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
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.
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.
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.
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
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!
