Question #1

  1. Suppose that X is a discrete random variable having probability function Pr(X = k) = c\(k^2\) for k = 1,2,3. Find c, Pr(X \(\le\) 2), E[X], and Var(X).

Answer #1

From the axioms of probability, we know that the probability of the entire sample space is 1. hence,

  1. find c Pr(X = 1) + Pr(X = 2)+ Pr(X = 3) = 1.

with: Pr(X = 1) = c \(k^2\) = c\(1^2\) = c
Pr(X = 2) = c \(k^2\) = c\(2^2\) = 4c
Pr(X = 3) = c \(k^2\) = c\(3^2\) = 9c

Therefore: \[ \begin{aligned} Pr(X = 1) + Pr(X = 2)+ Pr(X = 3) = & 1 \\ c + 4*c + 9*c = & 1 \\ 14*c = & 1 \\ c = & 1/14 \\ \end{aligned} \]

  1. Find Pr(X \(\le\) 2)

\[ \begin{aligned} Pr(X \le 2) = & Pr(X = 1) + Pr(X = 2) \\ Pr(X \le 2) = & c1^2 + c2^2 \\ Pr(X \le 2) = & (1/14) * 1 + (1/14) * 4 \\ \end{aligned} \]

Hence, Pr(X \(\le\) 2) = 0.3571429

  1. Find E(X)

E(X) = \(\sum_{i=1}^3 x_i\) * Pr(X=\(x_i\)) E(X) = 1(1(1/14)) + 2(4(1/14)) + 3(9(1/14))

Hence, the expected value E(X) = 2.5714286

  1. Find Var(X)

Var(X) = \(E[X^2]-(E[X])^2\) = 1^2 * (1/14) + 2^2 * (4/14) + 3^2 * (9/14) - (2.5714)^2

Hence the Var(X) = 0.387902

Question #2

  1. Suppose that X is a continuous random variable having p.d.f. f(x) = \(cx^2 for 1 \le x \le 2\) . Find c, \(Pr(X \ge 1), E[X], and Var(X)\) .

Answer #2

  1. Find c Since f(x) is p.d.f., then \(\int_{1}^{2} f(x) dx = 1\)
    \(\int_{1}^{2} cx^2 dx = 1\)
    \(\int_{1}^{2} cx^2 dx = \left[ \frac 1 3 cx^3 \right]_{1}^{2}= 1\) \(\int_{1}^{2} cx^2 dx = \frac 1 3 c*2^3 - \frac 1 3 c*1^1 = 1\) \(\int_{1}^{2} cx^2 dx = \frac 1 3 8c - \frac 1 3 c =1\) \(\int_{1}^{2} cx^2 dx = \frac 1 3 [8c - c] = 1\)

hence, \(\frac 1 3 [8c - c] = 1\) \([8c - c] = 3\) \(7c = 3\) Therefore, \(c= \frac 3 7\)

  1. Find Pr(X \(\ge\) 1) Since f(x) is a c.d.f. defined in the space \(1 \le x \le 2\). then Pr(X \(\ge\) 1) = \(\int_{1}^{2} f(x) dx= 1\)

  2. Find E[X] E[X] = \(\int_{1}^{2} xcx^2 dx\) E[X] = \(\int_{1}^{2} \frac 3 7 x^3 dx\) E[X] = \(\left[ \frac 3 {28} x^4 \right]_{1}^{2}\) E[X] = \(\frac 3 {28} 2^4 - \frac 3 {28} 1^4\) E[X] = \(\frac 3 {28} (16 - 1)\) E[X] = \(\frac 3 {28} (15)\) Hence, E(X) = 1.6071429

  3. Find Var(X)

Var(X) = \(E((X- \mu)^2)\)

we are going to use the Variance theorem: Var(X) = \(E(X^2) - E(X)^2 = E(X^2)- \mu^2\). hence, Var(X) = \(E(X^2) - \mu^2\) Var(X) = \(\int_{1}^{2} (cx^2)^2 dx - \mu^2\) Var(X) = \(\int_{1}^{2} \frac 9 {245} (x^2)^2 dx - \mu^2\) Var(X) = \(\int_{1}^{2} \frac 9 {245} x^4 dx - \mu^2\) Var(X) = \(\left[ \frac 9 {245} x^5 \right]_{1}^{2} - \mu^2\) Var(X) = \((\frac 9 {245} 2^5 - \frac 9 {245} 1^5) - \mu^2\) we have \(\mu\) = 1.6071429

Hence Var(X) = \((\frac 9 {245} 2^5 - \frac 9 {245} 1^5) - (1.6071429 )^2\) Var(X) = \((\frac 9 {245} (32 - 1)) - (1.6071429 )^2\) Var(X) = \((\frac 9 {245} (31)) - (1.6071429 )^2\)

Therefore,

Var(X) = -1.4441327

Question #3

Suppose that X and Y are jointly continuous random variables with

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

  1. Compute and plot \(f_x (x)\) and \(f_y (y)\).
  2. Are X and Y independent?
  3. Compute \(F_x (x)\) and \(F_y (y)\)
  4. Compute E[X], Var(X), E[Y], Var(Y), Cov(X,Y), Corr(X,Y)

Answer #3

  1. Compute and plot \(f_x (x)\) and \(f_y (y)\).

\[ f_X (x) = \int_1^2 f_{XY} (x,y) dy \] \[ f_X (x) = \int_1^2 (y-x) dy \] \[ f_X (x) = \left[ \frac {y^2} 2- xy\right]_{1}^{2} \] \[ f_X (x) = (\frac 4 2 - 2x ) - (\frac 1 2 - x) \] \[ f_X (x) = 2 - 2x - \frac 1 2 + x \] \[ f_X (x) = \frac 3 2 - x \]

now we will compute \(f_y (y)\)
\[ f_Y (y) = \int_0^1 f_{XY} (x,y) dx \] \[ f_Y (y) = \int_0^1 (y-x) dx \] \[ f_Y (y) = \left[ xy - \frac {x^2} 2 \right]_{0}^{1} \] \[ f_Y (y) = (y - \frac 1 2) - 0\] \[ f_Y (y) = y - \frac 1 2 \]

Plotting f(x) and f(y)

  1. Are X and Y independent?

We say X and Y are independent iif \(f(x,y) = f(x) f(y)\) We know that \(f(x,y) = y - x\). recall \(f_X (x) = \frac 3 2 - x\) and \(f_Y (y) = y - \frac 1 2\) Now let’s find out \(f(x) f(y)\) \(f(x) f(y) = (\frac 3 2 - x)(y - \frac 1 2)\) \(f(x) f(y) = y \frac 3 2 - \frac 3 4 - xy + \frac x 2\)

hence X and Y are Not independent as f(x) f(y) is not equal to \(f(x,y) =y - x\)

  1. Compute \(F_x (x)\) and \(F_y (y)\)

To find the joint CDF, we need to integrate the joint PDF.
recall our PDF’s are: \(f_X (x) = \frac 3 2 - x\) and \(f_Y (y) = y - \frac 1 2\) first, let’s find \(F_x (x)\) \[ F_x (x) = \int_{-\infty}^x f_x (u) du \] \[ F_x (x) = \frac{3}{2} x - \frac{1}{2} x^2 \] then, \(F_y (y)\) \[ F_y (y) = \int_{-\infty}^y f_y (u) du \] \[ F_y (y) = \frac{1}{2} y^2 - \frac{1}{2} y \]

  1. Compute E[X], Var(X), E[Y], Var(Y), Cov(X,Y), and Corr(X,Y).

\[ E[X] = \int_0^1 x f_x (x) dx \] \[ E[X] = \int_0^1 x (\frac 3 2 - x) dx \] \[ E[X] = \int_0^1 \frac{3}{2} x - x^2 \] \[ E[X] = \left[ \frac{3}{4} x^2 - \frac{1}{3} x^3 \right]_{0}^{1}\] \[ E[X] = \frac 3 4 1^2 - \frac 1 3 1^3 - 0 \] \[ E[X] = \frac 3 4 - \frac 1 3 \]

EX<- (3/4) - (1/3)
EXX<- EX^2

hence E(X) = 0.4166667

Var(X) \[ Var(X) = \int_0^1 x^2 f(x) dx - \mu^2\] \[ Var(X) = \int_0^1 x^2 (\frac{3}{2} - x) - \mu^2\] \[ Var(X) = \int_0^1 \frac{3}{2} x^2 - x^3 - \mu^2 \] \[ Var(X) = \left[ \frac{1}{2} x^3 - \frac{1}{4} x^4 \right]_{0}^{1} - \mu^2\] \[ Var(X) = \frac{1}{2} 1^3 - \frac{1}{4} 1^4 - 0 - \mu^2\] \[ Var(X) = \frac{1}{2} - \frac{1}{4} - 0.1736111\]

VarX<- (1/2) - (1/4) - EXX

hence Var(X) = 0.0763889

\[ E[Y] = \int_1^2 y f_y (y) dy \] \[ E[Y] = \int_1^2 y (y - \frac 1 2) dy \] \[ E[Y] = \int_1^2 (y^2 - \frac 1 2 y ) dy\] \[ E[Y] = \left[ \frac 1 3 y^3 - \frac 1 4 y^2 \right]_{1}^{2}\] \[ E[Y] = [ (\frac 1 3 2^3 - \frac 1 4 2^2) - (\frac 1 3 1^3 - \frac 1 4 1^2) \] \[ E[Y] = [ (\frac 8 3 - 1 ) - (\frac 1 3 - \frac 1 4 ) \]

EY<- ((8/3) - 1) - ( 1/3 - 1/4)
EYY<- EY^2

hence E(Y) = 1.5833333

\[ Var(Y) = \int_0^1 y^2 f(y) dy - \mu^2\] \[ Var(Y) = \int_0^1 y^2 (y-\frac{1}{2}) dy - \mu^2\] \[ Var(Y) = \int_0^1 (y^3 - \frac{1}{2} y^2) dy - \mu^2\] \[ Var(Y) = \left[ \frac{1}{4} y^4 - \frac{1}{6} y^3\right]_{1}^{2} - \mu^2 \] \[ Var(Y) = (\frac{1}{4} 2^4 - \frac{1}{6} 2^3) - (\frac{1}{4} - \frac{1}{6}) - \mu^2 \] \[ Var(Y) = (4 - \frac{4}{3} ) - (\frac{1}{4} - \frac{1}{6}) - 2.5069444 \]

VarY<- 4 - (4/3) - (1/4 - 1/6) - EYY

hence Var(Y) = 0.0763889

\[ Cov(X,Y) = E[XY] - E[X]E[Y] \]

\[ E[XY] = \int_{-\infty}^{\infty} \int_{-\infty}^{\infty} xyf_{xy}(x,y) dx dy \] \[ E[XY] = \int_{1}^{2} \int_{0}^{1} xy(y-x) dx dy = \int_{1}^{2} \int_{0}^{1} xy^2 - x^2 y dx dy \] \[ E[XY] = \int_1^2 \frac{x^2 y^2}{2} - \frac{x^3 y}{3} \vert_0^1 dy\] \[ E[XY] = \int_1^2 \frac{y^2}{2} - \frac{y}{3} dy = \frac{y^3}{6} - \frac{y^2}{6} \vert_1^2 = \frac{2}{3} \]

Next we compute E[X]E[Y]

\[ E[X]E[Y] = \int_{1}^{2} (\frac 3 2 - x) \int_{0}^{1} (y - \frac 1 2) \] \[ E[X]E[Y] = (\left[ \frac {3x} {2} - \frac {-x^2} 2\right]_{1}^{2}) ( \left[ \frac {y^2} 2 - \frac {1y} 2 \right]_{0}^{1}) \] \[ E[X]E[Y] = (\frac {6} {2} - \frac {4} 2 - \frac {3} {2} - \frac {1} 2) ( \frac {1} 2 - \frac {1} 2) \] \[ E[X]E[Y] = (\frac {6} {2} - \frac {4} 2 - \frac {3} {2} - \frac {1} 2) (0) \] \[ E[X]E[Y] = 0 \]

Therefore, \[Cov(X,Y) = E[XY] - E[X]E[Y] \] \[Cov(X,Y) = \frac{2}{3} - 0 \] \[Cov(X,Y) = \frac{2}{3} \]

Question #4

Suppose that the following 10 observations come from the same distribution (not highly skewed) with unknown mean \(\mu\).

7.3, 6.1, 3.8, 8.4, 6.9, 7.1, 5.3, 8.2, 4.9, 5.8

Compute \(\bar{X}\), \(S^2\), and an approximate 95% confidence interval for \(\mu\)

Answer #4

Compute \(\bar{X}\)

X<- c(7.3, 6.1, 3.8, 8.4, 6.9, 7.1, 5.3, 8.2, 4.9, 5.8)
n<- length(X)
Xbar <- mean(X)
VarX<- var(X)
sdX<- sd(X)
ciw <- qt(0.95, n) * (sdX / sqrt(n))

Our mean or \(\bar{X}\) is: 6.38

Our Variance or \(S^2\) is: 2.1617778

To approximate 95% confidence interval for \(\mu\), we will use the following:

\(\bar{X} \pm Z_{\frac a 2} * \frac {\sigma} {\sqrt n}\). where \(\bar{X}\) represents the mean. \(\sigma\) = standard deviation. n sample size. \(a\) confidence level. \(Z_{\frac a 2}\) is confidence coefficient.

Hence: our confidence interval is 0.8427024. Therefore, \(\mu\) (6.38) should be between 5.5372976 and 7.2227024

Question #5

A random variable X has the memoryless property if, for all s,t > 0,

\[ Pr(X > t+s | X > t) = Pr(X>s) \]

Show that the exponential distribution has the memoryless property.

Question #5

Recall We say that X has an exponential distribution with parameter \(\lambda\) if its probability density function is:

\[ f_X(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{x} \in Rx \\ 0, & \text{x} \notin Rx \end{cases} \] The parameter \(\lambda\) is called rate parameter.

\[Pr(X<t+s|X>t) = \frac {Pr(X< t+s and X>t)} {Pr(X>s)} \] \[Pr(X<t+s|X>t) = \frac {Pr( t < X < t+s)} {Pr(X>s)} \] \[Pr(X<t+s|X>t) = \frac {F(s+t) - F(s)} { 1 - F(x)} \] \[Pr(X<t+s|X>t) = \frac {1- e^{-\lambda (s+t)} - (1 - e^{-\lambda s})} {e^{-\lambda s}} \] \[Pr(X<t+s|X>t) = \frac {e^{-\lambda s} - e^{-\lambda (s+t)}} {e^{-\lambda s}} \] \[Pr(X<t+s|X>t) = \frac {e^{-\lambda s} - e^{-\lambda (s)} e^{-\lambda (t)} } {e^{-\lambda s}} \] \[Pr(X<t+s|X>t) = 1- e^{-\lambda (t)} \]

\[Pr(X<t+s|X>t) = P(X<s) \] using the complementary properties of probabilities we \[ Pr(X>t+s|X>t) = P(X>s) \]

Question #6

Suppose \(X_1,X_2,.,X_n\) are i.i.d \(Exp(\lambda =1)\). Use the Central Limit Theorem to find the approximate value of \(Pr(100 \leq \Sigma^{100}_{i=1} X_i \leq 110\)

Answer #6

Recall We say that X has an exponential distribution with parameter \(\lambda\) if its probability density function is:

\[ f_X(x) = \begin{cases} \lambda e^{-\lambda x}, & \text{x} \in Rx \\ 0, & \text{x} \notin Rx \end{cases} \]

in our example, we have \(\lambda = 1\), hence exponential distribution becomes:

\[ f_X(x) = \begin{cases} e^{-x}, & \text{x} \in Rx \\ 0, & \text{x} \notin Rx \end{cases} \]

We have n=100 samples. hence using Central Limit Theorem, we get:

\[ = (Pr(100 \leq \Sigma^{100}_{i=1} X_i \leq 110) = \] \[ = Pr( \frac {100 - E(X)} {\sqrt(nVar(x))} \leq Z \leq \frac {110 - E(X)} {\sqrt(nVar(x))} ) \] \[ = Pr( \frac {100 - 100} {\sqrt(100)} \leq Z \leq \frac {110 - 100} {\sqrt(100)} ) \] \[ = Pr( 0 \leq Z \leq 1) \] \[ = \phi(1) - \phi(0) \] \[ = 0.3413447 \]

Hence, the approximate value of \(Pr(100 \leq \Sigma^{100}_{i=1} X_i \leq 110\) = 0.3413447

Question DES 5.13

A random variable X that has a pmf given by \(p(x) = \frac{1}{n+1}\) over the range \(R_x = {0,1,2,...,n}\) is said to have a discrete uniform distribution.

  1. Find the mean and variance of this distribution.

hint: \(\sum_{i=1}^n i = \frac{n(n+1)} {2}\) and \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n +1)} {6}\)

\[mean = \frac {1} {n+1} \times \frac {(n+1)} {2} \]

\[mean = \frac{n}{2}\ \]

now we will find the variance:

We will use the below formula for the variance is:

\[Var(X) = E(X^2) - (E(X))^2\]

We already have \((E(X))^2\) which is \((\frac{n}{2})^2\)

Now we will need to compute \(E(X^2)\)

\[E(X^2) = \sum_{i=1}^n i^2\] from the hint, we have: \(\sum_{i=1}^n i^2 = \frac{n(n+1)(2n +1)} {6}\). hence, \[E(X^2) = \frac{1}{n+1} \frac{n(n+1)(2n +1)} {6}\]

Therefore,

\[Var(X) = E(X^2) - (E(X))^2\] \[Var(X) = \frac{1}{n+1} \frac{n(n+1)(2n +1)} {6} - (\frac{n}{2})^2\] \[Var(X) = \frac{n(2n +1)} {6} - (\frac{n}{2})^2\] \[Var(X) = \frac {2n(2n +1) - 3{n}^2} {12} \] \[Var(X) = \frac {4n^2 +2n - 3{n}^2} {12} \] \[Var(X) = \frac {n^2 +2n} {12} \] \[Var(X) = \frac {n(n +2)} {12} \]

  1. if \(R_x = {a,1+a,2+a,...,b}\) find mean and variance let’s expand the sum in our new Rx as follow:

\[ a+1+a+2+a+3+a...+a+n\] \[ a+a+a+a...a_n ..+.. 1+2+3+4...+n\] hence the sum in our new R_x is: \[ an+ \sum_{i=1}^n i\] \[ an+ \frac{n(n+1)} {2}\]

hence our mean is:

\[mean = \frac {1} {n+1} \times an \frac {(n+1)} {2} \] \[mean = \frac {an} {2} \]

Again, We will use the below formula for the variance is:

\[Var(X) = E(X^2) - (E(X))^2\] We already have \((E(X))^2\) which is \((\frac{an}{2})^2\)

Now we will need to compute \(E(X^2)\) using Algebra Identities, we have:

\[(a + b)^2 = a^2 + 2ab + b^2\] hence: \[(a + 1)^2 = a^2 + 2a + 1^2\] \[(a + 2)^2 = a^2 + 4a + 2^2\] \[(a + 3)^2 = a^2 + 6a + 3^2\] \[(a + 4)^2 = a^2 + 8a + 4^2\] \[...\] \[(a + n)^2 = a^2 + 2na + n^2\]

Which can be re-written as :

\[(a + 1)^2 = a^2 + 2a(1) + 1^2\] \[(a + 2)^2 = a^2 + 2a(2) + 2^2\] \[(a + 3)^2 = a^2 + 3a(3) + 3^2\] \[(a + 4)^2 = a^2 + 4a(4) + 4^2\] \[...\] \[(a + n)^2 = a^2 + 2a(n) + n^2\]

Hence the sum is \(X^2\) is: \[ n a^2 + 2a \sum_{i=1}^n i + \sum_{i=1}^n i^2 \] \[ n a^2 + 2a \frac{n(n+1)} {2} + \frac{n(n+1)(2n +1)} {6}\]

Now lets replace \(X^2\) in our Var(X) equation:

\[Var(X) = E(X^2) - (E(X))^2\] \[Var(X) = \frac{1}{n+1} (n a^2 + 2a \frac{n(n+1)} {2} + \frac{n(n+1)(2n +1)} {6}) - (\frac {an} {2})^2\] \[Var(X) = \frac{1}{n+1} (n a^2 + a {n(n+1)} + \frac{n(n+1)(2n +1)} {6}) - (\frac {an} {2})^2\] \[Var(X) = (\frac{n a^2}{n+1} + an + \frac{n(2n +1)} {6}) - (\frac {an} {2})^2\] \[Var(X) = \frac{n a^2}{n+1} + an - \frac {{an}^2} 4 + \frac{n(2n +1)} {6} \]

Question #5.14

The lifetime, in years, of a satellite placed in orbit is given by the following pdf:

\[ f(x) = \begin{cases} 0.4e^{-0.4x}, & \text{x} \geq 0 \\ 0, & \text{otherwise} \end{cases} \]

  1. What is the probability that this satellite is still “alive” after 5 years?
  2. What is the probability that the satellite dies between 3 and 6 years from the time it is placed in orbit?

Answer #5.14

  1. X is the lifetime of the satellite. X is exponentially distributed. The cumulative distribution of X is:

\[Pr(X \ge 5) = \int_{5}^\infty 0.4e^{-0.4x} dx\] \[Pr(X \ge 5) = \int_{5}^\infty \frac {2e^{- \frac {2x} 5}} 5 dx\]

now let’s subtitute u = \(-\frac {2x} {5}\)
wich imply \(\frac {du} {dx} = -\frac 2 5\)

now lets plug in our u and du/dx, we get:

\[Pr(X \ge 5) = - \int_{5}^\infty \frac {2e^{u}} {5} \frac {-5} 2 du\] \[Pr(X \ge 5) = - \int_{5}^\infty e^{u} du\] \[Pr(X \ge 5) = - e^{u} \] we replace u with \(-\frac {2x} {5}\) , we get:

\[Pr(X \ge 5) = \left[ - e^{-\frac {2x} {5}} \right]_{5}^\infty\]

as years goe to infinity the \(e^{-\frac {2x} {5}}\) goes to 0. hence:

\[Pr(X \ge 5) = 0 - (- e^{-\frac {2x5} {5}} )\] \[Pr(X \ge 5) = e^{-2} \]

Hence, the probability that this satellite is still “alive” after 5 years is: 0.1353353

  1. What is the probability that the satellite dies between 3 and 6 years from the time it is placed in orbit?

\[Pr( 3 \le X \le 6) = \int_{3}^{6} 0.4e^{-0.4x} dx\]

using the results from (a), we get:

\[Pr( 3 \le X \le 6) = \left[ - e^{-\frac {2x} {5}} \right]_{3}^{6}\] \[Pr( 3 \le X \le 6) = - e^{-\frac {12} {5}} - (- e^{-\frac {6} {5}}) \] \[Pr( 3 \le X \le 6) = - e^{-\frac {12} {5}} + e^{-\frac {6} {5}} \]

Hence, the probability that the satellite dies between 3 and 6 years from the time it is placed in orbit is: 0.2104763

Question 5.39

Three shafts are made and assembled into a linkage. The length of each shaft, in centimeters, is distributed as follows:

Shaft 1: N(60,0.09) Shaft 2: N(40,0.05) Shaft 3: N(50,0.11)

  1. What is the distribution of the length of the linkage?
  2. What is the probability that the linkage will be longer than 150.2 cm?
  3. The tolerance limits for the assembly are (149.83, 150.21). What proportion of assemblies are within the tolerance limits?

Answer 5.39

  1. What is the distribution of the length of the linkage?

the distribution of the length of the linkage is the sum all means and their respective standard deviation. hence, the distribution of the length of the linkage 60+40+50= 150 with standard deviation of sqrt(.09^2 + .05^2 + .11^2)= 0.1506652. Therefore distribution of the length of the linkage is: N(150, .15)

  1. What is the probability that the linkage will be longer than 150.2 cm?

\[ Z = \frac{X - E[X]}{\sigma(X)} \] \[ z = \frac {150.2 - 150} {.15} \] \[ z = \frac {.2} {.15} \] hence z = 1.3333333

therefore, the probability that the linkage will be longer than 150.2 cm is 0.0912112

  1. The tolerance limits for the assembly are (149.83, 150.21). What proportion of assemblies are within the tolerance limits?

\[ z1 = \frac {149.83 - 150} {.15} \] \[ z1 = \frac {150.21 - 150} {.15} \]

z1= -1.1333333 z2= 1.4

Therefore, the proportion of assemblies are within the tolerance limits of: 0.7907062 or 79%