Calculating the mean and variance from a moment generating function

  • #1

Homework Statement


Assume that X is squared-Chi-distributed, which means that the moment generating function is given by:

[itex]m(t)=(1-2t)^{-k/2}[/itex]

Use the mgf to find E(X) and var(X)

The Attempt at a Solution


I know that m'(0)=E(X), and m''(0)=var(X).

So I find:

[itex]m'(t)=k(1-2t)^{-(k/2)-1}[/itex]
which gives m'(0)=k

Similarily, I find

[itex]m''(t)=(k^{2}+2k)(1-2t)^{-(k/2)-2}[/itex]
which gives m''(0)=k^2+2k

However, in my textbook, it says that the variance of a square-chi distribution should be 2k, not k^2. Where do I go wrong?
 

Answers and Replies

  • #2
LCKurtz
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Homework Statement


Assume that X is squared-Chi-distributed, which means that the moment generating function is given by:

[itex]m(t)=(1-2t)^{-k/2}[/itex]

Use the mgf to find E(X) and var(X)

The Attempt at a Solution


I know that m'(0)=E(X), and m''(0)=var(X).
Your mistake is right there. m''(0) = E(X2), not var(X)

So I find:

[itex]m'(t)=k(1-2t)^{-(k/2)-1}[/itex]
which gives m'(0)=k

Similarily, I find

[itex]m''(t)=(k^{2}+2k)(1-2t)^{-(k/2)-2}[/itex]
which gives m''(0)=k^2+2k

However, in my textbook, it says that the variance of a square-chi distribution should be 2k, not k^2. Where do I go wrong?
 
  • #3
Of course. Then var(X)=E(X^2)-(E(X))^2 =k^2+2k-k^2=2k.

Thank you!!
 

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