# Math Help - Prove/disprove that a given moment generating funtion has a pron. distribution.

1. ## Prove/disprove that a given moment generating funtion has a pron. distribution.

Hello. Im trying to show if a distriubtion exists for a given moment generating function M(t)= t/(t-1) , ltl<1 . It says find one if it exists or prove there does not exist any. I dont know how to go about this question, but I feel its got a really easy solution. Thanks.

2. Originally Posted by kman320
Hello. Im trying to show if a distriubtion exists for a given moment generating function M(t)= t/(t-1) , ltl<1 . It says find one if it exists or prove there does not exist any. I dont know how to go about this question, but I feel its got a really easy solution. Thanks.
Doesn't exist because M(t) is not a moment generating function for a pdf. Look at the value of M(0) ....

3. Thanks alot for your help. I just figured that a little while ago. I knew it was a simple solution!
BTW, I really like the quotes in your sig.

4. Mind if I ask you another question Mr. F? Ive got a lognoraml dist. , its says show that it has no moment generating function. The pdf is (e^-((logx)^2/2)*(1/x*sqrt2pi) ,for 0<x<inf. I know that it blows up at one end, but how can I show that?

5. Originally Posted by kman320
Mind if I ask you another question Mr. F? Ive got a lognoraml dist. , its says show that it has no moment generating function. The pdf is (e^-((logx)^2/2)*(1/x*sqrt2pi) ,for 0<x<inf. I know that it blows up at one end, but how can I show that?
Show that $E[e^{tX}] = + \infty$ for any $t > 0$.

6. I know that just become the integral from 0 to inf ,of e^tx * the pdf, but I dont know why Im having such a hard time showing it is +inf for any t>0. Do I have to actually compute the integral? I was trying to find the limit as x->0 and also the limit when x->inf, and show it diverges at one end and converges at the other. But even then I am having trouble. Sorry for all the questions.

7. Originally Posted by kman320
I know that just become the integral from 0 to inf ,of e^tx * the pdf, but I dont know why Im having such a hard time showing it is +inf for any t>0. Do I have to actually compute the integral? I was trying to find the limit as x->0 and also the limit when x->inf, and show it diverges at one end and converges at the other. But even then I am having trouble. Sorry for all the questions.
If X is a random variable with normal distribution then the random variable $Y = e^X$ has a lognormal distribution.

I'll assume a standard normal distribution to make things simple. Then:

$E\left[ e^{tY} \right] = E\left[ e^{te^X} \right]$

$= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x} e^{-\frac{x^2}{2}} \, dx$

$= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x -\frac{x^2}{2}} \, dx$

$\geq \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{t \, \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) -\frac{x^2}{2}} \, dx$.

Since the exponent is a cubic for which the $x^3$ term is positive (if t > 0), the exponential must approach $+ \infty$ as $x \rightarrow \infty$.

Therefore $E\left[ e^{tY} \right] = + \infty$.

8. Originally Posted by mr fantastic
If X is a random variable with normal distribution then the random variable $Y = e^X$ has a lognormal distribution.

I'll assume a standard normal distribution to make things simple. Then:

$E\left[ e^{tY} \right] = E\left[ e^{te^X} \right]$

$= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x} e^{-\frac{x^2}{2}} \, dx$

$= \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} e^{t \, e^x -\frac{x^2}{2}} \, dx$

$\geq \frac{1}{\sqrt{2 \pi}} \int_{0}^{\infty} e^{t \, \left( 1 + x + \frac{x^2}{2} + \frac{x^3}{6}\right) -\frac{x^2}{2}} \, dx$.

Since the exponent is a cubic for which the $x^3$ term is positive (if t > 0), the exponential must approach $+ \infty$ as $x \rightarrow \infty$.

Therefore $E\left[ e^{tY} \right] = + \infty$.
It's interesting to note that while the moment generating function for Y doesn't exist, the moments of Y for all orders n = 1, 2, 3, ..... do exist (it's easy to prove this).

9. Oops, nevermind. i figured it out. Thank you very very much!