# Thread: moment generating

1. ## moment generating

If the moment generating function of X is $M(t)=\frac{1}{1-3t}$, $t<\frac{1}{3}$
(a) Find E(X)
(b) Find Var(X)
(c) P(6.1<X<6.7)

Thank you!

2. This has been posted already.
X is an EXP rv with parameter $\beta=3$.
Hence the mean and variance is 3 and 9.

Finally part (c) is ${1\over 3}\int_{6.1}^{6.7} e^{-x/3}dx$.

3. So E(X) is basically the same thing as the mean? Did you just square the parameter / mean to get the variance. For part c, using your equation, I got 0.023727. Thank you for your help.

4. Originally Posted by redpack
So E(X) is basically the same thing as the mean? Did you just square the parameter / mean to get the variance. For part c, using your equation, I got 0.023727. Thank you for your help.
mu=mean, just ask MOO

5. Originally Posted by redpack
Did you just square the parameter / mean to get the variance.
Thats because the distribution in question is exponential... For a general distribution the general method is to write the moment generating function as a power series and identifying the higher order moments.

6. Originally Posted by redpack
If the moment generating function of X is $M(t)=\frac{1}{1-3t}$, $t<\frac{1}{3}$
(a) Find E(X)
(b) Find Var(X)
(c) P(6.1<X<6.7)

Thank you!
First asked here: http://www.mathhelpforum.com/math-he...-function.html