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Chapter 7: Q.7.45 (page 362)

For a standard normal random variable Z,

Show that μn=0(2j)!2jj!

Hint: Start by expanding the moment generating function of Zinto a Taylor series about 0to obtain

EetZ=et2/2

=∑j=0∞t2/2jj!

Short Answer

Expert verified

EZn=M(n)(0)=n!2jj!

Step by step solution

01

Given Information

EetZ=et2/2

02

Explanation

We are given that Z~N(0,1). So, the MGF of Z is

MZ(t)=et22

Since the function above is exponential function, we can expand it into Taylor series easily. Using the definition of exponential function, we have that

MZ(t)=∑n=0∞t22nn!

=∑n=0∞t2n2nn!

=∑neven12n/2(n/2)!tn

03

Explanation

Therefore, MZ(t)is a series that contains only even summands. Thus

EZn=0

for odd n since there is no odd summands in the series. For even n=2j, we have that

EZn=M(n)(0)=n!2jj!

so we have proved the claimed.

04

Final Answer

EZn=M(n)(0)=n!2jj!

We have proved the claimed.

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