# Moment Generating Function

• November 4th 2009, 04:29 PM
xuyuan
Moment Generating Function
Can someone explain to me how this problem works?

Given the moment generating function Mx(t)=e^(3t+8t^2), find the moment generating function of the random variable Z=1/4(X-3) and use it to determine the mean and the variance of Z.

I don't understand why its asking you to find another mgf if one is given to you already. Thanks!
• November 4th 2009, 04:33 PM
mr fantastic
Quote:

Originally Posted by xuyuan
Can someone explain to me how this problem works?

Given the moment generating function Mx(t)=e^(3t+8t^2), find the moment generating function of the random variable Z=1/4(X-3) and use it to determine the mean and the variance of Z.

I don't understand why its asking you to find another mgf if one is given to you already. Thanks!

See 6.7 here: http://www.am.qub.ac.uk/users/g.gribakin/sor/Chap6.pdf
• November 4th 2009, 10:32 PM
matheagle
Mx(t)=e^(3t+8t^2) is the MGF of X

you are asked to get the MGF of Z=X/4-3/4 if I understand your ().
• November 7th 2009, 07:09 PM
xuyuan
Quote:

Originally Posted by matheagle
Mx(t)=e^(3t+8t^2) is the MGF of X

you are asked to get the MGF of Z=X/4-3/4 if I understand your ().

Hm, so I read through what was posted and my textbook, but I still don't quite understand. So I have the MGF of X, then to get the MGF of Z do I need to integrate Z from negative to positive infinity and somehow subsitute in X? This is really confusing me, any help is appreciated thanks!(Headbang)
• November 7th 2009, 07:24 PM
matheagle
Just substitute, or recognize the distribution of X and use that.

Z=X/4-3/4

So $M_Z(t)=E(e^{Zt})=E(e^{(X/4-3/4)t})=e^{(-3/4)t}E(e^{Xt/4})$

$=e^{(-3/4)t}E(e^{X(t/4)})=e^{(-3/4)t}M_X(t/4)$

Plug in t/4 for t in the mgf of X and examine what you have.
• November 7th 2009, 07:32 PM
mr fantastic
Quote:

Originally Posted by xuyuan
Hm, so I read through what was posted and my textbook, but I still don't quite understand. So I have the MGF of X, then to get the MGF of Z do I need to integrate Z from negative to positive infinity and somehow subsitute in X? This is really confusing me, any help is appreciated thanks!(Headbang)

You're probably expected to know and apply the result found at the bottom of page 1 here: http://web.as.uky.edu/statistics/use...320u04/mgf.pdf

In fact, it's the same result that I refered you to in my first reply. Did you in fact bother to click on the link and look at it?
• November 8th 2009, 10:30 AM
nnnikii
Hi, I've been working on the same problem and came to the answer
Mz(t)=e^[(4/5)t^2]

Is this the answer that you came to?
• November 8th 2009, 10:45 AM
mr fantastic
Quote:

Originally Posted by nnnikii
Hi, I've been working on the same problem and came to the answer
Mz(t)=e^[(4/5)t^2]

Is this the answer that you came to?

• November 8th 2009, 10:53 AM
nnnikii
Oopsie!! I just went over my calculation and realized I made a mistake. My new answer is

Mz(t)=e^[(1/2)t^2]

Is that correct?
• November 8th 2009, 11:04 AM
mr fantastic
Quote:

Originally Posted by nnnikii
Oopsie!! I just went over my calculation and realized I made a mistake. My new answer is

Mz(t)=e^[(1/2)t^2]

Is that correct?

Yes.
• November 8th 2009, 02:32 PM
matheagle
The point of this problem is that a linear transformation of a normal is a normal.
The mean and variance should be easy to calculate directly.