\def\problem#1{\goodbreak\bigskip\noindent{\bf #1)}\smallskip\penalty500}
\newfam\rsfs
\newfam\bbold
\def\scr#1{{\fam\rsfs #1}}
\def\bb#1{{\fam\bbold #1}}
\let\oldcal\cal
\def\cal#1{{\oldcal #1}}
\font\rsfsten=rsfs10
\font\rsfssev=rsfs7
\font\rsfsfiv=rsfs5
\textfont\rsfs=\rsfsten
\scriptfont\rsfs=\rsfssev
\scriptscriptfont\rsfs=\rsfsfiv
\font\bbten=msbm10
\font\bbsev=msbm7
\font\bbfiv=msbm5
\textfont\bbold=\bbten
\scriptfont\bbold=\bbsev
\scriptscriptfont\bbold=\bbfiv

\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}}

\problem{1}
\noindent{\bf (a)}
$$\E(Z) = \pmatrix{1\cr0}\qquad \var(Z) = \pmatrix{2&c\cr c&4}$$

\noindent{\bf (b)}
Covariance is a bilinear function, and $\var (kX) = \cov(kX,kX) =
k^2\var X.$ Also, $\cov(c, X) = 0,$ if $c$ is a constant, regardless of
$X.$ (because $\cov(c, X) = \E(cX) - \E(c)\E(X) = c\E(X) - c\E(X) = 0.$)
With linearity, this means $\cov(X+c, Y+b) = \cov(X, Y).$
$$\rho(2X-1, 5-Y) =
{\cov(2X-1, 5-Y)\over\sqrt{\var(2X-1)}\sqrt{\var(5-Y)}} =
{\cov(2X, -Y)\over\sqrt{\var(2X)}\sqrt{\var(-Y)}} = $$$$
{-2\cov(X, Y)\over2\sqrt{\var(X)}\sqrt{\var(Y)}} =
-{.5\over\sqrt{2}\sqrt{4}} =
-{\sqrt2\over8}.$$

\noindent{\bf (c)}

$X$ and $Y$ are not necessarily independent from each other, although
independence would give a covariance of 0 ($p_XY(x,y) = p_X(x)p_Y(y)
\to \E(XY) = \E(X)\E(Y).$)

Let $W \sim \cal N(0, 1),$ and $Z \sim 2\cal B(.5) - 1$ (i.e. it has a
.5 probability of being -1 or 1).
$X := \sqrt2 W + 1$ and $Y := 2ZW.$
These are strictly dependent because $Y = \sqrt2Z(X-1),$ so $Y$ has
conditional distribution $\sqrt2(x-1)(2\cal B(.5) - 1),$ which is
clearly not equal to its normal distribution (which can be fairly easily
verified by symmetry of $W$).
However, they have covariance 0:
$$\E(XY) - \E X\E Y = \E((\sqrt2 W + 1)2ZW) - \E(\sqrt 2 W + 1)\E(2ZW)
$$$$
= \E(2\sqrt2 ZW^2) - \E(\sqrt 2 W)\E(2ZW)
= 0 - 0\E(2ZW) = 0.
$$

% wikipedia says no

\problem{2}
\noindent{\bf (a)}

\def\idd#1#2{\dd{#1}{#2}^{-1}}
$Y_1 = 2X_2$ and $Y_2 = X_1 - X_2$ give us $X_2 = Y_1/2$ and
$X_1 = Y_2 + Y_1/2.$ This lets us compute Jacobian
$$J = \left|\matrix{\idd{x_1}{y_1} &\idd{x_1}{y_2}\cr
                    \idd{x_2}{y_1} &\idd{x_2}{y_2}}\right|
    = \left|\matrix{2&1\cr2&0}\right| = -2.$$
$$g(y_1, y_2) = \left\{\vcenter{\halign{\strut$#$,&\quad$#$\cr
                       |J|2e^{-(y_2+y_1/2)}e^{-y_1/2} & 0 < y_2+y_1/2 < y_1/2\cr
                       0 & {\rm elsewhere}\cr
                       }}\right.$$
$y_2+y_1/2 < y_1/2 \to y_2 < 0 \to -y_2 > 0.$
And $0 < y_2 + y_1/2 \to y_1 > -2y_2 > 0.$
$$ = \left\{\vcenter{\halign{\strut$#$,&\quad$#$\cr
            4e^{-y_2}e^{-y_1}&y_1 > -2y_2 > 0\cr
            0&{\rm elsewhere}\cr
            }}\right.$$

\noindent{\bf (b)}

$$g(y_1) = \int_{-\infty}^\infty g(y_1,y_2)dy_2 =
\bb I(y_1>0)\int_{-y_1/2}^0 4e^{-y_1}e^{-y_2} dy_2 =
\bb I(y_1>0)4e^{-y_1}(1-e^{y_1/2}).$$
$$g(y_2) = \int_{-\infty}^\infty g(y_1,y_2)dy_1 =
\bb I(y_2<0)\int_{-2y_2}^\infty 4e^{-y_1}e^{-y_2} dy_1 =
\bb I(y_2<0)4e^{-y_2}(-e^{2y_2})
.$$

\noindent{\bf (c)}

They are independent iff $g(y_1,y_2) \bb I(y_1 > -2y_2 > 0)e^{-y_1-y_2}
= g(y_1)g(y_2) = \bb I(y_1 > 0)\bb I(y_2 < 0) h(x),$
where $h(x)$ is the strictly non-zero product of exponents that would
result, showing that they are dependent (if $y_1 = -y_2 = 1,$ the right
indicators are satisfied but not the left indicator, and since $h(x)$ is
non-zero, we see a contradiction.)

\problem{3}
\noindent{\bf (a)}
We start determining the mgf from the pdf of $X,$
$p_X(x) =
\fr1{\sqrt{2\pi}}e^{-\fr12x^2}.$
$$\E(e^{tX}) =
\int_{-\infty}^\infty e^{tx}\fr1{\sqrt{2\pi}}e^{-\fr12x^2} dx =
\fr1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{tx-\fr12x^2} dx = $$$$
\fr1{\sqrt{2\pi}} \int_{-\infty}^\infty e^{tx-\fr12x^2-\fr12t^2 + \fr12t^2} dx =
e^{\fr12t^2} \int_{-\infty}^\infty \fr1{\sqrt{2\pi}}e^{-\fr12(x-t)^2} dx
= e^{\fr12t^2},
$$
by the final integrand being a normal pdf and therefore integrating to
1.

\noindent{\bf (b)}
$$M_Y(t) = \E(e^{t(aX+b)}) = \E(e^{bt}e^{atX}) = e^{bt}\E(e^{atX}) =
e^{bt}M_X(at) = e^{bt}e^{\fr12(at)^2}.$$

\noindent{\bf (c)}

Theorem 1.9.2 states that two probability distribution functions are
alike if and only if their moment generating functions are equal in some
vicinity of zero.
The mgf of $Y$ corresponds to $\cal N(b, a^2),$ which has the following
mgf (by computation from its pdf definition):
$$\int_{-\infty}^\infty e^{tx}\fr1{a\sqrt{2\pi}}e^{-(x-b)^2\over
2a^2} dx =
\fr1{a\sqrt{2\pi}}\int_{-\infty}^\infty e^{2a^2tx - x^2 + 2bx - b^2\over
2a^2} dx =$$$$
\fr1{a\sqrt{2\pi}}\int_{-\infty}^\infty e^{-(b+a^2t)^2 +
2(a^2t+b)x - x^2 - b^2 + (b+a^2t)^2\over 2a^2} dx =
e^{(b+a^2t)^2-b^2\over 2a^2}\int_{-\infty}^\infty
\fr1{a\sqrt{2\pi}}e^{-(b+a^2t+x)^2 \over 2a^2} dx =
e^{bt}e^{(at)^2\over2},
$$
proving $Y \sim \cal N(b, a^2),$ because this function is convergent
everywhere, and reusing the fact that the final integrand is a normal
pdf, so it must integrate to 1.

\bye