Stat 5101 (Geyer) Final Exam

Problem 1

The random vector (X, Y) is biivariate normal, hence so is the random vector (U, V) because it is a linear transformation of (X, Y). To show U and V are independent, we only need show they are uncorrelated by Theorem 4 of Chapter 12 in Lindgren. So we check
$$\begin{align*}\mathop{\rm cov}\nolimits(U, V) & = \mathop{\rm cov}\nolimits(X ... ...hop{\rm var}\nolimits(X) - \mathop{\rm var}\nolimits(Y) \\ & = 0 \end{align*}$$

Alternative Solution

The hard way to do this is to use the change of variable to find the joint density of U and V and see that it factors. The joint density of X and Y is

$$f(x, y) = \frac{1}{2 \pi \sigma^2} e^{- (x - \mu)^2 / 2 \sigma^2 - (y - \mu)^2 / 2 \sigma^2}$$

Solving for X and Y in terms of U and V gives
$$\begin{align*}X & = \frac{U + V}{2} \\ Y & = \frac{U - V}{2} \end{align*}$$
This transformation has derivative

$$\renewcommand{\arraystretch}{1.5} \begin{pmatrix} \frac{1}{2} & \frac{1}{2} \\ \frac{1}{2} & - \frac{1}{2} \end{pmatrix}$$

and Jacobian

$$\left( \frac{1}{2} \right) \left( - \frac{1}{2} \right) - \left( \frac{1}{2} \right)^2 = - \frac{1}{2}$$

Thus the joint density of U and V is
$$\begin{align*}g(u, v) & = f\left(\frac{u + v}{2}, \frac{u - v}{2}\right) \cdot... ...{ - \frac{1}{4 \sigma^2} (u^2 + v^2 - 4 \mu u + 4 \mu^2) \right\} \end{align*}$$
and we see that this indeed factors into a function of u times a function of v.

Problem 2

(a)

The BUP is $E(Y \mid X)$, which because functions of the conditioning variable behave like constants in conditional expectation is

$$E(Y \mid X) = E(X^2 + Z \mid X) = X^2 + E(Z \mid X)$$

and the last term is zero because X and Z are independent and E(Z) = 0. Thus the BUP is

g(X) = X²

(b)

The BLUP is

$$h(X) = \mu_Y + \beta (X - \mu_X)$$

where

$$\beta = \rho \frac{\sigma_Y}{\sigma_X} = \frac{\mathop{\rm cov}\nolimits(X, Y)}{\mathop{\rm var}\nolimits(X)}$$

Thus we need to know E(X), E(Y), $\mathop{\rm var}\nolimits(X)$, and $\mathop{\rm cov}\nolimits(X, Y)$. From the formulas for the exponential distribution
$$\begin{align*}E(X) & = \frac{1}{\lambda} \\ \mathop{\rm var}\nolimits(X) & = \frac{1}{\lambda^2} \end{align*}$$
Also
$$\begin{align*}E(Y) & = E(X^2 + Z) \\ & = E(X^2) + E(Z) \\ & = \frac{2}{\lambda^2} \end{align*}$$
where we used the hint to calculate E(X²) and the fact that E(Z) = 0. Finally
$$\begin{align*}\mathop{\rm cov}\nolimits(X, Y) & = E\{X (X^2 + Z)\} - E(X) E(Y) \... ...{1}{\lambda} \cdot \frac{2}{\lambda^2} \\ & = \frac{4}{\lambda^3} \end{align*}$$
where E(X³) is also calculated using the hint. Putting this all together gives

$$\beta = \frac{4 / \lambda^3}{1 / \lambda^2} = \frac{4}{\lambda}$$

and the BLUP is
$$\begin{align*}h(X) & = \frac{2}{\lambda^2} + \frac{4}{\lambda} \left( X - \fra... ...ambda} \right) \\ & = \frac{4 X}{\lambda} - \frac{2}{\lambda^2} \end{align*}$$

Problem 3

The relevant distribution theory is

$$X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n \sim \mathcal{N}\left(\mu, \frac{\sigma^2}{n}\right)$$

from which we get the answer
$$\begin{align*}P(X{\mkern -13.5 mu}\overline{\phantom{\text{X}}}_n < a) & = \Ph... ...t(\frac{105 - 100}{15 / \sqrt{16}}\right) \\ & = \Phi(1.333333) \end{align*}$$
Looking up in the normal table (Table I in Lindgren) we see that this is between .9082 and .9099, closer to the former, say .9088. (Any answer between .9082 and 0.9090 is accepted for full credit.)

Problem 4

The matrix $\mathbf{M} = \mathop{\rm var}\nolimits(\boldsymbol{W})$ has elements $m_{i j} = \mathop{\rm cov}\nolimits(W_i, W_j)$, where the W_i are the elements of W, that is,
$$\begin{align*}W_1 & = X \\ W_2 & = X + Y \\ W_3 & = X + Y + Z \end{align*}$$
and, as always, $m_{i i} = \mathop{\rm cov}\nolimits(W_i, W_i) = \mathop{\rm var}\nolimits(W_i)$. So
$$\begin{align*}m_{1 1} & = \mathop{\rm var}\nolimits(X) = \sigma^2 \\ m_{2 2} &... ...}\nolimits(Y) + \mathop{\rm cov}\nolimits(Y, Z) \\ & = 2 \sigma^2 \end{align*}$$
(all of the covariance terms are zero by the assumed independence of X, Y, and Z) and the rest of the components are defined by symmetry ( m_{i j} = m_{j i}). Hence

$$\boldsymbol{M} = \begin{pmatrix} \sigma^2 & \sigma^2 & \si... ...matrix} 1 & 1 & 1 \\ 1 & 2 & 2 \\ 1 & 2 & 3 \end{pmatrix}$$

Problem 5

There are several ways to do this. The simplest is to recognize that the unnormalized density is ``e to a quadratic'' and hence the conditional distribution of Y given X is normal. The only issue is then to figure out which normal distribution.

Having seen that the conditional distribution is normal, it must a normal distribution has unnormalized density of the form

$$\exp\left(- \frac{(y - \mu)^2}{2 \sigma^2} \right)$$

In order for this to be a conditional density, the mean $\mu$ and variance $\sigma^2$ are possibly functions of x.

In order for this to agree, up to constants, with the form given in the problem statement, the exponents of the two expressions must agree except for constants, which when we are conditioning on x includes functions of x. That is, the coefficients of y² and y in the two expressions
$$\begin{gather*}- x^2 - y^2 - x y + y \\ - \frac{(y - \mu)^2}{2 \sigma^2} = ... ...{2 \sigma^2} + \frac{\mu y}{\sigma^2} - \frac{\mu^2}{2 \sigma^2} \end{gather*}$$
must agree. Hence
$$\begin{gather*}- y^2 = - \frac{y^2}{2 \sigma^2} \\ - x y + y = \frac{\mu y}{\sigma^2} \end{gather*}$$
from which we infer
$$\begin{align*}\sigma^2 & = \frac{1}{2} \\ \mu & = \sigma^2 (1 - x) = \frac{1 - x}{2} \end{align*}$$
So the answer is $\mathcal{N}(\frac{1 - x}{2}, \frac{1}{2})$.

Problem 6

The inverse transformation is
$$\begin{align*}X & = U \\ Y & = U V \end{align*}$$
This transformation has derivative

$$\renewcommand{\arraystretch}{1.5} \begin{pmatrix} \frac{\pa... ...{pmatrix} = \begin{pmatrix} 1 & 0 \\ v & u \end{pmatrix}$$

and Jacobian $1 \cdot u - v \cdot 0 = u$.

The joint density of X and Y is

$$f(x, y) = \frac{1}{2 \pi} \exp\left(- \frac{x^2}{2} - \frac{y^2}{2} \right)$$

Thus the joint density of U and V is
$$\begin{align*}g(u, v) & = \frac{1}{2 \pi} \exp\left(- \frac{u^2}{2} - \frac{(u... ...\lvert u \rvert}{2 \pi} \exp\left(- \frac{u^2 (1 + v^2)}{2} \right) \end{align*}$$

Problem 7

The distribution of X is $\text{Poi}(\mu)$ with $\mu = \lambda A$, where $\lambda = 5.7$ per acre and A = 20 acres, so $\mu = 114$. Since $E(X) = \mathop{\rm var}\nolimits(X) = \mu$ by the formulas for the Poisson distribution

$$X \approx \mathcal{N}(\mu, \mu)$$

when $\mu$ is large (Section 4.7 of the handouts). So we approximate the probability $P(X > 100) = P(X \ge 101)$ by the probability P(Y > 100.5), where Y has this normal distribution. That probability is
$$\begin{align*}P(Y > 100.5) & = 1 - \Phi\left(\frac{100.5 - \mu}{\sqrt{\mu}}\ri... ... 1 - \Phi\left(-1.26439\right) \\ & = \Phi\left(1.26439\right) \end{align*}$$
Looking this up in the normal table (Table I in Lindgren) we see that the answer is between .8962 and .8980, a little closer to the former, say .8970 (any answer between .8962 and .8971 is accepted for full credit).

Not part of the problem, but the exact answer (calculated using a computer) is .8988, so the normal approximation is off by about .0018.

Problem 8

(a)

First, since $X \le 0$ is impossible, we have

$$F(x) = 0, \qquad x \le 0.$$

The other part of the c. d. f. is found by integrating the p. d. f..

$$F(x) = \int_0^x f(s) \, d s, \qquad x > 0$$

and this is
$$\begin{align*}F(x) & = \int_0^x \frac{2}{(1 + s)^3} \, d s \\ & = \left. \frac{1}{(1 + s)^2} \right\vert _0^x \\ & = 1 - \frac{1}{(1 + x)^2} \end{align*}$$

(b)

For a continuous random variable like this, the median is the unique solution to F(x) = 1 / 2, which is

$$1 - \frac{1}{(1 + x)^2} = \frac{1}{2}$$

or

$$\frac{1}{2} = \frac{1}{(1 + x)^2}$$

or

(1 + x)² = 2

or

$$1 + x = \sqrt{2}$$

or

$$x = \sqrt{2} - 1$$