Stat 5131 Final Exam

Problem 1

Let X be the number of wins. Then $X \sim \text{Bin}(n, p)$, where n = 100 and p = 18 / 38 = 9 / 19. The number of losses is n - X. Let W denote the gambler's net winnings. Then W = X - (n - X) = 2 X - n.

The mean.

By linearity of expectation and the mean of the binomial distribution being n p

$$E(W) = 2 E(X) - n = 2 n p - n = n (2 p - 1) = 100 \left(\frac{18}{19} - 1\right) = - 5.263$$

The standard deviation.

By the formula $\mathop{\rm var}\nolimits(a + b X) = b^2 \mathop{\rm var}\nolimits(X)$ and the variance of the binomial distribution being n p (1 - p)

$$\mathop{\rm var}\nolimits(W) = 2^2 \mathop{\rm var}\nolimits(... ... n p (1 - p) = 400 \frac{9}{19} \cdot \frac{10}{19} = 99.723$$

so the standard deviation is $\sqrt{99.723} = 9.986$.

Problem 2

The probability of rolling a six on each roll is 1 / 6 and the rolls are independent, the number of sixes X has a $\text{Bin}(n, p)$ distribution with n = 200 and p = 1 / 6. The mean is

E(X) = n p = 200 / 6 = 33.33333

The standard deviation is

$$\sqrt{\mathop{\rm var}\nolimits(X)} = \sqrt{n p (1 - p)} = \sqrt{200 \frac{1}{6} \cdot \frac{5}{6}} = 5.270463$$

The continuity correction says that we should standardize $40 - \frac{1}{2}$

$$z = \frac{39.5 - 33.33333}{5.270463} = 1.1700$$

The normal distribution table gives 0.1210 for the probability of the tail past z.

Just for the record (not part of the solution), the exact probability is 0.1223. The normal approximation gives almost three significant figures despite the skewness of the distribution. Without the continuity correction, the approximation is much worse.

Problem 3

By the memoryless property of the exponential distribution, the remaining service time for the first customer when I arrive has the same distribution as all the other service times. The time until I leave is the sum of five i. i. d. $\text{Exp}(\lambda)$ random variables with $\lambda = 1 / \text{5 minutes} = 12$ per hour. By the formulas for the exponential distribution, if $X \sim \text{Exp}(\lambda)$, then

$$E(X) = \frac{1}{\lambda} = \text{5 minutes}$$

and

$$\mathop{\rm var}\nolimits(X) = \frac{1}{\lambda^2} = \text{25 $\text{minutes}^2$}$$

(a)

The mean of the sum of n i. i. d. random variables with mean $\mu$ is $n \mu$. So the mean of the sum of five $\text{Exp}(\lambda)$ random variables is $5 \times 5$ minutes = 25 minutes.

(b)

The variance of the sum of n i. i. d. random variables with variance $\sigma^2$ is $n \sigma^2$. So the variance of the sum of five $\text{Exp}(\lambda)$ random variables is $5 \times \text{25 $\text{minutes}^2$ } = \text{125 $\text{minutes}^2$ }$, and the standard deviation is $\sqrt{125} = 11.18$ minutes.

(c)

The distribution of the sum of the five service times (the sum of five i. i. d. $\text{Exp}(\lambda)$ random variables) is $\text{Gam}(5, \lambda)$by the ``reproductive rule'' for gamma random variables. But this doesn't help since we don't have a table for the gamma distribution.

We have to do this another way using the Poisson distribution. The number of customers served in 30 minutes is $\text{Poi}(\lambda t)$ with t = 30minutes (assuming customers keep arriving so the server is always busy). So $\lambda t = (1 / 5) \times 30 = 6$. I get out in under 30 minutes if the number of customers served in this time is at least 5 (I'm the fifth). The probability of this is $P(X \ge 5)$ when $X \sim \text{Poi}(6)$. Clearly this argument also works if no one arrives after me. Whether I get out in under 30 minutes or not does not depend on what happens to people who arrive after me. Using the Poisson table, this is $1 - P(X \le 4) = 1 - .285 = 0.715$.

Problem 4

(a)

$$E(Y) = \int \frac{1}{x} f(x \vert \alpha, \lambda) \, d x... ...a}{\Gamma(\alpha)} \int x^{\alpha - 2} e^{- \lambda x} \, d x$$ (1)

Clearly this is just another gamma integral. The integrand is $x^{\beta - 1} e^{- \lambda x}$ with $\beta = \alpha - 1$. The integral exists when $\beta > 0$, which is when $\alpha > 1$.

(b)

From the fact that a gamma density integrates to one we have

$$\frac{\Gamma(\beta)}{\lambda^\beta} = \int x^{\beta - 1} e^{- \lambda x} \, d x$$

Hence

$$E(Y) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \cdot \frac{\... ...pha - 1)}{\lambda^{\alpha - 1}} = \frac{\lambda}{\alpha - 1}$$

Problem 5

Write W = X - Y - Z. Then W is normal (any linear function of a normal is normal, the sum of independent normals is normal, hence any linear combination of normals is normal). To figure out which normal, we only need calculate its mean and variance.

E(W) = E(X) - E(Y) - E(Z) = - 2

(expectation of sum is sum of expectations) and

$$\mathop{\rm var}\nolimits(W) = \mathop{\rm var}\nolimits(X) + \mathop{\rm var}\nolimits(Y) + \mathop{\rm var}\nolimits(Z) = 6$$

(variance of sum is sum of variances, if variables are independent). X > Y + Z if and only if W > 0, so the answer is

$$P(W > 0) = 1 - \Phi\left(\frac{0 - (- 2)}{\sqrt{6}}\right) = 1 - \Phi(0.8165) = \Phi(- 0.8165) = 0.2071$$

using linear interpolation in the normal table (which is not required for full credit, 0.2061 or 0.206 is also accepted for full credit).

Problem 6

The conditional p. d. f. of X given Y is

$$f(x \vert y) = \frac{f(x, y)}{f_Y(y)}$$

Start by figuring out the denominator, the marginal of Y,
$$\begin{align*}f_Y(y) & = \int f(x, y) \, d x \\ & = \frac{1}{6} \int_0^\in... ... y} \Gamma(1) \right) \\ & = \frac{(2 + 2 y + y^2) e^{- y}}{6} \end{align*}$$
Then the conditional is

$$f(x \vert y) = \frac{(x + y)^2 e^{- x - y}}{(2 + 2 y + y^2) e^{- y}} = \frac{(x + y)^2 e^{- x}}{2 + 2 y + y^2}$$

Problem 7

The regression function of X on Y is
$$\begin{align*}E(X \vert Y = y) & = \int x f(x \vert y) \, d x \\ & = \int_... ...\frac{1}{2 x^2} \right]_0^\infty \\ & = \frac{6 + 3 y}{3 + 2 y} \end{align*}$$

Problem 8

This uses the general change of variable formula for monotone but not linear change of variable (Theorem 8 of Chapter 3 in Lindgren): If X = h(Y) then

f_Y(y) = f_X[h(y)] |h'(y)|

Here solving Y = 1 / X gives X = 1 / Y so h(y) = 1 / y and h'(y) = - 1 / y². The p. d. f. of X is the $\text{Gam}(\alpha, \lambda)$density

$$f_X(x) = \frac{\lambda^\alpha}{\Gamma(\alpha)} x^{\alpha - 1} e^{- \lambda x}$$

Hence

$$f_Y(y) = f_X(1 / y) \frac{1}{y^2} = \frac{\lambda^\alpha}{\... ...pha)} \left(\frac{1}{y}\right)^{\alpha + 1} e^{- \lambda / y}$$