# Thread: Expectation of Discrete Random Variable

1. ## Expectation of Discrete Random Variable

Let X~Bin(n,p) be binomially distributed and define function f:R -->R by:

a --> E(exp(aX)) . (E is the expectation)

Find an explicit formula for f(a).

Any hints on how to start this one? I have no idea what is going on.

Thanks.

2. If you need to find $E(e^{aX})$ then isn't it just $M_X(a)$?

3. Yeah, we haven't been taught moment generating functions but after reading your post I did a quick search and know what to do now. Thanks

4. Oh, in that case you just use the Expected value definition for discrete distribution
$E(e^{aX})={\Sigma}_{x}e^{ax}f_X(x)$ where $f_X(x)$ is the density formula for the Binomial distribution.

5. Thanks, I've done that and got $f(a) = (pe^a+1-p)^n$ .

Now for the second part I have to show that $P(X>z)\leq\frac{f(a)}{e^{az}}$ for a>0, z>0 .

Any hints again?

6. if you haven't studied inequalities yet, I don't want to spoil you all the fun ))) Just go from the first principles, ie apply the definition of $P(X>z)$ given the distribution of X and you'll get there in no time!