Question

If X is a random variable with moment generating function
phi(t) = exp{ λ(e^t − 1)}, find the mean and variance of X.

Answer 1

@Chlorophyll

Answer 2

Expected value is just the \(1\)st moment.

Answer 3

what will the mean be? @wio :(

Answer 4

The \(n\)th moment is given by the \(n\)th derivative of the moment generating function.

Answer 5

\[
\text{E}[X] = \phi'(0)
\]\[
\text{Var}[X]=\text{E}[(X-\mu)^2] = \text{E}[X^2]-(\text{E}[X])^2 = \phi''(0)-(\phi'(0))^2
\]

Answer 6

@wio it seems the answer is 0!

Answer 7

Does that make sense in this case?

Answer 8

wait, after differentiating, we have put in the value 0 in t.

Answer 9

I'm getting \(\phi'(0) = \lambda \)

Answer 10

isn't differentiating that hard? if you do not mind, could you show me in details? Sorry! It would be of great help @wio

Answer 11

\[
\frac d{dt}\color{blue} \exp[ \color{red}{\lambda(e^t − 1)}] = \color{blue} \exp[\lambda(e^t − 1)]\cdot \color{red}{\lambda e^t}
\]

Answer 12

\[
\frac d{dt} \color{blue} {\exp[\lambda(e^t − 1)]}\cdot \color{red}{\lambda e^t}
= \color{blue} {\exp[\lambda(e^t − 1)] \cdot \lambda e^t} \cdot \color{red}{\lambda e^t} + \color{blue} {\exp[\lambda(e^t − 1)]}\cdot\color{red}{\lambda e^t}
\]

Answer 13

For the first one, I use chain rule. Red for inner function blue for outer.
For the second one, I use product rule.

Answer 14

So I'm getting \[
E[X] = \lambda\quad E[X^2]=\lambda^2+\lambda
\]

Answer 15

Heh, I knew something was up when the variance and mean were equal...
This is the MGF of the Poisson distribution.
http://en.wikipedia.org/wiki/Poisson_distribution

Answer 16

oh my. this is huge! Thanks a lot :) Var could be easily found by the formular. if we get E[X], den finding VAR wouldn't be tough. Hope I am right?

Answer 17

I already told you how to get variance...

Answer 18

right!