Stat 5101 (Geyer) Midterm 2

Problem 1

This is a multivariate change of variable problem (Section 12.1 in Lindgren).

The first step is to solve for the old variables in terms of the new variables. Clearly

X = U V.

Then

Y = U - X = U (1 - V).

The next step is to determine the Jacobian of the transformation

$$\begin{vmatrix} \frac{\partial x}{\partial u} & \frac{\part... ... u \\ 1 - v & - u \end{vmatrix} = - u v - u (1 - v) = u$$

Clearly U = X Y can be any positive number. Given U,

$$V = \frac{X}{U}$$

can be any number between 0 and 1 because X can be any number between Xand U. Thus the joint density is

$$f(u, v) = \tfrac{1}{2} u^2 e^{- u}, \qquad 0 < u < \infty,\ 0 < v < 1$$

Problem 2

This is a job for the ``recognizing the unnormalized density'' trick. First
$$\begin{align*}E(\sqrt{X}) & = E(X^{1 / 2}) \\ & = \int_0^\infty x^{1 / 2} ... ...a(\alpha)} \int_0^\infty x^{\alpha - 1 / 2} e^{- \lambda x} \, d x \end{align*}$$
If we recognize the integrand as a $\text{Gam}(\alpha + 1 / 2, \lambda)$density, we see the integral must give

$$E(\sqrt{X}) = \frac{\lambda^\alpha}{\Gamma(\alpha)} \frac{... ...lambda}} \cdot \frac{\Gamma(\alpha + 1 / 2)}{\Gamma(\alpha)}$$

No further simplification is possible. The recursion formula for gamma functions only allows cancellation of gamma functions with arguments differing by an integer.

Problem 3

(a)

Let X be the random variable in question. The count of points occurring in a fixed time interval of length t for a Poisson process with rate parameter $\lambda$ is $\text{Poi}(\lambda t)$. Here $\lambda = 0.2$ per day and t = 7 days. Thus the answer is $X \sim \text{Poi}(1.4)$.

(b)

The waiting and interarrival times T₁, T₂, $\ldots$ for a Poisson process with rate parameter $\lambda$ are i. i. d. $\text{Exp}(\lambda)$. The sum $S = T_1 + \cdots + T_n$ has a $\text{Gam}(n, \lambda)$ distribution (the ``reproductive rule'' for the exponential distribution). Here n = 10. Thus the answer is $S \sim \text{Gam}(10, 0.2)$ if we measure S in days.

(c)

For this part we have to recognize Bernoulli random variables. Let Y_i be a Bernoulli random variable that indicates the absence of accidents in the i-th week, that is,

$$Y_i = \begin{cases}1, & \text{if no accidents occur} \\ 0, & \text{if one or more accidents occur} \end{cases}$$

Let $W = Y_1 + \cdots + Y_n$. Then $W \sim \text{Bin}(n, p)$ by definition of the binomial distribution, where p = P(Y_i = 1). To calculate p we need to recognize that Y_i = 1 if and only if X_i = 0, where X_i is the number of accidents in the week defined in part (a). Thus

$$p = P(Y_i = 1) = P(X_i = 0) = e^{- \mu}$$

where $\mu = 1.4$ is the mean of X_i calculated in part (a).

Also n = 20. Thus the answer is $W \sim \text{Bin}(20, e^{- 1.4})$.

Problem 4

The distribution of X is $\text{Bin}(n, p)$ with n = 100and $p = \frac{1}{2}$. The mean and variance of X are
$$\begin{align*}E(X) & = n p = 50 \\ \mathop{\rm var}\nolimits(X) & = n p (1 - p) = 25 \end{align*}$$

The CLT applied to the ``reproductive rule'' for the Bernoulli distribution says that X is approximately normal for large n

$$X \approx \mathcal{N}(50, 25)$$

The continuity correction approximates $P(X \ge 60)$ by P(Y > 59.5), where $Y \sim \mathcal{N}(50, 25)$. We find this using a table for the standard normal distribution by standardizing Y. The approximation is

$$1 - \Phi\left(\frac{59.5 - 50}{5}\right) = 1 - \Phi\left(1.9\right) = 0.0287$$

Problem 5

(a)

The marginal p. d. f. of X is

$$f(x) = \lambda e^{- \lambda x}, \qquad 0 < x < \infty$$

just by definition of the notation $\text{Exp}(\lambda)$.

(b)

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

$$f(y \mid x) = x e^{- x y}, \qquad 0 < y < \infty$$

(the same as the above with x replaced by y and $\lambda$ replaced by x).

(c)

Joint equals marginal times conditional, hence

$$f(x, y) = \lambda x e^{- \lambda x - x y}, \qquad 0 < x < \infty,\ 0 < y < \infty.$$

(d)

To find the marginal of Y, integrate out X.
$$\begin{align*}f_Y(y) & = \int f(x, y) \, d x \\ & = \int_0^\infty \lambda x e^{- (\lambda + y) x} \, d x \end{align*}$$
This integral is done by recognizing that the integrand is an unnormalized $\text{Gam}(2, \lambda + y)$ density, which must integrate to

$$\frac{\Gamma(2)}{(\lambda + y)^2}$$

Hence

$$f_Y(y) = \frac{\lambda}{(\lambda + y)^2}, \qquad 0 < y < \infty.$$

because $\Gamma(2) = 1! = 1$.