If the moment generating function of X is $\displaystyle M(t)=\frac{1}{1-3t}$, $\displaystyle t<\frac{1}{3}$

A. Find E(X)

B. Find Var(X)

C. P(6.1 < x < 6.7)

Thank you in advance for your help!

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- May 3rd 2009, 01:47 PMlogitechmoment generating function
If the moment generating function of X is $\displaystyle M(t)=\frac{1}{1-3t}$, $\displaystyle t<\frac{1}{3}$

A. Find E(X)

B. Find Var(X)

C. P(6.1 < x < 6.7)

Thank you in advance for your help! - May 3rd 2009, 02:10 PMMoo
Hello,

By definition, $\displaystyle E(X^n)=M^{(n)}(0)$, where $\displaystyle M^{(n)}$ denotes the n-th derivative of the mgf.

Specifically, $\displaystyle E(X)=M'(0)$ and $\displaystyle E(X^2)=M''(0)$

For Var(X), recall the formula I put before : E(X²)-[E(X)]²

For question C., I don't know... Your mgf is the mgf of an exponential distribution with parameter 1/3, which would finish the question, by using the cumulative distribution function.

I'm looking for a way without "recognizing" the mgf, but it looks like there isn't any. - May 3rd 2009, 02:28 PMlogitech
This one is still difficult for me to understand. Thank you for trying to help me understand this one. I will keep working on it. I wish you could be my tutor! Taking an online class basically means there is no teacher, just a textbook and a list of work to do.

- May 3rd 2009, 02:46 PMmatheagle
MGF's are unique

This is a gamma with $\displaystyle \alpha=1$ and $\displaystyle \beta=3$

Hence the mean is $\displaystyle \alpha\beta=3$

and the variance is $\displaystyle \alpha\beta^2=9$

and yes, ms mathbeagle, now you can integrate the density to obtain the probabilities. - May 16th 2009, 03:02 AMmr fantastic
See this thread also: http://www.mathhelpforum.com/math-he...enerating.html

- Mar 6th 2016, 06:31 AMtallyogiRe: moment generating function
So if i understand it right t can go from 0 to 1/3. Not sure how to solve E(X) for 0 to 1/3