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\centerline{Mathematics 376 -- Probability and Statistics II}
\centerline{Problem Set 1 Solutions}
\centerline{January 30, 2004}
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\noindent
7.3/We have $n = 9$ samples from an $N(\mu,4)$ population,
so by Theorem 7.1, the sample mean $\overline{X}$ is 
$N(\mu,4/\sqrt{9}) = N(\mu,4/3)$.  Then
$$\eqalign{
P(-2 < \overline{X} - \mu < 2) &= P\left(-1.5 < 3 \left({\overline{X} - \mu \over 4}\right) < 1.5\right)\cr
&= 2(.5 - .0668)\cr
&= .8664\cr}$$
(by the standard normal table).  Note:  $3 \left({\overline{X} - \mu \over 4}\right)$
is $N(0,1)$.
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\noindent
7.4/Continuing from the previous problem, with an unknown 
number of samples $n$, we have $\sqrt{n}\left({\overline{X} - \mu\over 4}\right)$
is $N(0,1)$, so from the standard normal table again,
$$.90 = P(-1 < \overline{X} - \mu < 1) =  P\left({-\sqrt{n}\over 4} < \sqrt{n}\left({\overline{X} - \mu\over 4}\right) < {\sqrt{n}\over 4}\right)$$
when $\sqrt{n}/4 \ge 1.645$.  Rearranging and squaring, we get
$n \ge 43.3$, so $n \ge 44$.
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\noindent
7.7/a) By linearity of expected values
and Theorem 7.1, $E(\overline{X} - \overline{Y}) = \mu_1 - \mu_2$.
b)  Because the two samples are independent, $Cov(\overline{X},\overline{Y}) = 0$,
and 
$$V(\overline{X} - \overline{Y}) = (1)^2 V(\overline{X}) + (-1)^2 V(\overline{Y}) = 
{\sigma_1^2\over m} + {\sigma_2^2\over n}$$
c)  Using parts a, b and Theorem 7.1, we get that 
$\overline{X} - \overline{Y}$ is $N(\mu_1-\mu_2, \sqrt{ {\sigma_1^2\over m} + {\sigma_2^2\over n} })$.
Hence, we proceed as in 7.4.  With $\sigma_1^2 = 2, \sigma_2^2 = 2.5$, and 
$m = n$, we want to choose $n$ so that 
$$P\left({-\sqrt{n}\over \sqrt{4.5}} < Z < {\sqrt{n}\over \sqrt{4.5}}\right) = .95$$
This means we want the two ``tails'' of the standard normal to have 
probability .025 each, so 
$${\sqrt{n}\over \sqrt{4.5}} = 1.96 \Rightarrow n = 17.28$$
or $n = 18$ after going to the next whole number.
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\noindent
7.10/  a) By definition the $\chi^2(\nu)$ distribution is the 
same as the gamma distribution with $\alpha = \nu/2$ and $\beta = 2$.
Hence $E(U) = \alpha\cdot\beta = \nu$, and $V(U) = \alpha\cdot\beta^2 = 2\nu$
using our formulas from last semester.
b)  By Theorem 7.3, ${(n-1)S^2\over \sigma^2}$ is $\chi^2(n-1)$.
So ${n-1\over \sigma^2} E(S^2) = n-1$, or $E(S^2) = \sigma^2$.
Similarly,
$${(n-1)^2\over \sigma^4}V(S^2) = 2(n-1) \Rightarrow V(S^2) = {2\sigma^4\over n-1}$$
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\noindent
7.11/For this problem, you want to use Table 6 in the Appendix in the
text book (``Percentage points of the $\chi^2$ distributions'').
The entry marked $\chi^2_\alpha$ in the table is the probability 
in a {\it right tail} of the distribution.
a) Since $n = 20$, $\sigma^2 = 1.4$, from Theorem 7.3, we
know ${19S^2\over 1.4}$ is $\chi^2(19)$.  To get 
$P(S^2 \le b) = .975$, we want the probability of the right tail
to be $.025$, so we take the table entry $\chi^2_{.025}$ for 
$\nu = 19$ degrees of freedom and 
$${19b \over 1.4} = 32.8523 \Rightarrow b = 2.241$$
b) Similarly to get $P(a \le S^2) = .975$, we want 
$\chi^2_{.975}$ for $\nu = 19$ d.f.  We get
$${19a\over 1.4} = 8.90655 \Rightarrow a = .6563$$
c)  Using the sum rule, $P(A\cap B) = P(A) + P(B) - P(A \cup B)$.
Letting $A$ be the event $a \le S^2$ and $B$ be the event $S^2 \le b$, 
we get $P(a \le S^2 \le b) = P(A\cap B) = 2(.975) - 1 = .95$.
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\noindent
7.14/a) As always, $Z$ is $N(0,1)$, so $E(Z) = 0$ and $E(Z^2) = V(Z) + (E(Z))^2
= 1 + 0 = 1$.  b)  We want to use Theorem 5.9 here 
(independence $\Rightarrow$ expected values multiply).  Since
$T = Z \cdot {\sqrt{\nu}\over \sqrt{Y}}$, we have
$$E(T) = E(Z) \cdot E\left({\sqrt{\nu}\over \sqrt{Y}}\right) = 0$$
(unless $\nu = 1$ -- in that case, $\sqrt{Y}$ is also $N(0,1)$
and $E(T)$ is undefined.)  For the variance, we proceed as follows:
$$\eqalign{V(T) &= E(T^2) - (E(T))^2\cr
                &= E(T^2)\cr
                &= \nu E(Z^2\cdot {1\over Y})\cr
                &= \nu E(Z^2) E\left({1\over Y}\right) \quad {\rm\ using\ Th.\ 5.9}\cr}$$
We know $E(Z^2) = 1$ from above.  To get $E(Y^{-1})$ use the 
formula given with $a = -1$ (note: $\nu > 2 \Rightarrow \nu > -2a$), and 
the functional equation $\Gamma(x+1) = x\Gamma(x)$:
$$\eqalign{V(T) &= \nu E\left({1\over Y}\right)    \cr    
                &= \nu \cdot {\Gamma(\nu/2 - 1)\over \Gamma(\nu)}\cdot 2^{-1}\cr
                &= \nu\cdot{1\over \nu/2 - 1}\cdot {1\over 2}\cr
                &= {\nu\over \nu -2 }\cr}$$ 
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\noindent
7.19/a) $W$ is $\chi^2(5)$ using Theorem 7.2.  b) $U$ is $\chi^2(4)$ by 
Theorem 7.3. c) The sum of independent $\chi^2(\nu_1)$ and $\chi^2(\nu_2)$
random variables is $\chi^2(\nu_1+\nu_2)$, so this is $\chi^2(4+1) =\chi^2(5)$.
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\noindent
7/20/a)  This is $t(5)$ by Definition 7.2.  b)  $t(4)$ using 7.19 b.
c) By Theorem 7.1, $\overline{Y}$ is $N(0,1/\sqrt{5})$, so
$\sqrt{5}\cdot\overline{Y}$ is $N(0,1)$, so $5\overline{Y}^2$ is $\chi^2(1)$.
Then $     5\overline{Y}^2 + Y_6^2$ is $\chi^2(2)$.  Finally
the ratio is $F(2,4)$ by Definition 7.3.
                          
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