# Thread: Exponentially Distributed RV - MGF

1. ## Exponentially Distributed RV - MGF

I need help. The problem is this:

X is an exponentially distributed random variable with p.d.f.

f(x) = $\displaystyle \lambda{e}^{-\lambda*x}$
0<x<infinity

Show that the m.g.f. of X is $\displaystyle {(1 - t/\lambda)}$ for $\displaystyle {t<\lambda}$, and use this to find E(X) and Var(X).

I have no idea what to do with this problem or where to start. Please help point me somewhere? The only thing I could think of to potentially do is to undo the p.d.f. to get the original random variable, but...yeah. I'm really, really lost.

2. Originally Posted by notgod
I need help. The problem is this:

X is an exponentially distributed random variable with p.d.f.

f(x) = $\displaystyle \lambda{e}^{-\lambda*x}$
0<x<infinity

Show that the m.g.f. of X is $\displaystyle {(1 - t/\lambda)}$ for $\displaystyle {t<\lambda}$, and use this to find E(X) and Var(X).

I have no idea what to do with this problem or where to start. Please help point me somewhere? The only thing I could think of to potentially do is to undo the p.d.f. to get the original random variable, but...yeah. I'm really, really lost.

just combine the exponential terms.

$\displaystyle \lambda\int_0^{\infty}e^{-x(\lambda-t)}dx$

You can now integrate BUT it's better to notice another exponential density

$\displaystyle ={\lambda\over \lambda-t} (\lambda-t) \int_0^{\infty}e^{-x(\lambda-t)}dx$

$\displaystyle ={\lambda\over \lambda-t}$

since that is another valid density, just let $\displaystyle \lambda^* = \lambda-t$

3. ## problem from my book

Let X and Y be two Statistically independent Exponentially distributive random variables with parameters $\displaystyle \lambda1$ and $\displaystyle \lambda2$ respectively.

a) find the CDF (commulative density function) of Z = X/X+Y , (use property of exponential pdf to evaluate the integrals in this question).

b) find the expected value of Z when $\displaystyle \lambda1$=1, and $\displaystyle \lambda2$= 2

So please can any one solve this, with all the steps of integration , Thanks a lot.

4. I think Laurent solved this a couple of days ago here.
Z is a beta.