Chapter 5, pages 18, 19 - MGF coefficients of t, t^2,t^3, etc..

[stylz]

Hi,

Near the bottom of page 18, Chapter 5, the core reading states that E[X^2] = coefficient of t^2/2! = o^2 + u^2. This is taken from the expansion immediatly preceding.

I don't see how they got this. Which part of the expansion do we use? Is it the whole formula in the brackets? The coefficient seems to me to just be 1/4o^2.

Following on from that I can't work out how they have got E[Z], E[Z^2] etc on the top of the next page (19).

I don't understand what part to use as the coefficient, whether it is just the part in front of t^2 or the whole amount in the brackets.

Can someone enlighten me please.
Cheers

 

[deepakraomore]

Hi,
Instead of using series expansion, use the derivative of the expansion, as
1) derive 1st order derivative series expansion of Mx(t) and then put t=0, will get E[X]= Mu.
Now, derive 2nd order derivative of series expansion Mx(t) and then put
t=0, will get 6^2+u^2
2) for Standard normal Z dist. follow the same method.

Regards,
Deepak,
IND.

 

[stylz]

Thanks Deepak.

Yes I do know how to work it out using derivatives, but I was trying to follow the core reading where they have used the coefficients. I can't quite understand exactly how they have got those coefficients.

Cheers.

 

[deepakraomore]

Hi,
After calculations what i observed is as
While using Series Expansion,
1) Expand the third term of Mx(t) i.e. term with 1/2! Coefficient and then get the parameters u^2+6^2 in one bracket which is the coefficient of t^2/2!. Will give the required result for moment.
We can’t use the bracket as it is, since we have to find the term t with the same power.
2) Since for normal Z
There is parameter with coefficient 1/2! So E[Z]=0
The coefficient of t^2/2! is 1 so E[Z^2]=1
There is parameter with coefficient 1/3! So E[Z^3]=0
3) By my understanding the part with t^r/r! To be use as coefficient.

Regards,
Deepak

 

[stylz]

Hi,
After calculations what i observed is as
While using Series Expansion,
1) Expand the third term of Mx(t) i.e. term with 1/2! Coefficient and then get the parameters u^2+6^2 in one bracket which is the coefficient of t^2/2!. Will give the required result for moment.
We can’t use the bracket as it is, since we have to find the term t with the same power.
2) Since for normal Z
There is parameter with coefficient 1/2! So E[Z]=0
The coefficient of t^2/2! is 1 so E[Z^2]=1
There is parameter with coefficient 1/3! So E[Z^3]=0
3) By my understanding the part with t^r/r! To be use as coefficient.

Regards,
Deepak

Hi Deepak,
Thanks again for your help. Using the numbers in your response:

1) Can you give me your workings when expanding the brackets for the third term. I did not end up with u^2 + 6^2.
2) I don't see how for norma Z, the coefficient of t^2/2! is 1. This is the same method as above isn't it where I expand the brackets of the third term, which is (1/2t^2)^2. How does this expand to 1? Sorry, I'm still not following.

 

[deepakraomore]

Hi,
The expansion is as,
(ut+(1/2)6^2t^2))(1/2^4!)=(u^2t^2+u6^2t^3+(1/4)6u^4t^4)(1/2!)
=u^2t^2/2!+u6^2t^3/2!+(1/4)6^4t^4/2!
Here 1/2! will remain as it is for all terms in the bracket.
Now we have, Mx(t)=1+ut+6^2t^2/2+!)+(u^2t^2+u6^2t^3+(1/4)6u^4)(1/2!)
Since the we have to use coefficient of t^2/2!. And by expanding the third term we get only one coefficient for t^2/2!. The rest having t^3/2! and
t^4/2!
Now taking out the terms having 1/2 common (i.e 1/2!)
we get,
Mx(t)=1+ut+(6^2t^2/2!)+(u^2t^2/2!)
=1+ut+t^2/2!(u^2+6^2)
The coefficient of t^2/2! is th erequired value for E[X^2]

For using series expansion and coefficient of t^r/r!, refer the page no 12, chapter 05-Generating Functions.

Regards,
deepak

Hi Deepak,
Thanks again for your help. Using the numbers in your response:

1) Can you give me your workings when expanding the brackets for the third term. I did not end up with u^2 + 6^2.
2) I don't see how for norma Z, the coefficient of t^2/2! is 1. This is the same method as above isn't it where I expand the brackets of the third term, which is (1/2t^2)^2. How does this expand to 1? Sorry, I'm still not following.

 

[stylz]

Oh yes I see now. You have actually ended up combining the second part of the second term with the third term to get the coefficient. I was only expanding the third term and wondering why the coefficient ended up being only u^2, rather than u^2 + 6^2 when combined with the last part of the second term.

Very good. Cheers!