\magnification=\magstep1
\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 5 Solutions}
\centerline{March 19, 2004}
\bigskip
\noindent
Extra Problem A.  1)  Recall that a random variable $Y$ has an
$F(n,m)$ distribution if and only if $Y$ has the form
$Y = {W_1/n\over W_2/m}$, where $W_1$ has a $\chi^2(n)$ 
distribution and $W_2$ has a $\chi^2(m)$ 
distribution.  Hence $1/Y = Z$ has an $F(m,n)$ distribution, and 
hence taking reciprocals, then reversing the inequality (replacing
$\alpha$ by $1 - \alpha$) we get:
$$\eqalign{ P(Y \ge f_{\alpha}(n,m)) &= \alpha\cr
\Rightarrow\ P\left({1\over Y} \le {1\over f_{\alpha}(n,m)}\right) &= \alpha\cr
\Rightarrow\ P\left(Z \ge {1\over f_{\alpha}(n,m)}\right) &= 1 - \alpha\cr}$$
By definition, this means 
$${1\over f_{\alpha}(n,m)} = f_{1-\alpha}(m,n)$$
as claimed.
\medskip
\noindent
2)  By part 1, and the F-table in the book's appendix,
$$f_{.99}(7,10) = {1\over f_{1-.99}(10,7)} = {1\over f_{.01}(10,7)} = {1\over 6.62} 
\doteq .15106$$
\bigskip
\noindent
8.73/ Since the sample sizes $n_1 = 16$ and $n_2 = 20$ are relatively
small, we use the $t$-distribution formula to find the confidence
interval.  Using the given information,
the pooled estimator for the variance (assumed equal) 
is 
$$S_p^2 = {15\cdot 36 + 19\cdot 64\over 34} \doteq 51.647$$
so the 95\% confidence interval is
$$\mu_1 - \mu_2 = 
\overline{X}_1 - \overline{X}_2 \pm t_{.025}(34) \sqrt{51.647} \sqrt{{1\over 16} + {1\over 20}}
= -1 \pm 4.7245$$
(using $t_{.025}(34) = 1.96$ from the $t$-table in the book).
\bigskip
\noindent
8.76 a) Similar to 8.73:  The interval is
$$446-534 \pm t_{.025}(28)\sqrt{14\cdot 42^2+14\cdot 45^2\over 28}\cdot
\sqrt{2\over 15} = -88 \pm 32.55$$
b)  Also similar:
$$548 - 517 \pm t_{.025}(28) \sqrt{14\cdot 57^2+14\cdot 52^2\over 28}\cdot
\sqrt{2\over 15} = 31 \pm 40.79$$
c)  The data supports the conclusion that the verbal scores of 
language/literature students are higher than the verbal scores
of engineering students (the interval in part a consists entirely
of negative numbers).  But the data does not support the conclusion
that the math scores are higher for engineering students than
for language/literature students.
\medskip
\noindent
d)  We assume that the scores are independently and randomly selected
from the two groups, and that the scores in each case have 
normal distributions with {\it equal variances}.
\bigskip
\noindent
8.81/Computing $\overline{X}$ and $S^2$ from the given data, we 
get
$$\overline{X} = 85.73\quad S^2 \doteq .5027$$
Then the 90\% confidence interval for $\sigma^2$ is (since $n = 6$),
$$\left[{5 S^2\over \chi^2_{.05}(5)}, {5 S^2\over \chi^2_{.95}(5)}\right]
= \left[{5\cdot .5027 \over 11.0705}, {5\cdot .5047\over 1.145476}\right]
= [.2276,2.194]$$
\bigskip
\noindent
8.83/a) Since ${(n-1)S^2\over \sigma^2}$ has a $\chi^2(n-1)$ distribution, 
we have
$$P\left({(n-1)S^2\over \sigma^2} \ge \chi^2_{1-\alpha}\right) = 1 - \alpha,$$
so multiplying by $\sigma^2$, dividing by $\chi^2_{1-\alpha}$, we get
$$P\left(\sigma^2 \le {(n-1)S^2\over \chi^2_{1-\alpha}}\right) = 1 - \alpha .$$
So the $(1-\alpha)\times 100\%$ upper confidence bound is
$${(n-1)S^2\over \chi^2_{1-\alpha}},$$
(using $n - 1$ degrees of freedom).
b) Similarly, to get a lower confidence bound,
$$P\left({(n-1)S^2\over \sigma^2} \le \chi^2_{\alpha}\right) = 1 - \alpha,$$
so multiplying by $\sigma^2$, dividing by $\chi^2_{1-\alpha}$, we get
$$P\left(\sigma^2 \ge {(n-1)S^2\over \chi^2_{\alpha}}\right) = 1 - \alpha .$$
So the $(1-\alpha)\times 100\%$ upper confidence bound is
$${(n-1)S^2\over \chi^2_{\alpha}},$$
(using $n - 1$ degrees of freedom again).
\bigskip
\noindent
8.84/Since 
$$P\left({(n-1)S^2\over \chi^2_{\alpha/2}}\le \sigma^2 \le {(n-1)S^2\over \chi^2_{1-\alpha/2}}\right) = 1-\alpha,$$
if we take square roots everywhere on the ``inside'', we get:
$$P\left(\sqrt{{(n-1)S^2\over \chi^2_{\alpha/2}}} \le \sigma \le \sqrt{{(n-1)S^2\over \chi^2_{1-\alpha/2}}}\right) = 1-\alpha,$$
So the $(1-\alpha)\times 100\%$ confidence interval for $\sigma$
is
$$\left[\sqrt{{(n-1)S^2\over \chi^2_{\alpha/2}}}, \sqrt{{(n-1)S^2\over \chi^2_{1-\alpha/2}}}\right].$$
\bigskip
\noindent
9.13/For each sample $Y_i$, we have
$$E(Y_i) = \int_0^1 y\cdot \theta y^{\theta-1}\ dy = \theta\int_0^1 y^{\theta}\ dy = {\theta\over \theta+1}$$
and 
$$E(Y_i^2) = \int_0^1 y^2\cdot \theta y^{\theta-1}\ dy = \theta\int_0^1 y^{\theta+1}\ dy = {\theta\over \theta+2}$$
Hence 
$$V(Y_i) = E(Y_i^2) - (E(Y_i))^2 = {\theta\over \theta+2} - \left({\theta\over \theta+1}\right)^2= 
{\theta\over (\theta+1)^2(\theta+2)}.$$
Now, $\overline{Y} = {1\over n}(Y_1 + \cdots + Y_n)$ satisfies
$$E(\overline{Y}) = {1\over n}(E(Y_1) + \cdots + E(Y_n)) = {\theta\over \theta+1}$$
so $\overline{Y}$ is {\it unbiased} for ${\theta\over \theta+1}$.  Moreover,
by independence,
$$V(\overline{Y}) = {1\over n^2}(V(Y_1) + \cdots + V(Y_n)) ={1\over n}\cdot{\theta\over (\theta+1)^2(\theta+2)}$$
As $n \to \infty$, this clearly shows
$$\lim_{n\to\infty} V(\overline{Y}) = 0$$
so $\overline{Y}$ is consistent by Theorem 9.1.  
\bigskip
\noindent
9.19/a)  Since $Y_1$ has an $N(\mu,1)$ distribution, $E(Y_1) = \mu$.  Hence
$Y_1$ is unbiased for $\mu$.  \medskip
\noindent
b) Using the standard normal table,
$P(|Y_1 - \mu| \le 1) = P(-1 \le Y_1 - \mu \le 1) = 1 - 2(.1587) = .6824$.\medskip
\noindent
c)  No, $Y_1$ is not a consistent estimator for $\mu$.  Let $\epsilon = 1$.
As $n \to \infty$, the probability that $P(|Y_1 - \mu| \le 1)$ does not 
change (since the estimator does not depend on the additional samples
after the first).  In particular it does not tend to 1.
\bigskip
\noindent
9.61/To find the method of moments estimator, we first compute the 
expected value a random variable with this pdf:
$$\mu_1' = E(Y) = \int_0^1 y\cdot (\theta+1)y^\theta\ dy = {\theta+1\over \theta+2}$$
(similar to 9.13 above).  Set 
$\mu_1' = {\theta+1\over \theta+2} = \overline{Y} = m_1'$ and solve for $\theta$:
$$\theta+1 = \overline{Y}\theta + 2\overline{Y}$$
This is the method of moments estimator:
$$\widehat{\theta} = {2\overline{Y} - 1\over 1 - \overline{Y}}.$$

To see this is consistent, we can use Theorem 9.2.  As in 9.13, $n \to \infty$ 
$\lim_{n\to \infty} \overline{Y} = {\theta+1\over\theta+2}$ ``in probability''
(i.e. intuitively, the probability that they are equal tends to 1, or
more precisely, for all $\epsilon > 0$, the probability
that the difference is $\le \epsilon$ goes to 1 as $n \to \infty$).  Then
$$\eqalign{\lim_{n\to\infty} {2\overline{Y} - 1\over 1 - \overline{Y}} &= 
{\lim_{n\to\infty} 2\overline{Y} - 1 \over \lim_{n\to\infty} 1 - \overline{Y}}\cr
&={2{\theta+1\over\theta+2} - 1\over 1 - {\theta+1\over\theta+2}}\cr
&=\theta\cr}$$

Finally, the given sufficient statistic $U = -\sum_{i=1}^n \ln(Y_i)$ satisfies
$\exp(-U) = Y_1\cdots Y_n$ (the product of the sample values.  The sample mean 
is not a function of $U$ or $\exp(-U)$.  (If it were, then we would have that
the sum of the sample values was determined by the product, but that is 
clearly false.  For instance, if $n = 2$, 
$$Y_1 = 1,\ Y_2 = 1\quad \Rightarrow\quad Y_1Y_2 = 1,\ Y_1 + Y_2 = 2$$
but
$$Y_1 = 2,\ Y_2 = .5\quad \Rightarrow\quad Y_1Y_2 = 1,\ Y_1 + Y_2 = 2.5)$$
By the consequence of the Rao-Blackwell theorem we mentioned in class,
this says that $\widehat{\theta}$ cannot be the most efficient
(minimum variance) unbiased estimator for this problem.
\bigskip
\noindent
9.67/The given pdf has the form of the {\it beta-distribution} pdf with 
$\alpha = \beta = \theta$.  This is a slightly tricky example for the 
method of moments, though.  Setting $\mu_1' = E(Y) = \overline{Y} = m_1'$,
we get
$${\theta\over \theta + \theta} = \overline{Y} \Rightarrow {1\over 2} = \overline{Y}$$
But there is no $\theta$ to solve for in this equation!  This means that
we can go on and set $\mu_2' = m_2'$ and try to use that to solve
for $\theta$, though.  Using the formula for the variance of a
beta-distributed random variable we get:
$$\mu_2' = E(Y^2) = V(Y) + (E(Y))^2 = {\theta^2\over 2\theta^2(2\theta+1)} +{1\over 4}
= m_2'$$
After simplifying
$$m_2' = {2\theta+2\over 8\theta+4} \Rightarrow \widehat{\theta} = {1 - 2m_2'\over 4m_2' -1}$$
\bigskip
\noindent
9.75/a)  The likelihood function is
$$L(y_1,\ldots,y_n|\theta) = \left({1\over 2\theta + 1}\right)^n$$
(if $0 < y_i < 2\theta + 1$ for all $i$ and zero otherwise).  This function
is {\it decreasing} for all $\theta > -1/2$ (the vertical asymptote for $L$).
Hence to maximize the likelihood, subject to the given sample
values, we want to choose $\theta$ so that the right hand endpoint
of the interval $[0,2\theta]$ occurs as far to the {\it left} as possible,
consistent with the sample values.  This occurs at the 
sample maximum: $Y_{(n)} = 2\theta + 1$.
Hence
$$\widehat{\theta}_{ML} = {Y_{(n)} - 1\over 2}$$
b)  Use the {\it invariance property} of ML estimators (see page 452).
The variance is 
$$t(\theta) = {(2\theta + 1)^2\over 12}$$
which is a 1-1 function on $\theta > -1/2$.  Hence
$$\widehat{t(\theta)}_{ML} = t(\widehat{\theta}_{ML}) = {\left(2\left({Y_{(n)} - 1\over 2}\right) + 1\right)^2\over 12}
={(Y_{(n)})^2\over 12}$$
\bigskip
\noindent
9.80/The logarithm of the likelihood function is
$$\ln(L) = \ln((\theta+1)^n(y_1\cdots y_n)^\theta) = n\ln(\theta+1) + \theta\cdot \sum_{i=1}^n\ln(y_i)$$
To maximize $\ln(L)$, we differentiate with respect to $\theta$ and set equal to zero:
$${d\ln(L)\over d\theta} = {n\over \theta+1} + \sum_{i=1}^n\ln(y_i) = 0$$
Solving for $\theta$, we get
$$\widehat{\theta}_{ML} = -{n + \sum_{i=1}^n\ln(y_i) \over \sum_{i=1}^n\ln(y_i)} = {-n\over \sum_{i=1}^n\ln(y_i)} -1$$
This is a maximum by the second derivative test: ${d^2\ln(L)\over d\theta^2} = {-n\over (\theta+1)^2} < 0$
for all $\theta$.
This is quite different from the method of moments estimator from 9.61!




\bye