# Aside

He will post last assignment today

# Continous distribution

1. $\mu (a,b)$

$E(X) = \int\limits_a^b x \frac 1 {b-a} dx = \frac {a+b} 2$

In similar fashion, we can find $E(X^2) = \int\limits_a^b x^2 \frac 1 {b-a} dx$. And hence $Var(X) = E(X^2) - (E(X))^2 = \frac {(a-b)^2} {12}$

# Gamma distribution

$E(X) = \int\limits_0^\infty x \frac 1 {\Gamma(x)\beta^\alpha} x^{\alpha-1} e ^{-\frac x \beta}dx$

The idea is to reduce the integral to a gamma function.

E(X) = frac 1 {Gamma(x)beta\^alpha} intlimits_0\^infty x\^{(alpha+1)-1}e \^{-frac x beta}dx

let $y=\frac x \beta$ , let $dx=\beta dy$ and let $x=\beta y$. As $x$ goes from $0$ to $\infty$, so does $y$. Then we get:

begin{aligned} E(X) &= frac 1 {Gamma(x)beta\^alpha} intlimits_0\^infty y\^{(alpha+1)-1} beta\^alpha beta e\^{-y} dy \ &= frac 1 {Gamma(x)} beta Gamma(alpha+1) \ &= frac 1 {Gamma(x)} beta Gamma(alpha) \ &= alpha beta end{aligned}

In similar fashion, we find $E(X^2)$ and hence $Var(X) = \alpha\beta^2$. A special case we have immediately.

For the exponential distribution ( $\alpha = 1$ )

$E(X) = \beta$ $Var(X) = \beta^2$

If $X\sim\chi^2$, then by a setting we get $E(X) = k$ and $Var(X) = 2k$

# The normal distribution

By definiton

E(X) = intlimits_{-infty}\^infty x frac{1}{sqrt{2pisigma\^2}}operatorname{exp}left{-frac{left(x-muright)\^2}{2sigma\^2}right} = mu

This is calculated using integration by part, which he didn’t show and will not ask on the exam. :D

Similarily for $E(X^2)$ and hence $Var(X) = \sigma^2$

# Geometric distribution

E(X) = sumlimits_{x=1}\^infty x (1-p)\^{x-1}p

(Which is kinda difficult to evaluate directly), However, note that $-\frac d {dp} (1-p)^x = x(1-p)^{x-1}$

Therefore

begin{aligned} E(X) &= p sumlimits_{x=1}\^infty frac d {dp} (1-p)\^x\ &= -p frac d {dp} sumlimits_{x=1}\^infty (1-p)\^x\ &=-p frac d {dp} frac {(1-p)\^1} {1-(1-p)}\ &=-p frac d {dp} frac {1-p} {p} \ &=frac 1 p end{aligned}

To find $E(X^2)$, we first need to find $E(X(X-1))$ so that the diffetiation trick works. Thus we find $Var(X) = \frac {(1-p)^2} p$

# Moment Generating Fucntion

Let $X$ be a r.v, then we define the moment generating function (mgf) of $X$ to be $E(e^{tx})$, donted by $M_X(t)$

# Note

1. $M_X(t)$ is only well defined if $E(e^{tx})$ is finite in an interval around $t=0$

So there are indeed distributions that do not have a mgf

2. We have $M_X(t) = \int\limits_{-\infty}^{\infty} e^{tx} f_X(x)dx$ for continous case

For discrete case $M_X(t) = \sum\limits_{all\, x} e^{tx} P(X=x)$

3. It can be shown ( by The Uniqueness Theorem) that there is a one-to-one correspondance between an mgf and its probability distribution. i.e once you know $M_X(t)$ there can be only one $f_X$ or $P(X)$ that correspond to $M_X$

Because of this correspondance it is sometimes useful to work with the mgf rather than with the distribution and vice versa.

4. It is easy to see that if a and b are constants then

M_{aX+b}(t) = e\^{tb} M_X(at)

lhs

E[-e\^{t(aX+b)}] &= E[e\^{atX}e\^{bt}]\ &=e\^{bt} Eleft[ e\^{atX} right] \ &=e\^{bt}M_X(at)

5. Reason for the name

THM: $M^{(k)}(0) = E\left[X^k\right]$

We need only carry out one integration (summation) to find $M_X(t)$ and then we can obtain the moment by differenciation.

Proof (continous case)

M_X(t) = intlimits_{-infty}\^infty e\^{tx} f_X(x)dx

can’t typeset this math, will copy from somewhere else later.

# Application

1. Find the mgf of a Binomial r.v. and use it to find $E(X)$ and $Var(X)$

# Solution

M_X(t) &= sumlimits_{x=0}\^n e\^{tx} {n choose x} p\^x (1-p)\^{n-x}\ &= sumlimits_{x=0}\^n {n choose x} (e\^t p)\^x (1-p)\^{n-x} \ &= (pe\^t + (1-p))\^n

We have

E(X) = M\^prime (0) = n(pe\^t + 1-p)\^{n-1} pe\^t = np

Now we find $E(X^2)$, we find $M^{\prime\prime} (0)$

We get

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