path: root/zhilova/final.tex

\newfam\rsfs
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\def\cal#1{{\oldcal #1}}
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\textfont\rsfs=\rsfsten
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\def\E{\bb E}
\def\P{\bb P}
\newcount\qnum
\def\fr#1#2{{#1\over #2}}
\def\var{\mathop{\rm var}\nolimits}
\def\cov{\mathop{\rm cov}\nolimits}
\def\dd#1#2{\fr{\partial #1}{\partial #2}}
\def\bmatrix#1{\left[\matrix{#1}\right]}

\def\problem#1{\vskip0pt plus 2in\goodbreak
\vskip0pt plus -2in\medskip\noindent{\bf #1)}\smallskip\penalty500}
\def\part#1{\goodbreak\smallskip\noindent{\bf (#1)}\penalty500\par}

\problem{1}
\part{a}
\let\truebeta\beta
\def\beta{\hat\truebeta}
$${\cal L}(\beta;X_1,\ldots,X_n) = \prod_{i=1}^{n} {1\over
\Gamma(\alpha)\beta^\alpha}X_i^{\alpha-1}e^{-X_i/\beta}$$
$$\ell(\beta) = \ln{\cal L}(\beta) = -n\ln(\Gamma(\alpha)) - \alpha n\ln\beta +
    \sum_{i=1}^n [(\alpha-1)\ln(X_i) - X_i/\beta].$$
$${d\ell(\beta)\over d\beta} = -{\alpha n\over\beta} + \sum_{i=1}^n
{X_i\over\beta^2} = 0 \to \sum_{i=1}^n X_i = \beta \alpha n \to \beta
= \overline X/\alpha = \overline X/4,$$
where $\overline X = \fr1n \sum_{i=1}^n X_i.$

\part{b}
\let\beta\truebeta
$$\E(\overline X) = \fr1n\sum_{i=1}^n \E X = \E X,$$
so $\E\hat\beta = \E X / \alpha.$
$$\E X = \int_0^\infty {x^\alpha e^{-x/\beta}\over\Gamma(\alpha)\beta^\alpha} dx
= \alpha\beta\int_0^\infty {x^\alpha e^{-x/\beta}\over
\Gamma(\alpha+1)\beta^{\alpha+1}} dx
= \alpha\beta,$$
because the integrand is the gamma distribution for $\alpha' =
\alpha+1.$ Therefore, the estimator is unbiased because
$\E\hat\beta = \alpha\beta/\alpha = \beta.$

\part{c}
By the central limit theorem, as $n\to\infty,$ $\hat\beta \sim
{\cal N}(\E (\hat\beta), \var(\hat\beta)/n) = {\cal N}(\beta,
\var(\hat\beta)/n),$ so if
$\var \hat\beta<\infty,$ the estimator is consistent.

The mgf of $X_k$ is $(1-\beta t)^\alpha,$ so the mgf of
$Y = \sum_{i=1}^n X_i$ is $(1-\beta t)^{n\alpha}.$
$$Y \sim \Gamma(\alpha n, \beta) \to \overline X \sim \Gamma(\alpha n,
\beta/n) \to \hat\beta \sim \Gamma(\alpha n, \beta/n\alpha),$$
by change of variable in the gamma distribution.%
\footnote{*}{$X\sim \Gamma(\alpha, \beta)$ has pdf
${1\over\Gamma(\alpha)\beta^\alpha}x^{\alpha-1}e^{-x\over\beta},$ so
$kX$ has pdf
${1\over\Gamma(\alpha)\beta^\alpha}(kx)^{\alpha-1}e^{-kx\over\beta},$
giving $kX \sim \Gamma(\alpha, k\beta).$}
The gamma function has variance $\alpha\beta^2,$ so the variance of
$\hat\beta$ is $\beta^2/n\alpha,$ which is less than infinity, so the
estimator is consistent (and it will tend to zero).
% to prove, and to find variance. actually, we don't need to find
% variance for this problem, but we do for (d). find the lecture!!

\part{d}

Since we've already proven $\E(\hat\beta) = \beta$ and obtained a value
for $\var\hat\beta,$ we get the following result by the Central Limit
Theorem,
$\sqrt n(\hat\beta - \beta) \to {\cal N}(0, \var(\hat\beta)) =
{\cal N}(0, \beta^2/n\alpha)$

\problem{2}

\part{a}

With $\theta = 0,$ by Student's Theorem,
$${\overline X\over S/\sqrt{n}} \sim t(n-1).$$
Taking $t_{1-\alpha}(n-1)$ to be the $1-\alpha$ quantile of the
t-distribution with $n-1$ degrees of freedom, we have a test ${\overline
X\over S/\sqrt{n}} \geq t_{1-\alpha}(n-1).$

Substituting in our given values ($\overline X = .8,$ $n = 25,$ $S^2 =
2.56$),
$${.8\over 1.6/\sqrt{25}} \geq 1.711 \to 2.5 \geq 1.711,$$
so our data does pass the test.

\part{b}

\def\cdf{{\bf T}}
Letting $\cdf$ be the cdf of a t-distribution with $n-1$ degrees of
freedom, we use the following result of student's theorem
$${\overline X - \theta\over S/\sqrt n} \sim t(n-1)$$
to get the probability this static is in the critical set as
$$\gamma_C(\theta) = \P({\overline X\over S/\sqrt n} > 1.711) =
\P({\overline X-\theta\over S/\sqrt n} > 1.711-{\theta\over S/\sqrt n})
$$$$
= 1-\cdf(1.711-{\theta\over S/\sqrt n}) =
1-\cdf(1.711-{\theta\over .32})
= \cdf({25\theta\over8} - 1.711).
$$

\problem{3}

\part{a}

The cdf of $X_i$ on the support of its pdf $[0,1],$ is $x,$ (zero on
$x<0$ and one on $x>1$)
With $X_{(1)} = \min(X_1,\ldots,X_6)$ and $X_{(6)} =
\max(X_1,\ldots,X_6),$
$$\P(X_{(1)} \geq x) = \P(X_1 \geq x)\P(X_2 \geq x)\cdots\P(X_6 \geq x)
= (1-x)^6 \to \P(X_{(1)} \leq x) = 1-(1-x)^6.$$
$$\P(X_{(6)} \leq x) = \P(X_1 \leq x)\P(X_2 \leq x)\cdots\P(X_6 \leq x)
= x^6.$$

These give pdfs for $X_{(1)}$ and $X_{(6)},$ respectively, $6(1-x)^5$
and $6x^5.$

\part{b}

$$\E(X_{(1)} + X_{(6)}) = \int_0^1 x(6(1-x)^5 + 6x^5) dx =$$$$
\left[ x(-(1-x)^6 + x^6) \right]_0^1 - \int_0^1 -(1-x)^6 + x^6 dx =
1 - \fr17\left[-(1-x)^7 + x^7\right]_0^1 = 1.
$$

This makes sense by symmetry. The maximum value of the set has the same
pdf as the minimum value reflected about .5, so we expect $\E X_{(6)} =
1 - \E X_{(1)}.$

\problem{4}

\part{a}

$$Y := 2X_2 - X_1 = \pmatrix{-1&2&0}{\bf X} = t^T{\bf X}.$$
$$\E Y = t^T\E{\bf X} = -1(-1) + 2(2) = 5.$$
$$\var Y = \E(t^TXX^Tt) - \E(t^TX)\E(X^Tt) = t^T(\var X)t =
\bmatrix{-1&2&0}\bmatrix{4&-1&0\cr-1&1&0\cr0&0&2}\bmatrix{-1\cr2\cr0} =
12.
$$

\def\cdf{{\bf\Phi}}
Where $\cdf$ is the cdf of the standard normal distribution, we
normalize $2X_2-X_1,$ to reach the standard normal and get
$$\P(2X_2 > X_1 + 3) = \P(2X_2 - X_1 - 5 > -2) = \P((2X_2 - X_1 -
5)/\sqrt{12} > -2/\sqrt{12}) = 1-\cdf(-1/\sqrt{3}) \approx .793.$$

\part{b}

If $X,Y \sim {\cal N}(0,1)$ and $X\perp Y,$ $X^2 + Y^2 \sim \chi^2(2)$
by the definition of $\chi^2.$

For our problem, $Y^TY \sim \chi^2(2)$ if $Y$ is a multivariate normal
with
$$\E Y = \pmatrix{0\cr0} \qquad \var Y = \pmatrix{1&0\cr0&1}.$$

$$\var (X_1,X_3)^T = \pmatrix{4&0\cr0&2} \to Y =
\pmatrix{1/2&0\cr0&1/\sqrt2}(X_1,X_3)^T + \mu,$$
$A$ given by $(\var (X_1,X_3)^T)^{-1/2},$ from $A = \Sigma^{-1/2}$
(3.5.12) in the textbook.

$$\E Y = \bmatrix{-1/2\cr\sqrt2} + \mu = 0 \to \mu =
\bmatrix{1/2\cr -\sqrt2}\qquad A = \bmatrix{1/2&0\cr0&1/\sqrt2}$$

\bye