Exponentially Distributed RV - MGF

notgod · #1 $Old$ November 1st, 2009, 03:04 PM

I need help. The problem is this:

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

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

Show that the m.g.f. of X is ${(1 - t/\lambda)}$ for ${t<\lambda}$ , 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.

matheagle · #2 $Old$ November 1st, 2009, 03:45 PM

Quote:

Originally Posted by notgod $View Post$

I need help. The problem is this:

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

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

Show that the m.g.f. of X is ${(1 - t/\lambda)}$ for ${t<\lambda}$ , 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.

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

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

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

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

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

shabbas77 · #3 $Old$ November 4th, 2009, 02:46 PM

Let X and Y be two Statistically independent Exponentially distributive random variables with parameters $\lambda1$ and $\lambda2$ 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 $\lambda1$ =1, and $\lambda2$ = 2

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

matheagle · #4 $Old$ November 4th, 2009, 04:00 PM

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