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sanctifier

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## Homework Statement

The probability density function (p.d.f.) of a random variable X is

[itex] P(X=k)= p^{k} (1-p) \;\;\;\; where \;\;\;\; 0<p<1 \;\;\;\; and \;\;\;\; k=0,1,2,... [/itex]

Question 1: what is the moment generating function (m.g.f.) of X?

Question 2: What are expectation E(x) and variance Var(x) of X?

Question 3: if [itex] X_{1},\; X_{2},\;\;...,\;\; X_{n} [/itex] are sampled from X, what is the distribution of [itex] \sum_{i=1}^{n} X_i [/itex]

## Homework Equations

Nothing special.

## The Attempt at a Solution

Answer 1:

[itex] E[e^{tx} ] = \sum_{x=0, 1, ...}^{ \infty } e^{tx} p^x (1-p)=(1-p) \sum_{x=0, 1, ...}^{ \infty } {e^tp}^x=(1-p)\lim_{n \to \infty} \frac{1-{e^tp}^n}{1-e^tp} = \frac{1-p}{1-e^tp} [/itex]

[itex] when \;\;0<e^tp<1,\;\;i.e.,\;\;t<ln \frac{1}{p} [/itex]

Hence, m.g.f. is [itex] \psi (t) = \frac{1-p}{1-e^tp} [/itex]

Answer 2:

[itex] \begin{cases} \psi' (t) = \frac{d\psi (t)}{dt} = - \frac{1-p}{(1-e^tp)^2} (-e^tp) = \frac{(1-p)e^tp}{(1-e^tp)^2} \\ \psi'' (t) = \frac{d\psi' (t)}{dt} =(1-p)p \frac{e^t(1-e^tp)^2+e^t2(1-e^tp)e^tp}{(1-e^tp)^4}=(1-p)pe^t \frac{1+e^tp}{(1-e^tp)^3} \end{cases} [/itex]

[itex] \begin{cases} \psi' (0) = \frac{(1-p)p}{(1-p)^2}= \frac{p}{1-p} \\ \psi'' (0) = \frac{(1-p)p(1+p)}{(1-p)^3} = \frac{p(1+p)}{(1-p)^2} \end{cases} [/itex]

[itex] E(X)= \psi' (0) \;\; and\;\; Var(X) = E[X^2]-E[X]^2 = \psi'' (0) - ( \psi' (0))^2 = \frac{p}{(1-p)^2} [/itex]

Answer 3:

[itex] {\overline{X}} _n = \frac{\sum_{i=1}^{n} X_i }{n} [/itex] can be approximated by a normal distribution of [itex] \mu =E(x) [/itex] and [itex] \sigma ^2 =\frac{Var(X)}{n} [/itex], because [itex] \sum_{i=1}^{n} X_i = n {\overline{X}} _n [/itex], hence [itex] \sum_{i=1}^{n} X_i [/itex] has a normal distribution of [itex] \mu =nE(X) [/itex] and [itex] \sigma ^2 =nVar(X) [/itex]

Are these answers correct? Thank you in advance!