Statistics 5101, Fall 2000, Geyer

Homework Solutions #10

Problem L5-3

Let Y_i = X_i^k, then Y₁, Y₂, $\ldots$ is a sequence of independent identically distributed random variables (functions of independent random variables are independent by Theorem 13 of Chapter 3 in Lindgren) with expectation

$$\mu_Y = E(Y_i) = E(X^k).$$

Then the LLN says

$$Y{\mkern -14.0 mu}\overline{\phantom{\text{Y}}}_n \stackrel{P}{\longrightarrow}\mu_Y$$

but this is just other notation for

$$\frac{1}{n} \sum_{i = 1}^n X_i^k \stackrel{P}{\longrightarrow}E(X^k).$$

Problem L5-6

Write Y for the weight of the 100 booklets. Then

$$\begin{split} E(Y) & = 100 \\ \mathop{\rm var}\nolimits(Y) & = 100 \times .02^2 = .04 \end{split}$$

so

$$P(Y > 100.5) = 1 - P(Y < 100.5) = 1 - \Phi \left( \frac{1... ...100}{ \sqrt{100} \times .02} \right) = 1 - \Phi(2.5) = .0062$$

Problem L5-9

Let $Y \sim \mathcal{U}(-0.5, 0.5)$ be one error, then from the appendix on brand name distributions

$$\begin{split} E(Y) & = 0 \\ \mathop{\rm var}\nolimits(Y) & = \frac{1}{12} \end{split}$$

If W is the sum of n i. i. d. such errors then

$$\begin{split} E(W) & = 0 \\ \mathop{\rm var}\nolimits(W) & = \frac{n}{12} \end{split}$$

Thus

$$\begin{split} P\left(\lvert W \rvert < \sqrt{n}/2 \right) &... ...\\ & = 1 - 2 \Phi(- \sqrt{3}) \\ & = 0.9167355 \end{split}$$

Problem L6-13

By direct count, the probability of a sum of 5 or less rolling a pair of dice is 5/18. Thus, if Y is the number of such rolls in 72 tries, $Y \sim \text{Bin}(72, 5/18)$, and

$$\begin{split} E(Y) & = 72 \times \frac{5}{18} = 20 \\ \ma... ...hop{\rm sd}\nolimits(Y) & = \sqrt{14.4444} = 3.8006 \end{split}$$

So, using a continuity correction,

$$P(Y \ge 28) = 1 - \Phi \left( \frac{27 + 0.5 - 20}{3.8006} \right) = .0242$$

Problem L6-86

From a picture of the triangular density, the two inside intervals have three times the probability of the outside intervals. Thus the probabilities of the intervals are $\frac{1}{8}$, $\frac{3}{8}$, $\frac{3}{8}$, and $\frac{1}{8}$.

Let X₁, X₂, X₃, and X₄ be the counts in the cells (1, 2, 2, 1), then this is a multinomial random vector and the probability of these counts is

$$\begin{split} \binom{n}{x_1, x_2, x_3, x_4} p_1^{x_1} p_2^{x... ...& = 180 \cdot \frac{3^4}{8^6} \\ & = 0.0556183 \end{split}$$

Problem L12-12

Since it is a linear transformation of a multivariate normal random vector, (X, Y) is also multivariate normal with mean vector zero because

$$\begin{split} E(X) & = E(U) + 2 E(V) = 0 \\ E(Y) & = 3 E(U) - E(V) = 0 \end{split}$$

and variance matrix M with components

$$\begin{split} m_{1 1} & = \mathop{\rm var}\nolimits(X) \\... ...olimits(V) \\ & = 1 \\ m_{2 1} & = m_{1 2} \end{split}$$

Problem N5-7

From the variance formula for the multinomial in the appendix on brand name distributions

$$\begin{split} \mathop{\rm var}\nolimits(X_i - X_j) & = \ma... ... p_i p_j \\ & = n [ p_i + p_j - (p_i - p_j)^2 ] \end{split}$$

Problem N5-10

The problem is to specialize the formula

$$f_{\mathbf{X}}(\mathbf{x}) = \frac{1}{(2 \pi)^{n / 2} \det(... ...{\mu})' \mathbf{M}^{-1} (\mathbf{x}- \boldsymbol{\mu}) \right)$$

for the density of the multivariate normal to the two-dimensional case, when the mean vector is

$$\boldsymbol{\mu}= \begin{pmatrix}\mu_X \\ \mu_Y \end{pmatrix}$$

and the variance matrix is

$$\mathbf{M}= \begin{pmatrix}\sigma^2_X & \rho \sigma_X \sigma_Y \\ \rho \sigma_X \sigma_Y & \sigma^2_Y \end{pmatrix}$$

Using the hints

$$\det(\mathbf{M}) = \sigma^2_X \sigma^2_Y (1 - \rho^2)$$

and

$$\renewcommand{\arraystretch}{1.25} \begin{split} \mathbf{M}^... ...gma_X \sigma_Y} & \frac{1}{\sigma^2_Y} \end{pmatrix}\end{split}$$

The constant part of the density is now done

$$\frac{1}{(2 \pi)^{n / 2} \det(\mathbf{M})^{1 / 2}} = \frac{1}{2 \pi \sigma_X \sigma_Y \sqrt{1 - \rho^2}}$$

because n = 2. So the only thing left is to match up the quadratic form in the exponent.

In general a quadratic form is written out explicitly in terms of components as

$$\begin{split} \mathbf{z}' \mathbf{A}\mathbf{z} & = \sum_{i... ...i = 1}^{n - 1} \sum_{j = i + 1}^n a_{i j} z_i z_j \end{split}$$

In this case the quadratic form in the exponent is
$$\begin{multline*}(\mathbf{x}- \boldsymbol{\mu})' \mathbf{M}^{-1} (\mathbf{x}- \b... ... \frac{\rho (x - \mu_X) (y - \mu_Y)}{\sigma_X \sigma_Y} \right) \end{multline*}$$
which is the quadratic form in the formula to be proved. So we're done.

Problem N5-11

In this case the elements of the partitioned variance matrix are all scalars

$$\begin{split} \mathbf{M}_{1 1} & = \sigma^2_X \\ \mathbf{... ...\\ \mathbf{M}_{2 2}^{-1} & = \frac{1}{\sigma^2_Y} \end{split}$$

Hence

$$\begin{split} E(X \mid Y) & = \boldsymbol{\mu}_1 + \mathbf... ...sigma_X \sigma_Y \\ & = \sigma^2_X (1 - \rho^2) \end{split}$$

Problem N5-12

We are to calculate $P\{ q(\mathbf{X}) < d \}$ for given d, where

$$q(\mathbf{x}) = (\mathbf{x}- \boldsymbol{\mu})' \mathbf{M}^{-1} (\mathbf{x}- \boldsymbol{\mu})$$

and

$$\mathbf{X}\sim \mathcal{N}(\boldsymbol{\mu}, \mathbf{M})$$

Now Problem 12-32 in Lindgren referred to in the hint says almost the same what we want

$$q_2(\mathbf{Y}) = \mathbf{Y}' \mathbf{M}^{-1} \mathbf{Y}\sim \text{chi}^2(p)$$

where

$$\mathbf{Y}\sim \mathcal{N}(0, \mathbf{M})$$

The only differences are (1) we have no means subtracted off in q₂and (2) Y has mean zero. However,

$$q(\mathbf{X}) = q_2(\mathbf{X}- \boldsymbol{\mu})$$

and

$$\mathbf{X}- \boldsymbol{\mu}\sim \mathcal{N}(0, \mathbf{M})$$

so we can apply the 12-32 to this problem obtaining

$$q(\mathbf{X}) \sim \text{chi}^2(p)$$

Thus

$$P\{ q(\mathbf{X}) < d \} = F(d),$$

where F is the the c. d. f. of the $\text{chi}^2(p)$ distribution.

Problem N5-13

(a)

Write

$$\mathbf{Z}= \begin{pmatrix}U - V \\ V - W \end{pmatrix}$$

Then

$$\mathbf{Z} = \begin{pmatrix}1 & -1 & 0 \\ 0 & 1 & -1 \end{pmatrix} \begin{pmatrix}U \\ V \\ W \end{pmatrix}$$

thus is a linear transformation of multivariate normal, hence multivariate normal with

$$E(\mathbf{Z}) = \begin{pmatrix}1 & -1 & 0 \\ 0 & 1 & -1 \en... ... \\ 0 \\ 0 \end{pmatrix} = \begin{pmatrix}0 \\ 0 \end{pmatrix}$$

and

$$\mathop{\rm var}\nolimits(\mathbf{Z}) = \begin{pmatrix}1 & ... ... \end{pmatrix} = \begin{pmatrix}2 & -1 \\ -1 & 2 \end{pmatrix}$$

(b)

From the formula for the variance,

$$\mathop{\rm var}\nolimits(Z_1) = \mathop{\rm var}\nolimits(Z_2) = 2$$

and

$$\mathop{\rm cor}\nolimits(Z_1, Z_2) = - \frac{1}{2}$$

Thus the conditional distribution of Z₁ given Z₂is normal with mean

$$E(Z_1 \mid Z_2) = - \frac{1}{2} \cdot Z_2$$

and variance

$$\mathop{\rm var}\nolimits(Z_1 \mid Z_2) = 2 \left[1 - \left(- \frac{1}{2}\right)^2 \right] = \frac{3}{2}$$