1. ## Calculating the MGF

This a pretty weird question... because:

$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$

But the limit: $\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$ is undefined?

How am I meant to compute the MGF then?

Thanks

2. ## Re: Calculating the MGF

Hi, usagi_killer.

I'm not familiar with what MGF stand for, but it looks like you may want to re-check your antiderivative step. I think there needs to be a negative in the exponent and a negative out front of everything; doing this, then evaluating the limit in k (assuming t>1) should give the correct result.

Does this help?

Good luck!

3. ## Re: Calculating the MGF

Originally Posted by usagi_killer

This a pretty weird question... because:

$E(e^{tX}) = M(t) = \int_0^{\infty} e^{xt} e^{-x} dx = \int_0^{\infty} e^{-x(1-t)}dx = \lim_{k \to \infty} \left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$

But the limit: $\lim_{k \to \infty} \left[\frac{e^{k(t-1)}}{t-1}\right]$ is undefined?

How am I meant to compute the MGF then?

Thanks
Hi GJA: MGF means moment generating function

you can write $\left[\frac{e^{x(t-1)}}{t-1}\right]_0^k$ as $\frac{1}{-(1-t)} \bigg[e^{-x(1-t)}\bigg]_0^\infty$ and proceed

4. ## Re: Calculating the MGF

Thanks for that,

However, the above still diverges for t>1 doesn't it?

5. ## Re: Calculating the MGF

I think you are meant to assume t < 1, or deduce that this is a necessary assumption.