We concern ourselves with the problem of finding the probability distributions or densities of functions of one or more random variables. Given a set of random variables \(X_1, X_2, \ldots, X_n\) and their joint probability or density, we are interested in finding the joint probability distribution of some random variable:
\[Y = u(X_1, X_2, \ldots, X_n)\]
This means the values of \(Y\) are related to those of the \(X\)’s by the equation \(y = u(x_1, x_2, \ldots, x_n)\).
Several methods are available for solving this kind of problem. The ones we discuss are:
A straightforward method of obtaining the probability density of a function of continuous random variables consists of:
Thus, if \(X_1, X_2, \ldots, X_n\) are continuous random variables with a given joint probability density, the probability density of \(Y = u(X_1, X_2, \ldots, X_n)\) is obtained by first determining:
\[F_Y(y) = \Pr(Y \leq y) = \Pr\!\left[u(X_1, X_2, \ldots, X_n) \leq y\right]\]
and then differentiating to get:
\[f(y) = \frac{d}{dy} F(y)\]
Problem: The probability density of \(X\) is given by:
\[f(x) = \begin{cases} 6x(1 - x), & 0 < x < 1 \\ 0, & \text{elsewhere} \end{cases}\]
Find the probability density of \(Y = X^3\).
Solution:
Letting \(G(y)\) denote the distribution function of \(Y\):
\[G(y) = \Pr(Y \leq y) = \Pr(X^3 \leq y) = \Pr\!\left(X \leq y^{1/3}\right)\]
\[= \int_0^{y^{1/3}} 6x(1-x)\, dx = 3y^{2/3} - 2y\]
Differentiating:
\[g(y) = \frac{d}{dy} G(y) = 2y^{-1/3} - 2 = \frac{2}{3}\left(y^{-1/3} - 1\right) \cdot 3 = 2\left(y^{-1/3} - 1\right)\]
More precisely:
\[g(y) = \frac{1}{3}\left(2y^{-1/3} - 2\right)\]
\[\boxed{g(y) = \begin{cases} \frac{1}{3}(2y^{-1/3} - 2), & 0 < y < 1 \\ 0, & \text{elsewhere} \end{cases}}\]
Problem: If \(Y = |X|\), show that:
\[g(y) = \begin{cases} f(y) + f(-y), & y > 0 \\ 0, & \text{elsewhere} \end{cases}\]
where \(f(x)\) is the probability density of \(X\).
Solution:
For \(y > 0\):
\[G(y) = \Pr(Y \leq y) = \Pr(|X| \leq y) = \Pr(-y \leq X \leq y) = F(y) - F(-y)\]
Differentiating:
\[g(y) = \frac{d}{dy}\left[F(y) - F(-y)\right] = f(y) \cdot 1 - f(-y) \cdot (-1) = f(y) + f(-y)\]
Since \(X\) cannot be negative, \(g(y) = 0\) for \(y < 0\). Letting \(g(0) = 0\):
\[\boxed{g(y) = \begin{cases} f(y) + f(-y), & y > 0 \\ 0, & \text{elsewhere} \end{cases}}\]
Problem: Given the joint density of \(X_1\) and \(X_2\):
\[f(x_1, x_2) = \begin{cases} 6e^{-x_1 - 2x_2}, & x_1 > 0,\ x_2 > 0 \\ 0, & \text{elsewhere} \end{cases}\]
Find the probability density of \(Y = X_1 + X_2\).
Solution:
Integrating the joint density over the region where \(x_1 + x_2 \leq y\):
\[F(y) = \int_0^y \int_0^{y - x_1} 6e^{-x_1 - 2x_2}\, dx_2\, dx_1\]
After integration:
\[F(y) = 1 - 2e^{-y} + 3e^{-2y} - 2e^{-3y} \quad \text{(after simplification)}\]
Differentiating gives the density:
\[\boxed{f(y) = \begin{cases} 6e^{-2y} - 6e^{-3y}, & y > 0 \\ 0, & \text{elsewhere} \end{cases}}\]
If \(f(x) = 2xe^{-x^2}\) for \(x > 0\) (0 elsewhere) and \(Y = X^2\), find:
If \(X\) has an exponential distribution with parameter \(\lambda\), use the distribution function technique to find the p.d.f. of \(Y = \ln X\).
If \(X\) has the uniform density with \(\alpha = 0\) and \(\beta = 1\), find the p.d.f. of \(Y = \sqrt{X}\): \[g(y) = \begin{cases} 2y, & 0 < y < 1 \\ 0, & \text{elsewhere} \end{cases}\]
If the joint p.d.f. of \(X\) and \(Y\) is \(f(x,y) = 4xye^{-(x^2+y^2)}\) for \(x,y > 0\) (0 elsewhere) and \(Z = X^2 + Y^2\), find:
If \(X_1\) and \(X_2\) are independent exponential random variables with parameters \(\lambda_1\) and \(\lambda_2\), find the p.d.f. of \(Y = X_1 + X_2\) when:
With reference to Exercise 5 (when \(\lambda_1 \neq \lambda_2\)), show that \(Z = \dfrac{X_1}{X_1 + X_2}\) has a uniform density with \(\alpha = 0\), \(\beta = 1\).
If the joint density of \(X\) and \(Y\) is \(f(x,y) = e^{-(x+y)}\) for \(x,y > 0\) (0 elsewhere) and \(Z = \dfrac{X+Y}{2}\), find the p.d.f. of \(Z\) using the distribution function technique.
The percentages of copper and iron in a certain ore are \(X_1\) and \(X_2\) respectively. Their joint density is: \[f(x_1, x_2) = \begin{cases} 5, & 0 \leq x_1 \leq 1,\ 0 \leq x_2 \leq \frac{2-x_1}{2} \\ 0, & \text{elsewhere} \end{cases}\] Find the p.d.f. of \(Y = X_1 + X_2\) and \(E(Y)\).
\[g(y) = \begin{cases} \tfrac{11}{9}\cdot\tfrac{y}{2}, & 0 \leq y < 1 \\ \tfrac{3(2-y)(7-4y)}{22}, & 1 \leq y \leq 2 \\ 0, & \text{elsewhere} \end{cases}\]
We can determine the probability distribution or density of a function of a random variable without first finding its distribution function.
In the discrete case, when the relationship between the values of \(X\) and \(Y = u(X)\) is one-to-one, we simply make the appropriate substitution.
Problem: If \(X\) is the number of heads in four tosses of a fair coin, find the probability distribution of:
\[Y = \frac{1}{X + 1}\]
Solution:
Using the binomial distribution with \(n = 4\), \(p = \frac{1}{2}\):
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \(f(x)\) | \(\frac{1}{16}\) | \(\frac{4}{16}\) | \(\frac{6}{16}\) | \(\frac{4}{16}\) | \(\frac{1}{16}\) |
Applying \(y = \frac{1}{x+1}\):
| \(y\) | 1 | \(\frac{1}{2}\) | \(\frac{1}{3}\) | \(\frac{1}{4}\) | \(\frac{1}{5}\) |
|---|---|---|---|---|---|
| \(g(y)\) | \(\frac{1}{16}\) | \(\frac{4}{16}\) | \(\frac{6}{16}\) | \(\frac{4}{16}\) | \(\frac{1}{16}\) |
Substituting \(x = \frac{1}{y} - 1\) into the binomial formula directly:
\[g(y) = \binom{4}{\frac{1}{y}-1}\left(\frac{1}{2}\right)^4, \quad y = 1, \tfrac{1}{2}, \tfrac{1}{3}, \tfrac{1}{4}, \tfrac{1}{5}\]
Note: The probabilities remain unchanged — only the variable changes from \(X\) to \(Y\).
Problem: Find the probability distribution of \(Z = (X - 2)^2\) where \(X\) is as in Example 4.
Solution:
| \(x\) | 0 | 1 | 2 | 3 | 4 |
|---|---|---|---|---|---|
| \((x-2)^2 = z\) | 4 | 1 | 0 | 1 | 4 |
Solving \(x = 2 \pm \sqrt{z}\) and summing probabilities for each \(z\):
\[h(0) = f(2) = \frac{6}{16} = \frac{3}{8}\]
\[h(1) = f(1) + f(3) = \frac{4}{16} + \frac{4}{16} = \frac{8}{16} = \frac{1}{2}\]
\[h(4) = f(0) + f(4) = \frac{1}{16} + \frac{1}{16} = \frac{2}{16} = \frac{1}{8}\]
| \(z\) | 0 | 1 | 4 |
|---|---|---|---|
| \(h(z)\) | \(\frac{3}{8}\) | \(\frac{1}{2}\) | \(\frac{1}{8}\) |
Theorem 1: Let \(f(x)\) be the probability density of the continuous random variable \(X\). If \(y = u(x)\) is differentiable and either strictly increasing or strictly decreasing for all values where \(f(x) > 0\), then the equation \(y = u(x)\) can be uniquely solved for \(x = w(y)\), and the probability density of \(Y = u(X)\) is:
\[g(y) = f[w(y)] \cdot \left|w'(y)\right|, \quad \text{provided } u'(x) \neq 0\]
and \(g(y) = 0\) elsewhere.
Proof sketch: For the increasing case, \(\Pr(a < Y < b) = \Pr(w(a) < X < w(b))\). Changing variables \(x = w(y)\), \(dx = w'(y)\,dy\) in the integral yields \(g(y) = f[w(y)] \cdot |w'(y)|\). The decreasing case follows similarly, with the absolute value ensuring positivity. \(\square\)
Problem: If \(X\) has the exponential distribution \(f(x) = e^{-x}\) for \(x > 0\) (0 elsewhere), find the p.d.f. of \(Y = \sqrt{X}\).
Solution:
The inverse is \(x = y^2\), so \(w(y) = y^2\) and \(w'(y) = 2y\).
\[g(y) = f(y^2) \cdot |2y| = e^{-y^2} \cdot 2y = 2ye^{-y^2}\]
\[\boxed{g(y) = \begin{cases} 2ye^{-y^2}, & y > 0 \\ 0, & \text{elsewhere} \end{cases}}\]
Problem: A spinner gives angle \(\Theta\) with uniform density \(f(\theta) = \frac{1}{2\pi}\) for \(-\frac{\pi}{2} < \theta < \frac{\pi}{2}\) (0 elsewhere). Find the density of \(X = a\tan\Theta\), the abscissa on the \(x\)-axis.
Solution:
The inverse is \(\theta = \arctan(x/a)\), so:
\[\frac{d\theta}{dx} = \frac{a}{a^2 + x^2}\]
\[g(x) = \frac{1}{2\pi} \cdot \frac{a}{a^2 + x^2} \cdot 2\pi = \frac{a}{\pi(a^2 + x^2)} \quad \text{for } -\infty < x < \infty\]
This is the Cauchy distribution.
Given the joint distribution of \(X_1\) and \(X_2\), we want the distribution of \(Y = u(X_1, X_2)\).
Strategy: Introduce an auxiliary variable \(Z\) (e.g., \(Z = X_2\)), find the joint density of \((Y, Z)\), then integrate out \(Z\) to obtain the marginal density of \(Y\).
Theorem 2: Let \(f(x_1, x_2)\) be the joint probability density of the continuous random variables \(X_1\) and \(X_2\). If:
\[y_1 = u_1(x_1, x_2) \quad \text{and} \quad y_2 = u_2(x_1, x_2)\]
are partially differentiable and define a one-to-one transformation, with inverse \(x_i = w_i(y_1, y_2)\), then the joint density of \(Y_1 = u_1(X_1, X_2)\) and \(Y_2 = u_2(X_1, X_2)\) is:
\[g(y_1, y_2) = f[w_1(y_1, y_2),\ w_2(y_1, y_2)] \cdot |J|\]
where the Jacobian is:
\[J = \begin{vmatrix} \dfrac{\partial x_1}{\partial y_1} & \dfrac{\partial x_1}{\partial y_2} \\[8pt] \dfrac{\partial x_2}{\partial y_1} & \dfrac{\partial x_2}{\partial y_2} \end{vmatrix}\]
Problem: \(X_1\) and \(X_2\) are independent Poisson random variables with parameters \(\lambda_1\) and \(\lambda_2\). Find the distribution of \(Y = X_1 + X_2\).
Solution:
Their joint distribution is:
\[f(x_1, x_2) = \frac{e^{-\lambda_1}\lambda_1^{x_1}}{x_1!} \cdot \frac{e^{-\lambda_2}\lambda_2^{x_2}}{x_2!}\]
Setting \(y = x_1 + x_2\) (so \(x_1 = y - x_2\)) and summing over \(x_2 = 0, 1, \ldots, y\):
\[h(y) = \sum_{x_2=0}^{y} \frac{e^{-(\lambda_1+\lambda_2)}\lambda_1^{y-x_2}\lambda_2^{x_2}}{(y-x_2)!\, x_2!} = \frac{e^{-(\lambda_1+\lambda_2)}(\lambda_1+\lambda_2)^y}{y!}\]
using the binomial expansion \((\lambda_1 + \lambda_2)^y = \sum_{x_2=0}^{y}\binom{y}{x_2}\lambda_1^{y-x_2}\lambda_2^{x_2}\).
Conclusion: The sum of two independent Poisson random variables with parameters \(\lambda_1\) and \(\lambda_2\) is Poisson with parameter \(\lambda_1 + \lambda_2\).
Problem: Given joint density \(f(x_1, x_2) = e^{-(x_1+x_2)}\) for \(x_1, x_2 > 0\) (0 elsewhere), find:
Solution (a):
Solving: \(x_1 = y_1 y_2\) and \(x_2 = y_1(1 - y_2)\).
\[J = \begin{vmatrix} y_2 & y_1 \\ 1-y_2 & -y_1 \end{vmatrix} = -y_1 y_2 - y_1(1-y_2) = -y_1\]
\[|J| = y_1\]
\[g(y_1, y_2) = e^{-y_1 y_2 - y_1(1-y_2)} \cdot y_1 = y_1 e^{-y_1}\]
\[\boxed{g(y_1, y_2) = \begin{cases} y_1 e^{-y_1}, & y_1 > 0,\ 0 < y_2 < 1 \\ 0, & \text{elsewhere} \end{cases}}\]
Solution (b):
\[h(y_2) = \int_0^\infty y_1 e^{-y_1}\, dy_1 = \Gamma(2) = 1! = 1\]
\[\boxed{h(y_2) = \begin{cases} 1, & 0 < y_2 < 1 \\ 0, & \text{elsewhere} \end{cases}}\]
This is the uniform distribution on \((0, 1)\).
Problem: \(f(x_1, x_2) = 1\) for \(0 < x_1 < 1\), \(0 < x_2 < 1\) (0 elsewhere). Find the marginal density of \(Y = X_1 + X_2\).
Solution:
Let \(Y = X_1 + X_2\) and \(Z = X_2\). Then \(x_1 = y - z\), \(x_2 = z\), \(|J| = 1\).
The region maps to \(z < y < z+1\), \(0 < z < 1\), i.e., \(z < y < z+1\).
Integrating out \(z\):
\[h(y) = \begin{cases} \int_0^y dz = y, & 0 \leq y < 1 \\[4pt] \int_{y-1}^{1} dz = 2 - y, & 1 \leq y \leq 2 \\[4pt] 0, & \text{elsewhere} \end{cases}\]
This is the triangular distribution.
Problem: \(f(x_1, x_2, x_3) = e^{-(x_1+x_2+x_3)}\) for \(x_i > 0\) (0 elsewhere). Find the marginal density of \(Y_1 = X_1 + X_2 + X_3\).
Solution:
Let \(Y_2 = X_2\) and \(Y_3 = X_3\). Then \(x_1 = y_1 - y_2 - y_3\), \(x_2 = y_2\), \(x_3 = y_3\). The \(3\times 3\) Jacobian equals 1.
\[g(y_1, y_2, y_3) = e^{-y_1}, \quad y_1 > y_2 + y_3,\ y_2, y_3 > 0\]
Integrating out \(y_2\) and \(y_3\):
\[h(y_1) = \int_0^{y_1}\int_0^{y_1 - y_3} e^{-y_1}\, dy_2\, dy_3 = \frac{1}{2}y_1^2 e^{-y_1}\]
\[\boxed{h(y_1) = \begin{cases} \dfrac{1}{2}y_1^2 e^{-y_1}, & y_1 > 0 \\ 0, & \text{elsewhere} \end{cases}}\]
This is a Gamma distribution with \(\alpha = 3\) and \(\beta = 1\).
If \(X\) has a hypergeometric distribution with \(m=3\), \(N=6\), \(n=2\), find the p.d. of \(Z = (X-1)^2\). \[h(0) = \frac{3}{5}, \quad h(1) = \frac{2}{5}\]
If \(X\) has a binomial distribution with \(n=3\), \(p=\frac{1}{3}\), find the p.d. of:
If \(X = \ln(Y)\) has a normal distribution with mean \(\mu\) and std dev \(\sigma\), find the p.d.f. of \(Y\) (the log-normal distribution): \[g(y) = \frac{1}{\sqrt{2\pi}\,\sigma y} \exp\!\left(-\frac{(\ln y - \mu)^2}{2\sigma^2}\right), \quad y > 0\]
If \(f(x) = kx^3(1-x^2)^6\) for \(0 < x < 1\) (0 elsewhere), find the p.d.f. of \(Y = X^2\). Show it is a Beta distribution and find \(k\).
The distribution is Beta with \(\alpha = 4\), \(\beta = 2\); \(k = 320\).
If \(X\) has a uniform density with \(\alpha=0\), \(\beta=1\), show that \(Y = -2\ln X\) has a gamma distribution. Find its parameters.
If \(X\) has a uniform density with \(\alpha=0\), \(\beta=1\), show that \(Y = X^{-1/\alpha}\) (for \(\alpha > 0\)) has the Pareto distribution.
Consider \(X \sim \text{Uniform}(-1, 3)\):
If \(f(x_1, x_2) = \dfrac{x_1 x_2}{36}\) for \(x_1, x_2 = 1, 2, 3\), find the p.d. of:
With the same joint distribution as in Exercise 8, find:
If \(X_1\), \(X_2\), \(X_3\) have the multinomial distribution with \(n=2\), \(\theta_1=\frac{1}{4}\), \(\theta_2=\frac{1}{2}\), \(\theta_3=\frac{5}{12}\), find the joint p.d. of \(Y_1 = X_1 + X_2\), \(Y_2 = X_1 - X_2\), \(Y_3 = X_3\).
If \(X_1\) and \(X_2\) are independent binomial with parameters \((n_1, \theta)\) and \((n_2, \theta)\), show that \(Y = X_1 + X_2\) is binomial with parameters \((n_1 + n_2, \theta)\).
If \(X\) and \(Y\) are independent standard normal, show that \(Z = X + Y\) is normally distributed. Find the mean and variance. \(\mu = 0\), \(\sigma^2 = 2\)
Given \(f(x, y) = 12y(1-x-y)\) for \(0 < x < 1\), \(0 < y < 1-x\) (0 elsewhere):
Two independent Cauchy random variables \(X_1\) and \(X_2\) with density \(f(x) = \dfrac{1}{\pi(1+x^2)}\):
Given \(f(x,y) = 24xy\) for \(0 < x < 1\), \(0 < y < 1\), \(x + y < 1\), find the joint density of \(Z = X+Y\) and \(W = X\).
Let \(X\) and \(Y\) be independent gamma random variables. Find the joint density of \(U = \dfrac{X}{X+Y}\) and \(V = X+Y\), and identify the marginal density of \(U\).
The Maxwell–Boltzmann velocity distribution is \(f(v) = kv^2 e^{-\beta v^2}\) for \(v > 0\). Show that kinetic energy \(E = \frac{1}{2}mV^2\) has a gamma distribution.
Moment-generating functions provide an elegant tool for finding the distribution of linear combinations of independent random variables.
The method is based on the uniqueness of the MGF: if \(M_Y(t) = M_Z(t)\) for all \(t\) in some interval, then \(Y\) and \(Z\) have the same distribution.
Theorem 3: If \(X_1, X_2, \ldots, X_n\) are independent random variables and \(Y = X_1 + X_2 + \cdots + X_n\), then:
\[M_Y(t) = \prod_{i=1}^{n} M_{X_i}(t)\]
Proof:
\[M_Y(t) = E\!\left[e^{tY}\right] = E\!\left[e^{t(X_1 + \cdots + X_n)}\right] = E\!\left[e^{tX_1}\right] \cdots E\!\left[e^{tX_n}\right] = \prod_{i=1}^{n} M_{X_i}(t)\]
(using independence). \(\square\)
Problem: \(X_1, \ldots, X_n\) are independent Poisson with parameters \(\lambda_1, \ldots, \lambda_n\). Find the distribution of \(Y = X_1 + \cdots + X_n\).
Solution:
The MGF of a Poisson\((\lambda_i)\) is \(M_{X_i}(t) = e^{\lambda_i(e^t - 1)}\). By Theorem 3:
\[M_Y(t) = \prod_{i=1}^n e^{\lambda_i(e^t-1)} = e^{(\lambda_1 + \cdots + \lambda_n)(e^t - 1)}\]
Conclusion: \(Y \sim \text{Poisson}(\lambda_1 + \lambda_2 + \cdots + \lambda_n)\).
Problem: \(X_1, \ldots, X_n\) are i.i.d. Exponential\((\lambda)\) (equivalently Gamma\((1, \lambda)\)). Find the distribution of \(Y = \sum X_i\).
Solution:
The MGF of Gamma\((\alpha, \beta)\) is \(M_X(t) = \left(1 - \frac{t}{\beta}\right)^{-\alpha}\). For Exponential\((\lambda)\):
\[M_{X_i}(t) = \left(1 - \frac{t}{\lambda}\right)^{-1}\]
\[M_Y(t) = \left(1 - \frac{t}{\lambda}\right)^{-n}\]
Conclusion: \(Y \sim \text{Gamma}(n, \lambda)\) with p.d.f.:
\[g(y) = \frac{\lambda^n}{\Gamma(n)} y^{n-1} e^{-\lambda y}, \quad y > 0\]
Setup: The number of fish caught per hour at Lake Victoria follows \(\text{Poisson}(\lambda = 1.6)\).
Let \(Y = X_1 + X_2 \sim \text{Poisson}(3.2)\):
\[P(Y = 4) = \frac{e^{-3.2}(3.2)^4}{4!} = \frac{0.04076 \times 104.86}{24} \approx \mathbf{0.1781}\]
Let \(Y = X_1 + X_2 + X_3 \sim \text{Poisson}(4.8)\):
\[P(Y \geq 2) = 1 - P(Y = 0) - P(Y = 1)\]
\[P(Y = 0) = e^{-4.8} \approx 0.00823, \quad P(Y = 1) = 4.8e^{-4.8} \approx 0.03950\]
\[P(Y \geq 2) = 1 - 0.04773 \approx \mathbf{0.9523}\]
Let \(Y = X_1 + X_2 + X_3 + X_4 \sim \text{Poisson}(6.4)\):
\[P(Y \leq 3) = \sum_{y=0}^{3} \frac{e^{-6.4}(6.4)^y}{y!}\]
\[= 0.00166 + 0.01063 + 0.03403 + 0.07253 \approx \mathbf{0.1189}\]
Chebyshev’s Theorem: If \(\mu\) and \(\sigma\) are the mean and standard deviation of a random variable \(X\), then for any \(k > 0\):
\[\Pr(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2}\]
Equivalently: \(\Pr(\mu - k\sigma < X < \mu + k\sigma) \geq 1 - \dfrac{1}{k^2}\).
This provides a universal lower bound on the probability that \(X\) lies within \(k\) standard deviations of its mean, valid for any distribution.
By definition:
\[\sigma^2 = E[(X - \mu)^2] = \int_{-\infty}^{\infty}(x-\mu)^2 f(x)\,dx\]
Split the integral into three regions:
\[\sigma^2 \geq \int_{-\infty}^{\mu - k\sigma}(x-\mu)^2 f(x)\,dx + \int_{\mu + k\sigma}^{\infty}(x-\mu)^2 f(x)\,dx\]
Since \((x - \mu)^2 \geq k^2\sigma^2\) in the outer regions:
\[\sigma^2 \geq k^2\sigma^2 \cdot \Pr(|X - \mu| \geq k\sigma)\]
\[\Pr(|X - \mu| \geq k\sigma) \leq \frac{1}{k^2}\]
\[\Pr(|X - \mu| < k\sigma) \geq 1 - \frac{1}{k^2} \quad \square\]
Problem: \(X\) has p.d.f. \(f(x) = 630x^4(1-x)^4\) for \(0 < x < 1\). Compare the exact probability within two standard deviations of the mean to Chebyshev’s bound.
Solution:
Comparing to the Beta\((\alpha, \beta)\) density, we identify \(\alpha = 5\), \(\beta = 5\).
\[\mu = E(X) = \frac{\alpha}{\alpha + \beta} = \frac{5}{10} = 0.5\]
\[\sigma^2 = \text{Var}(X) = \frac{\alpha\beta}{(\alpha+\beta)^2(\alpha+\beta+1)} = \frac{25}{100 \times 11} = \frac{1}{44}\]
\[\sigma = \frac{1}{\sqrt{44}} \approx 0.1508\]
The interval within 2 standard deviations: \([0.5 - 2(0.1508),\ 0.5 + 2(0.1508)] = [0.20, 0.80]\).
Exact probability:
\[P(0.20 < X < 0.80) = 630\int_{0.20}^{0.80} x^4(1-x)^4\, dx \approx \mathbf{0.96}\]
Chebyshev bound (with \(k = 2\)):
\[1 - \frac{1}{k^2} = 1 - \frac{1}{4} = 0.75\]
The exact probability \(0.96\) is much stronger than Chebyshev’s lower bound \(0.75\).
Problem: \(X \sim \text{Uniform}(-3, 3)\). With \(k = \frac{3}{2}\), find:
Solution:
\[\mu = \frac{-3 + 3}{2} = 0, \qquad \sigma^2 = \frac{(3 - (-3))^2}{12} = \frac{36}{12} = 3, \qquad \sigma = \sqrt{3}\]
With \(k = \frac{3}{2}\): \(k\sigma = \frac{3}{2}\sqrt{3}\)
Exact probability:
\[P\!\left(|X| < \frac{3\sqrt{3}}{2}\right) = P\!\left(-\frac{3\sqrt{3}}{2} < X < \frac{3\sqrt{3}}{2}\right) = \frac{1}{6} \cdot 3\sqrt{3} \approx 0.866\]
Actually, since \(\frac{3\sqrt{3}}{2} \approx 2.598 < 3\):
\[P = \frac{2 \cdot 2.598}{6} = \frac{5.196}{6} \approx \mathbf{0.866}\]
Chebyshev bound:
\[1 - \frac{1}{(3/2)^2} = 1 - \frac{4}{9} = \frac{5}{9} \approx 0.556\]
MGF Technique: If \(X_1\) and \(X_2\) are independent binomial with parameters \((n_1, \theta)\) and \((n_2, \theta)\), show using MGFs that \(Y = X_1 + X_2 \sim \text{Binomial}(n_1 + n_2, \theta)\).
If \(n\) independent random variables each have Gamma\((\alpha, \beta)\) distributions, find the MGF of their sum and identify the distribution. Answer: Gamma\((n\alpha, \beta)\).
If \(X_i \sim N(\mu_i, \sigma_i^2)\) independently, find the MGF of \(Y = \sum X_i\) and identify the distribution. \[Y \sim N\!\left(\sum_{i=1}^n \mu_i,\ \sum_{i=1}^n \sigma_i^2\right)\]
Prove the generalization of Theorem 3: if \(Y = a_1 X_1 + \cdots + a_n X_n\) with \(X_i\) independent, then \(M_Y(t) = \prod_{i=1}^n M_{X_i}(a_i t)\).
A lawyer receives calls on an unlisted number at rate \(2.1\)/half-hour and on a listed number at rate \(10.9\)/half-hour (independent Poisson). Find the probabilities that in half an hour she receives:
(Repeat of fish problem — see Section @ref(ex-fish) above.)
If the time a doctor spends per patient is Exponential\((\lambda = \frac{1}{9})\) (minutes), find the probability the doctor spends at least 20 minutes with:
Chebyshev: Find the smallest \(k\) such that \(P(|X - \mu| < k\sigma) \geq\):
The thiamine content in a bread slice has \(\mu = 0.260\) mg, \(\sigma = 0.005\) mg. By Chebyshev’s theorem, between what values must the thiamine content lie for:
If \(E(X) = 3\) and \(E(X^2) = \frac{13}{2}\), use Chebyshev’s theorem to find a lower bound for \(P(2 < X < 8)\). Answer: 0.84
End of STA227 Weeks 4–6 Notes