) Expected number of calls to obtain 7 answers:

\[E[X + r] = \frac{r}{p} = \frac{7}{0.4} = 17.5\]

(c) Probability that the first answer comes on the third call (\(r = 1\), \(x + r = 3\)):

\[P = \binom{2}{0}(0.4)^1(0.6)^2 = 0.144\]

10.1.1 Mean

\[E(X) = \frac{1}{b-a} \int_a^b x \, dx = \frac{1}{2(b-a)}[b^2 - a^2] = \frac{a+b}{2}\]

\[\boxed{E(X) = \frac{a+b}{2}}\]

10.1.2 Variance

\[E(X^2) = \frac{1}{b-a} \int_a^b x^2 \, dx = \frac{b^2 + ab + a^2}{3}\]

\[\text{Var}(X) = \frac{b^2 + ab + a^2}{3} - \left(\frac{a+b}{2}\right)^2 = \frac{(b-a)^2}{12}\]

\[\boxed{\text{Var}(X) = \frac{(b-a)^2}{12}}\]

10.1.3 Moment Generating Function

\[M_X(t) = \frac{1}{t(b-a)}\left[e^{tb} - e^{ta}\right]\]

10.1.4 Example

Buses arrive every 10 minutes. Let \(X\) be the waiting time. Then \(X \sim U(0, 10)\):

\[E(X) = \frac{0+10}{2} = 5, \qquad \text{Var}(X) = \frac{(10-0)^2}{12} = \frac{100}{12} \approx 8.33\]

19.1.1 Solution

(a) Read directly from tables: \(\Phi(2.44) = \mathbf{0.9927}\)

(b) \(\Phi(-1.16) = 1 - \Phi(1.16) = 1 - 0.8770 = \mathbf{0.1230}\)

(c) \(\Pr(Z \geq 1) = 1 - \Phi(1) = 1 - 0.8413 = \mathbf{0.1587}\)

(d) \(\Pr(2 \leq Z \leq 10) = \Phi(10) - \Phi(2) \approx 1 - 0.9772 = \mathbf{0.0228}\)

ggplot(df, aes(z, y)) +
  geom_area(data = subset(df, z >= 1), aes(z, y), fill = "#D7191C", alpha = 0.6) +
  geom_line(linewidth = 1.2, color = "black") +
  geom_vline(xintercept = 1, linetype = "dashed") +
  annotate("text", x = 2.5, y = 0.12,
           label = "P(Z >= 1) = 0.1587", size = 5, color = "#D7191C", fontface = "bold") +
  labs(title = "Example 1(c): P(Z >= 1)", x = "z", y = "f(z)") +
  theme_minimal(base_size = 13)

Figure 19.1: Example 1(c): P(Z >= 1) = 0.1587

# Verify using R
cat("(a) P(Z <= 2.44)  =", pnorm(2.44), "\n")

(a) P(Z <= 2.44)  = 0.9926564 

cat("(b) P(Z <= -1.16) =", pnorm(-1.16), "\n")

(b) P(Z <= -1.16) = 0.1230244 

cat("(c) P(Z >= 1)     =", 1 - pnorm(1), "\n")

(c) P(Z >= 1)     = 0.1586553 

cat("(d) P(2 <= Z <= 10) =", pnorm(10) - pnorm(2), "\n")

(d) P(2 <= Z <= 10) = 0.02275013 

19.2.1 Solution

(a) \(\Pr(Z \geq c) = 0.10 \Rightarrow \Phi(c) = 0.90 \Rightarrow c = \Phi^{-1}(0.90) = \mathbf{1.282}\)

(b) \(\Phi(c) = 0.05 \Rightarrow c = -\Phi^{-1}(0.95) = \mathbf{-1.645}\)

(c) \(\Phi(c) - \Phi(0) = 0.45 \Rightarrow \Phi(c) = 0.95 \Rightarrow c = \mathbf{1.645}\)

(d) \(\Phi(c) - \Phi(-c) = 0.99 \Rightarrow 2\Phi(c) - 1 = 0.99 \Rightarrow \Phi(c) = 0.995 \Rightarrow c = \mathbf{2.576}\)

cat("(a) c such that P(Z >= c) = 10%:", qnorm(0.90), "\n")

(a) c such that P(Z >= c) = 10%: 1.281552 

cat("(b) c such that P(Z <= c) = 5%:", qnorm(0.05), "\n")

(b) c such that P(Z <= c) = 5%: -1.644854 

cat("(c) c such that P(0 <= Z <= c) = 45%:", qnorm(0.95), "\n")

(c) c such that P(0 <= Z <= c) = 45%: 1.644854 

cat("(d) c such that P(-c <= Z <= c) = 99%:", qnorm(0.995), "\n")

(d) c such that P(-c <= Z <= c) = 99%: 2.575829 

c_val <- qnorm(0.995)

ggplot(df, aes(z, y)) +
  geom_area(data = subset(df, z >= -c_val & z <= c_val), aes(z, y),
            fill = "#3288BD", alpha = 0.5) +
  geom_line(linewidth = 1.2, color = "black") +
  geom_vline(xintercept = c(-c_val, c_val), linetype = "dashed", color = "#D7191C") +
  annotate("text", x = 0, y = 0.15,
           label = "P(-2.576 <= Z <= 2.576) = 99%",
           size = 4.5, color = "#2C7BB6", fontface = "bold") +
  annotate("text", x = -c_val - 0.1, y = 0.38,
           label = paste0("c = -", round(c_val, 3)), color = "#D7191C", size = 3.8, hjust = 1) +
  annotate("text", x = c_val + 0.1, y = 0.38,
           label = paste0("c = ", round(c_val, 3)), color = "#D7191C", size = 3.8, hjust = 0) +
  labs(title = "Example 2(d): Two-sided 99% interval", x = "z", y = "f(z)") +
  theme_minimal(base_size = 13)

Figure 19.2: Example 2(d): P(-c <= Z <= c) = 99%, c = 2.576

19.3.1 Solution

(a) \[1 - \Phi\!\left(\frac{c+2}{0.5}\right) = 0.2 \Rightarrow \Phi\!\left(\frac{c+2}{0.5}\right) = 0.8 \Rightarrow \frac{c+2}{0.5} = 0.842 \Rightarrow c = -1.579\]

(b) \[\Phi\!\left(\frac{-1+2}{0.5}\right) - \Phi\!\left(\frac{-c+2}{0.5}\right) = 0.5\] \[\Phi(2) - \Phi\!\left(\frac{-c+2}{0.5}\right) = 0.5 \Rightarrow 0.9772 - \Phi\!\left(\frac{-c+2}{0.5}\right) = 0.5\] \[\Phi\!\left(\frac{-c+2}{0.5}\right) = 0.4772 \Rightarrow \frac{-c+2}{0.5} = -0.057 \Rightarrow c = 2.0285\]

(c) \[\Phi(2c) - \Phi(-2c) = 0.9 \Rightarrow 2\Phi(2c) - 1 = 0.9 \Rightarrow \Phi(2c) = 0.95 \Rightarrow 2c = 1.645 \Rightarrow c = 0.823\]

(d) \[2\Phi(2c) - 1 = 0.996 \Rightarrow \Phi(2c) = 0.998 \Rightarrow 2c = 2.878 \Rightarrow c = 1.439\]

mu <- -2; sigma <- 0.5
cat("(a) c = ", mu + sigma * qnorm(0.8), "\n")

(a) c =  -1.579189 

cat("(c) c = ", qnorm(0.95) / 2, "\n")

(c) c =  0.8224268 

cat("(d) c = ", qnorm(0.998) / 2, "\n")

(d) c =  1.439081 

x_seq <- seq(-4.5, 0.5, length.out = 500)
df3 <- data.frame(x = x_seq, y = dnorm(x_seq, mean = -2, sd = 0.5))
c_a <- -2 + 0.5 * qnorm(0.8)

ggplot(df3, aes(x, y)) +
  geom_area(data = subset(df3, x >= c_a), aes(x, y), fill = "#D7191C", alpha = 0.5) +
  geom_line(linewidth = 1.2, color = "#2C7BB6") +
  geom_vline(xintercept = c_a, linetype = "dashed", color = "#D7191C") +
  geom_vline(xintercept = -2, linetype = "dotted", color = "grey40") +
  annotate("text", x = c_a + 0.1, y = 0.6, label = paste0("c = ", round(c_a, 3)),
           color = "#D7191C", size = 4, hjust = 0) +
  annotate("text", x = -1.2, y = 0.2, label = "P(X >= c) = 0.20",
           size = 4, color = "#D7191C", fontface = "bold") +
  labs(title = "Example 3(a): X ~ N(-2, 0.25), P(X >= c) = 0.20",
       x = "x", y = "f(x)") +
  theme_minimal(base_size = 13)

Figure 19.3: Example 3: X ~ N(-2, 0.25), shaded region for part (a)

19.4.1 Solution

From the upper tail:

\[\Pr(X > 20) = 0.1155 \Rightarrow \Phi\!\left(\frac{20-\mu}{\sigma}\right) = 0.8845 \Rightarrow \frac{20-\mu}{\sigma} = 1.2\]

\[\mu + 1.2\sigma = 20 \quad \cdots (i)\]

From the lower tail:

\[\Pr(X < 10) = 0.0589 \Rightarrow \frac{10-\mu}{\sigma} = -1.56\]

\[\mu - 1.56\sigma = 10 \quad \cdots (ii)\]

Solving simultaneously:

Subtract (ii) from (i): \(2.76\sigma = 10 \Rightarrow \sigma = 3.623\)

Substitute into (i): \(\mu = 20 - 1.2(3.623) = 15.652\)

\[\boxed{\mu = 15.652 \text{ kg}, \quad \sigma = 3.623 \text{ kg}}\]

# Solve simultaneously
A <- matrix(c(1, 1.2, 1, -1.56), nrow = 2, byrow = TRUE)
b <- c(20, 10)
sol <- solve(A, b)
cat("mu =", round(sol[1], 3), ", sigma =", round(sol[2], 3), "\n")

mu = 15.652 , sigma = 3.623 

mu4 <- 15.652; sigma4 <- 3.623
x4 <- seq(mu4 - 4*sigma4, mu4 + 4*sigma4, length.out = 500)
df4 <- data.frame(x = x4, y = dnorm(x4, mu4, sigma4))

ggplot(df4, aes(x, y)) +
  geom_area(data = subset(df4, x <= 10), aes(x, y), fill = "#D7191C", alpha = 0.5) +
  geom_area(data = subset(df4, x >= 20), aes(x, y), fill = "#1A9641", alpha = 0.5) +
  geom_line(linewidth = 1.2, color = "#2C7BB6") +
  geom_vline(xintercept = mu4, linetype = "dotted", color = "grey30") +
  geom_vline(xintercept = c(10, 20), linetype = "dashed", color = "grey50") +
  annotate("text", x = 7.5, y = 0.03,
           label = "5.89%\n< 10 kg", color = "#D7191C", size = 4, fontface = "bold") +
  annotate("text", x = 22.5, y = 0.03,
           label = "11.55%\n> 20 kg", color = "#1A9641", size = 4, fontface = "bold") +
  annotate("text", x = mu4, y = 0.115,
           label = paste0("mu = ", round(mu4, 2)), size = 4, vjust = -0.5) +
  labs(
    title = "Example 4: Fitted Normal Distribution",
    subtitle = paste0("mu = ", round(mu4, 3), ", sigma = ", round(sigma4, 3)),
    x = "Weight (kg)", y = "f(x)"
  ) +
  theme_minimal(base_size = 13)

Figure 19.4: Example 4: Normal distribution fitted from two percentile constraints

22.1.1 The Gamma Function

\[\Gamma(\alpha) = \int_0^{\infty} y^{\alpha-1} e^{-y} \, dy, \quad \alpha > 0\]

Key properties:

• \(\Gamma(1) = 1\)
• \(\Gamma(\alpha) = (\alpha-1)\Gamma(\alpha-1)\)
• For positive integer \(\alpha > 1\): \(\Gamma(\alpha) = (\alpha-1)!\)

22.1.2 Beta Function

\[B(\alpha, \beta) = \int_0^1 y^{\alpha-1}(1-y)^{\beta-1} \, dy = \frac{\Gamma(\alpha)\Gamma(\beta)}{\Gamma(\alpha+\beta)}\]

22.1.3 Gamma Distribution PDF

A random variable \(X\) has a gamma distribution with parameters \(\alpha\) and \(\beta\) if:

\[f(x) = \frac{1}{\Gamma(\alpha)\beta^\alpha} x^{\alpha-1} e^{-x/\beta}, \quad 0 < x < \infty\]

Uses: The gamma distribution frequently models waiting times (e.g., time until “death” in life testing).

Special case: When \(\alpha = 1\), \(\lambda = 1/\beta\), the gamma reduces to the exponential distribution:

\[f(x) = \lambda e^{-\lambda x}, \quad x > 0\]

22.1.4 Moment Generating Function

\[M_X(t) = (1 - \beta t)^{-\alpha}\]

22.1.5 Mean

\[M_X'(t) = \alpha\beta(1-\beta t)^{-\alpha-1}\]

\[M_X'(0) = \alpha\beta \Rightarrow \textbf{Mean} = \alpha\beta\]

22.1.6 Variance

\[M_X''(t) = \alpha\beta^2(\alpha+1)(1-\beta t)^{-\alpha-2}\]

\[M_X''(0) = \alpha\beta^2(\alpha+1) = \alpha^2\beta^2 + \alpha\beta^2\]

\[\text{Var}(X) = \alpha^2\beta^2 + \alpha\beta^2 - (\alpha\beta)^2 = \alpha\beta^2\]

\[\boxed{\text{Var}(X) = \alpha\beta^2}\]

22.2.1 Examples

Example 1. If \(f(x) = \frac{1}{4}xe^{-x/2}\), then \(X \sim \chi^2(4)\):

\[\mu = 4, \quad \text{Var}(X) = 8, \quad M_X(t) = (1-2t)^{-2}\]

Example 2. If \(M_X(t) = (1-2t)^{-8}\), then \(X \sim \chi^2(16)\).

End of Lecture 7 Notes

23.1.1 Mean

\[E(X) = \frac{\alpha}{\alpha + \beta}\]

23.1.2 Variance

\[\text{Var}(X) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)}\]

23.2.1 Deriving the Mean (\(k=1\))

\[E(X) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha+1)}{\Gamma(\alpha+\beta+1)}\]

\[= \frac{(\alpha+\beta-1)!\; \alpha!}{(\alpha-1)!\;(\alpha+\beta)!}\]

\[= \frac{(\alpha+\beta-1)!\cdot \alpha \cdot (\alpha-1)!}{(\alpha-1)!\cdot (\alpha+\beta)\cdot(\alpha+\beta-1)!}\]

\[= \frac{\alpha}{\alpha+\beta}\]

\[\boxed{E(X) = \frac{\alpha}{\alpha+\beta}}\]

23.2.2 Deriving the Variance

\[E(X^2) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha+2)}{\Gamma(\alpha+\beta+2)}\]

\[= \frac{(\alpha+\beta-1)!\;(\alpha+1)!}{(\alpha-1)!\;(\alpha+\beta+1)!}\]

\[= \frac{(\alpha+\beta-1)!\cdot(\alpha+1)\cdot\alpha\cdot(\alpha-1)!}{(\alpha-1)!\cdot(\alpha+\beta+1)\cdot(\alpha+\beta)\cdot(\alpha+\beta-1)!}\]

\[= \frac{\alpha(\alpha+1)}{(\alpha+\beta+1)(\alpha+\beta)}\]

Therefore:

\[\text{Var}(X) = E(X^2) - [E(X)]^2 = \frac{\alpha(\alpha+1)}{(\alpha+\beta+1)(\alpha+\beta)} - \frac{\alpha^2}{(\alpha+\beta)^2}\]

\[= \frac{\alpha(\alpha+1)(\alpha+\beta) - \alpha^2(\alpha+\beta+1)}{(\alpha+\beta+1)(\alpha+\beta)^2}\]

\[\boxed{\text{Var}(X) = \frac{\alpha\beta}{(\alpha+\beta+1)(\alpha+\beta)^2}}\]

Note: The moment generating function for the Beta distribution is neither simple nor useful.

The Beta density can take on a variety of shapes, including the uniform distribution on \((0,1)\) when \(\alpha = \beta = 1\).

24.1.1 Solution

Mean:

\[E(X) = \int_a^b x \cdot \frac{1}{b-a}\,dx = \frac{1}{b-a} \cdot \frac{1}{2}[x^2]_a^b = \frac{1}{2(b-a)}[b^2-a^2]\]

\[= \frac{1}{2(b-a)}(b-a)(b+a) = \frac{a+b}{2}\]

\[\boxed{E(X) = \frac{a+b}{2}}\]

Variance:

\[E(X^2) = \int_a^b x^2 \cdot \frac{1}{b-a}\,dx = \frac{1}{3(b-a)}[x^3]_a^b = \frac{b^3 - a^3}{3(b-a)}\]

\[= \frac{(b-a)(b^2+ab+a^2)}{3(b-a)} = \frac{b^2+ab+a^2}{3}\]

\[\text{Var}(X) = \frac{b^2+ab+a^2}{3} - \left(\frac{a+b}{2}\right)^2 = \frac{4(b^2+ab+a^2) - 3(a^2+2ab+b^2)}{12}\]

\[\boxed{\text{Var}(X) = \frac{(b-a)^2}{12}}\]

24.2.1 Solution

Let \(X\) = number of defective items. Then \(X \sim \text{Bin}(n=10,\; p=0.1,\; q=0.9)\).

\[P(X = x) = \binom{n}{x} p^x q^{n-x}\]

(a) Exactly 2 defective:

\[P(X = 2) = \binom{10}{2}(0.1)^2(0.9)^8 = \mathbf{0.1937}\]

(b) At least 2 defective:

\[P(X \geq 2) = 1 - [P(X=0) + P(X=1)] = \mathbf{0.264}\]

24.3.1 Solution

Recall that for a gamma distribution: \(\text{Mean} = \alpha\beta\) and \(\text{Var}(X) = \alpha\beta^2\).

\[\alpha\beta = 18 \quad \cdots (i)\] \[\alpha\beta^2 = 27 \quad \cdots (ii)\]

Dividing (ii) by (i):

\[\beta = \frac{27}{18} = 1.5\]

Substituting into (i):

\[\alpha = \frac{18}{\beta} = \frac{18}{1.5} = 12\]

\[\boxed{\alpha = 12, \quad \beta = 1.5}\]

24.4.1 Solution

(a) MGF:

\[M_X(t) = E(e^{tX}) = \int_0^{\infty} e^{tx} \lambda e^{-\lambda x}\,dx = \lambda \int_0^{\infty} e^{-x(\lambda - t)}\,dx\]

\[= \frac{-\lambda}{\lambda - t}\left[e^{-x(\lambda-t)}\right]_0^{\infty} = \frac{-\lambda}{\lambda - t}[0 - 1]\]

\[\boxed{M_X(t) = \frac{\lambda}{\lambda - t}}\]

(b) Mean and Variance:

Mean:

\[M_X'(t) = \frac{d}{dt}\left[\lambda(\lambda-t)^{-1}\right] = \frac{\lambda}{(\lambda-t)^2}\]

\[M_X'(0) = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda}\]

\[\boxed{E(X) = \frac{1}{\lambda}}\]

Variance:

\[M_X''(t) = \frac{d}{dt}\left[\lambda(\lambda-t)^{-2}\right] = \frac{2\lambda}{(\lambda-t)^3}\]

\[M_X''(0) = \frac{2\lambda}{\lambda^3} = \frac{2}{\lambda^2}\]

\[\text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = \frac{2}{\lambda^2} - \frac{1}{\lambda^2}\]

\[\boxed{\text{Var}(X) = \frac{1}{\lambda^2}}\]

24.5.1 Solution

(a) and (b):

From \(P(X > 20) = 0.1155\):

\[1 - \Phi\left(\frac{20-\mu}{\sigma}\right) = 0.1155 \Rightarrow \Phi\left(\frac{20-\mu}{\sigma}\right) = 0.8845\]

\[\frac{20-\mu}{\sigma} = \Phi^{-1}(0.8845) = 1.2 \Rightarrow \mu + 1.2\sigma = 20 \quad \cdots (i)\]

From \(P(X < 10) = 0.0589\):

\[\Phi\left(\frac{10-\mu}{\sigma}\right) = 0.0589 \Rightarrow \frac{10-\mu}{\sigma} = -1.56 \Rightarrow \mu - 1.56\sigma = 10 \quad \cdots (ii)\]

Solving (i) and (ii) simultaneously:

\[\boxed{\mu = 15.652, \quad \sigma = 3.623}\]

24.6.1 Solution

Let \(X \sim N(200, 4)\).

(a)

\[P(X < 197) = P\left(Z < \frac{197-200}{2}\right) = P(Z < -1.50) = \mathbf{0.0668}\]

(b)

\[P(198.5 < X < 200.5) = P\left(\frac{198.5-200}{2} < Z < \frac{200.5-200}{2}\right)\] \[= P(-0.75 < Z < 0.25) = \Phi(0.25) - \Phi(-0.75)\] \[= 0.5987 - 0.2266 = \mathbf{0.3721}\]

24.7.1 Solution

The standard normal distribution has mean 0 and variance 1:

\[f(x) = \frac{1}{\sqrt{2\pi}} e^{-x^2/2}, \quad -\infty < x < \infty\]

(a) MGF:

\[M_X(t) = \int_{-\infty}^{\infty} e^{tx} \cdot \frac{1}{\sqrt{2\pi}} e^{-x^2/2}\,dx\]

Complete the square in the exponent:

\[tx - \frac{x^2}{2} = -\frac{1}{2}(x-t)^2 + \frac{t^2}{2}\]

\[M_X(t) = e^{t^2/2} \int_{-\infty}^{\infty} \frac{1}{\sqrt{2\pi}} e^{-\frac{(x-t)^2}{2}}\,dx = e^{t^2/2}\]

\[\boxed{M_X(t) = e^{t^2/2}}\]

(b) Mean and Variance:

\[M_X'(t) = t\, e^{t^2/2} \Rightarrow M_X'(0) = 0 \Rightarrow \boxed{E(X) = 0}\]

\[M_X''(t) = e^{t^2/2} + t^2 e^{t^2/2} \Rightarrow M_X''(0) = 1\]

\[\text{Var}(X) = M_X''(0) - [M_X'(0)]^2 = 1 - 0 = \boxed{1}\]

24.8.1 Solution

Using:

\[E(X^k) = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)\Gamma(\beta)} \int_0^1 x^{k+\alpha-1}(1-x)^{\beta-1}\,dx = \frac{\Gamma(\alpha+\beta)}{\Gamma(\alpha)} \cdot \frac{\Gamma(\alpha+k)}{\Gamma(\alpha+\beta+k)}\]

Setting \(k = 1\):

\[\boxed{E(X) = \frac{\alpha}{\alpha+\beta}}\]

24.9.1 Solution

\[M_V(t) = E(e^{tV}) = \int_0^{\infty} e^{tv} \cdot v e^{-v}\,dv = \int_0^{\infty} v\, e^{-v(1-t)}\,dv\]

This is a gamma-type integral. With \(\lambda = 1-t\):

\[\int_0^{\infty} v\,e^{-\lambda v}\,dv = \frac{1}{\lambda^2}\]

\[\boxed{M_V(t) = \frac{1}{(1-t)^2}}\]

25.2.1 Introduction

When two discrete random variables \(X\) and \(Y\) are considered together, we define their joint probability mass function \(f(x,y) = P(X=x, Y=y)\).

Example. Suppose we toss a pair of fair, four-sided dice — one RED and one BLACK. Let:

• \(X\) = outcome on the RED die \(\in \{1,2,3,4\}\)
• \(Y\) = outcome on the BLACK die \(\in \{1,2,3,4\}\)

The joint sample space has \(4 \times 4 = 16\) equally likely outcomes:

\[S = \{(1,1),(1,2),\ldots,(4,4)\}\]

Since the dice are fair:

\[P(X = x,\; Y = y) = \frac{1}{16} \quad \text{for all } (x,y) \in S\]

25.2.2 Example 2: Joint Distribution of Discrete \(X\) and \(Y\)

Consider the following joint probability distribution:

Verification that probabilities sum to 1:

\[\frac{1}{12}+\frac{1}{6}+\frac{1}{12}+\frac{1}{8}+\frac{1}{4}+\frac{1}{24}+\frac{1}{12}+\frac{1}{24}+\frac{1}{12} = 1 \checkmark\]

Find: (a) the marginal distribution of \(X\); (b) the marginal distribution of \(Y\); (c) whether \(X\) and \(Y\) are independent; (d) \(\text{Cov}(X,Y)\); (e) the correlation coefficient.

25.3.1 Joint Probability Density Function

Let \(X\) and \(Y\) be two continuous random variables with two-dimensional sample space \(S\). The function \(f(x,y)\) is a joint probability density function if it satisfies:

1. \(f(x,y) \geq 0\) for all \((x,y)\)
2. \(\displaystyle\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} f(x,y)\,dx\,dy = 1\)
3. For any region \(A\) in the sample space: \(P[(X,Y) \in A] = \displaystyle\iint_A f(x,y)\,dx\,dy\)

Key definitions:

• Marginal PDF of \(X\): \(\displaystyle f_X(x) = \int_{-\infty}^{\infty} f(x,y)\,dy\)
• Marginal PDF of \(Y\): \(\displaystyle f_Y(y) = \int_{-\infty}^{\infty} f(x,y)\,dx\)
• Independence: \(X\) and \(Y\) are independent if \(f(x,y) = f_X(x)\,f_Y(y)\)
• Covariance: \(\text{Cov}(X,Y) = E(XY) - E(X)E(Y)\)
• Correlation: \(\rho_{XY} = \dfrac{\text{Cov}(X,Y)}{\sigma_X \sigma_Y}\)

25.3.2 Example 1: \(f(x,y) = 4xy\)

\[f(x,y) = \begin{cases} 4xy, & 0 < x < 1,\; 0 < y < 1 \\ 0, & \text{elsewhere} \end{cases}\]

25.3.3 Example 2: \(f(x,y) = kx + y\)

\[f(x,y) = \begin{cases} kx + y, & 0 < x < 1,\; 0 < y < 1 \\ 0, & \text{elsewhere} \end{cases}\]

25.3.4 Example 3: \(f(x,y) = cx + 1\) over a Triangular Region

\[f_{X,Y}(x,y) = \begin{cases} cx + 1, & x \geq 0,\; y \geq 0,\; x+y < 1 \\ 0, & \text{otherwise} \end{cases}\]