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\def\Pr{\bb P}
\def\E{\bb E}
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\def\q{\afterassignment\qq\qnum=}
\def\qq{\qqq{\number\qnum}}
\def\qqq#1{\bigskip\goodbreak\noindent{\bf#1)}\smallskip}
\def\fr#1#2{{#1\over #2}}
\def\var{\mathop{\rm var}\nolimits}
\def\infint{\int_{-\infty}^\infty}

\q1

The mean of this distribution $\E(X)$ is
$$\E(X) = \infint xf(x) dx.$$
This integral evaluates to 0 by symmetry because $f(x) =
{1\over2}ce^{-c|x|}$ is an even function, so $xf(x)$ is odd.

The variance of this distribution is $\var(X) = \E(X^2) - \E(X)^2 =
\E(X^2).$ By theorem 5.58,
$$\E(X^2) = \infint x^2f(x) dx = \infint x^2{1\over2}ce^{-c|x|} dx
= \int_0^\infty x^2ce^{-cx} dx
= ce^0{2\over c^3} = 2c^{-2}.$$
by repeated integration by parts

\q2

$$\Pr(X\geq w) = \sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda}.$$
$$\Pr(Y\leq \lambda) = \int_0^\lambda {1\over\Gamma(w)} x^{w-1}e^{-x}dx
= \int_0^\lambda {x^{w-1}e^{-x}\over (w-1)!} dx,$$
because $\Gamma(w) = (w-1)!$

We are going to prove this equality by induction on $w$.
It is true for $w=1$ because the Poisson distribution will sum to
$1-e^{-\lambda}$ (because $\Pr(X\geq0) = 1$ and $\Pr(X<1) = \Pr(X=0) =
e^{-\lambda}\lambda^0/0!.$)
The Gamma distribution is ${1\over\Gamma(1)}\int_0^\lambda x^0e^{-x}dx =
-e^{-x}\big|^\lambda_0 = -e^{-\lambda} + 1.$

Assuming the following equality holds for $w,$ it will be shown to
hold for $w+1$:
$$\sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda} = \int_0^\lambda
{x^{w-1}e^{-x}\over (w-1)!} dx.$$
$$\sum_{k=w}^\infty {1\over k!}\lambda^ke^{-\lambda}
= {1\over w!}\lambda^we^{-\lambda} + \sum_{k=w+1}^\infty
{\lambda^ke^{-\lambda}\over k!}.$$
$${1\over (w-1)!}\int_0^\lambda x^{w-1}e^{-x}dx
= {1\over (w-1)!}\left({x^we^{-x}\over w}\big|^\lambda_0 +
\int_0^\lambda {x^we^{-x}\over w} dx\right) =
{\lambda^we^{-\lambda}\over w!} + \int_0^\lambda {x^we^{-x}\over
\Gamma(w+1)} dx$$
by integration by parts.
$$\Longrightarrow \sum_{k=n+1}^\infty {\lambda^ke^{-\lambda}\over k!} =
\int_0^\lambda {x^we^{-x}\over \Gamma(w+1)} dx.$$

QED

\q3

The density function $f(x)$ is proportional, so there is some constant
$c$ such that $$1 = \infint f(x)dx = c\infint g(x)dx = 2c\int_1^\infty
x^{-n}dx = 2c\left(-{x^{1-n}\over n-1}\right)\big|^\infty_1 =
2c\left({1\over n-1}\right) \to c = {n-1\over2}.$$
$$f(x) = {n-1\over2}g(x),$$
and this will look like a vertically stretched $1/x^2$ graph.

The mean and variance of $X$ exist when $\E(X)$ and $\E(X^2)$ exist,
respectively. The first exists when $n>2$ because for $n=2,$
$$\E(X) = \int_1^\infty xx^{-n}dx = \ln(x)\big|^\infty_1,$$ and,
similarly for $\E(X^2)$ under $n\leq3.$ $n>3$ is the condition that it
exists because $$\int_1^\infty x^2x^{-n}dx = \ln(x)\big|^\infty_1,$$ for
$n=3$ (and $\int\ln(x)dx$ for $n=2$).

\q4

The density function of $Y=|X|$ is:
$$\{x\geq0: {2\over\sqrt{2\pi}}\exp(-{1\over2}x^2).\hbox{ 0
otherwise.}\}$$

$$\E(Y) = \int_0^\infty {2x\over \sqrt{2\pi}}\exp(-{1\over2}x^2)dx =
{1\over\sqrt{2\pi}}\int_0^\infty\exp(-\fr u2)du,$$
by u-substitution, becoming
$$-{2\over\sqrt{2\pi}}e^{-u\over2}\big|_0^\infty =
{\sqrt{2}\over\sqrt{\pi}}.$$

Similarly, $\var(x) = \E(Y^2) - \E(Y)^2.$
$$\E(Y^2) = \int_0^\infty {2x^2\over \sqrt{2\pi}}\exp(-\fr12 x^2)dx
= -{2\over\sqrt{2\pi}}x\exp(-\fr12 x^2)\big|^\infty_0
- {2\over\sqrt{2\pi}}\int_0^\infty -e^{-\fr12 x^2}dx,$$
by integration by parts, turning into
$$0 + 1,$$
because the first evaluates to zero at both extrema and the second is
the distribution function of the normal distribution so integrates to 1.

\bye
