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Subject: Statistics
Category: Reference, Education and News > Homework Help
Asked by: brkyrhrt99_00-ga
List Price: $5.00

Posted: 04 Oct 2002 11:18 PDT
Expires: 03 Nov 2002 10:18 PST
Question ID: 72511

if Z~N(0,1), compute f(t)=E(e^tZ) and use the result to compute E(Z^4).
Anyone please help.

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Subject: Re: Statistics
Answered By: rbnn-ga on 04 Oct 2002 12:15 PDT
Rated: 5 out of 5 stars

Thank you for posing this question. It is an interesting question with a surprisingly simple and elegant solution. The function f(t) you define is known as the moment generating function for a normal distribution. http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html . Once you compute the moment generating function, it turns out you can easily compute all the moments of a distribution. The actual computation of the moment generating function for a normal distribution is performed in the course notes: http://www.qmw.ac.uk/~ugte133/courses/MATHSTAT/6normgf.pdf . (You will need a pdf reader to view this URL, which you can get from this URL: http://www.adobe.com/products/acrobat/readstep.html . Please see equations 2, 3, and 4 on page 1 of that document. If you have any questions about their derivation, I would be happy to answer them. I am not redoing their work here since actually integrals and quotients can be a little bit hard to read in text. The answer turns out to be, by the way, f(t)= e^( (t^2)/2) To use this value to compute the expectation of Z^4, we follow the derivation on the presentation of moment generating functions at: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html and specifically: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/node4.html . Recall that () e^x= 1 + x + x^2/2! + x^3 /3! ... Hence, f(t) = E(e^(tZ)) = E( 1 + tZ + t^2/2! Z^2 + ...) = E(1) + E(tZ) + E(t^2/2! Z^2 + ...) () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... But we know already that f(t)= e^( (t^2)/2) = 1 + t^2/2 + (t^2/2)^2/2! + ... = 1 + t^2/2 + t^4/8 + ... from the defintion of exponetial () above. We now have two expressions for f(t), the above and that of (**) . Equating the coefficients of t^4, we see that E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 . Search Strategy: moment generating function normal distribution moment generating function probability density "normal distribution" "moment generating function" normal distribution
Request for Answer Clarification by brkyrhrt99_00-ga on 05 Oct 2002 08:25 PDT I am rerally unclear at how you derive at 3. I understand how you find the moment generating function. But from then on I am lost
Clarification of Answer by rbnn-ga on 05 Oct 2002 10:15 PDT I'd be happy to clarify, but first, I don't know what you mean by the comment: "I am rerally unclear at how you derive at 3." What is 3? Also, do you mean that you are ok with the derivation that f(t)=E(e^(t^2)/2) ? If so, just tell me where things start to get unclear, i.e. which line. (By the way when I first saw this proof it also seemed somehow magical and strange to me; all this machinery that doesn't have anything to do with E(Z^4) somehow E(Z^4) just falls out.
Request for Answer Clarification by brkyrhrt99_00-ga on 05 Oct 2002 10:52 PDT this is where I am lost: () e^x= 1 + x + x^2/2! + x^3 /3! ... Hence, f(t) = E(e^(tZ)) = E( 1 + tZ + t^2/2! Z^2 + ...) = E(1) + E(tZ) + E(t^2/2! Z^2 + ...) () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... But we know already that f(t)= e^( (t^2)/2) = 1 + t^2/2 + (t^2/2)^2/2! + ... = 1 + t^2/2 + t^4/8 + ... from the defintion of exponetial () above. We now have two expressions for f(t), the above and that of (**) . Equating the coefficients of t^4, we see that E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 .
Clarification of Answer by rbnn-ga on 05 Oct 2002 13:44 PDT I will begin by recalling some basic facts about Taylor series. [Taylor Series fact] Every infinitely differentiable function g(x) can be written in a Taylor series: g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... where a_i is constant value equal to the i'th derivative of g, evaluated at 0, divided by i! (i!, of course, is 1 * 2 * 3 ... i ): a_0 = g(0) a_1 = g'(0)/1! a_2 = g''(0)/2! a_3 = g'''(0)/3! ... [Power series fact] If it is also the case that g(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ... then a_i = b_i everywhere . The above are some facts from calculus that we will use in the following. I will now clarify my original derivation, with new comments in brackets. () e^x= 1 + x + x^2/2! + x^3 /3! ... [This is the taylor series expansion of e^x . It holds because the i'th derivative of e^x equals e^x; when evaluated at 0, therefore, the i'th derivative of e^x is therefore equal to e^0 which is 1. It follows from the formula after "Taylor series fact" above] Hence, f(t) = E(e^(tZ)) [This is the definition of f(t) ] = E( 1 + tZ + t^2/2! Z^2 + ...) [We are simply rewriting e^(tZ) using the formula (), where "x" is replaced by "tZ"] = E(1) + E(tZ) + E(t^2/2! Z^2) + ... [We are using the fact that the sum of the expectations of random variables is equal to the expectation of the sum of these variables] () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... [Here, remember that t is a a constant. So we are using that the fact that, if Y is a random variable, and if a is any constant, then E( aY ) = a E(Y) .] But we know already that f(t)= e^( (t^2)/2) [We now have two expressions for f(t). The one in and the one we proved, this one, in the first part of the problem ] = 1 + t^2/2 + (t^2/2)^2/2! + ... [This uses the expression () above, with x replaced by t^2/2 .] [Let's call the following equation (*)] = 1 + t^2/2 + t^4/8 + ... [This just writes out the first few terms. We are only interested in the coefficient of t^4.] from the definition of exponetial () above. We now have two expressions for f(t), the above and that of () . [The first expression for f(t) that we found is expression () and the second is expression (*): () f(t)= 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... () f(t)= 1 + 0t +t^2/2 + 0t^3 + t^4/8 + ... ] Equating the coefficients of t^4, we see that [Recall from "power series fact" above that if two power series equal the same funtion, the coefficients of each term must be equal. Hence, the coefficients of t^4 in the two expressions for f(t) above must be equal. These coefficients are E(Z^4)/4! and 1/8 for () and (**) respectively. Hence: ] E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 . [We just multiplied [That is since 4!/8 = 4 3 * 2 * 1 / 8 = 3]

brkyrhrt99_00-ga rated this answer: 5 out of 5 stars

thanks a lot

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Thank you for posing this question. It is an interesting question with a surprisingly simple and elegant solution. The function f(t) you define is known as the moment generating function for a normal distribution. http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html . Once you compute the moment generating function, it turns out you can easily compute all the moments of a distribution. The actual computation of the moment generating function for a normal distribution is performed in the course notes: http://www.qmw.ac.uk/~ugte133/courses/MATHSTAT/6normgf.pdf . (You will need a pdf reader to view this URL, which you can get from this URL: http://www.adobe.com/products/acrobat/readstep.html . Please see equations 2, 3, and 4 on page 1 of that document. If you have any questions about their derivation, I would be happy to answer them. I am not redoing their work here since actually integrals and quotients can be a little bit hard to read in text. The answer turns out to be, by the way, f(t)= e^( (t^2)/2) To use this value to compute the expectation of Z^4, we follow the derivation on the presentation of moment generating functions at: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/moments.html and specifically: http://www.plmsc.psu.edu/~www/matsc597/probability/moments/node4.html . Recall that () e^x= 1 + x + x^2/2! + x^3 /3! ... Hence, f(t) = E(e^(tZ)) = E( 1 + tZ + t^2/2! Z^2 + ...) = E(1) + E(tZ) + E(t^2/2! Z^2 + ...) () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... But we know already that f(t)= e^( (t^2)/2) = 1 + t^2/2 + (t^2/2)^2/2! + ... = 1 + t^2/2 + t^4/8 + ... from the defintion of exponetial () above. We now have two expressions for f(t), the above and that of (**) . Equating the coefficients of t^4, we see that E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 . Search Strategy: moment generating function normal distribution moment generating function probability density "normal distribution" "moment generating function" normal distribution
Request for Answer Clarification by brkyrhrt99_00-ga on 05 Oct 2002 08:25 PDT I am rerally unclear at how you derive at 3. I understand how you find the moment generating function. But from then on I am lost
Clarification of Answer by rbnn-ga on 05 Oct 2002 10:15 PDT I'd be happy to clarify, but first, I don't know what you mean by the comment: "I am rerally unclear at how you derive at 3." What is 3? Also, do you mean that you are ok with the derivation that f(t)=E(e^(t^2)/2) ? If so, just tell me where things start to get unclear, i.e. which line. (By the way when I first saw this proof it also seemed somehow magical and strange to me; all this machinery that doesn't have anything to do with E(Z^4) somehow E(Z^4) just falls out.
Request for Answer Clarification by brkyrhrt99_00-ga on 05 Oct 2002 10:52 PDT this is where I am lost: () e^x= 1 + x + x^2/2! + x^3 /3! ... Hence, f(t) = E(e^(tZ)) = E( 1 + tZ + t^2/2! Z^2 + ...) = E(1) + E(tZ) + E(t^2/2! Z^2 + ...) () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... But we know already that f(t)= e^( (t^2)/2) = 1 + t^2/2 + (t^2/2)^2/2! + ... = 1 + t^2/2 + t^4/8 + ... from the defintion of exponetial () above. We now have two expressions for f(t), the above and that of (**) . Equating the coefficients of t^4, we see that E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 .
Clarification of Answer by rbnn-ga on 05 Oct 2002 13:44 PDT I will begin by recalling some basic facts about Taylor series. [Taylor Series fact] Every infinitely differentiable function g(x) can be written in a Taylor series: g(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ... where a_i is constant value equal to the i'th derivative of g, evaluated at 0, divided by i! (i!, of course, is 1 * 2 * 3 ... i ): a_0 = g(0) a_1 = g'(0)/1! a_2 = g''(0)/2! a_3 = g'''(0)/3! ... [Power series fact] If it is also the case that g(x) = b_0 + b_1 x + b_2 x^2 + b_3 x^3 + ... then a_i = b_i everywhere . The above are some facts from calculus that we will use in the following. I will now clarify my original derivation, with new comments in brackets. () e^x= 1 + x + x^2/2! + x^3 /3! ... [This is the taylor series expansion of e^x . It holds because the i'th derivative of e^x equals e^x; when evaluated at 0, therefore, the i'th derivative of e^x is therefore equal to e^0 which is 1. It follows from the formula after "Taylor series fact" above] Hence, f(t) = E(e^(tZ)) [This is the definition of f(t) ] = E( 1 + tZ + t^2/2! Z^2 + ...) [We are simply rewriting e^(tZ) using the formula (), where "x" is replaced by "tZ"] = E(1) + E(tZ) + E(t^2/2! Z^2) + ... [We are using the fact that the sum of the expectations of random variables is equal to the expectation of the sum of these variables] () = 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... [Here, remember that t is a a constant. So we are using that the fact that, if Y is a random variable, and if a is any constant, then E( aY ) = a E(Y) .] But we know already that f(t)= e^( (t^2)/2) [We now have two expressions for f(t). The one in and the one we proved, this one, in the first part of the problem ] = 1 + t^2/2 + (t^2/2)^2/2! + ... [This uses the expression () above, with x replaced by t^2/2 .] [Let's call the following equation (*)] = 1 + t^2/2 + t^4/8 + ... [This just writes out the first few terms. We are only interested in the coefficient of t^4.] from the definition of exponetial () above. We now have two expressions for f(t), the above and that of () . [The first expression for f(t) that we found is expression () and the second is expression (*): () f(t)= 1 + tE(Z) + t^2/2! E(Z^2) + t^3/3! E(Z^3) + t^4/4! E(Z^4) + .... () f(t)= 1 + 0t +t^2/2 + 0t^3 + t^4/8 + ... ] Equating the coefficients of t^4, we see that [Recall from "power series fact" above that if two power series equal the same funtion, the coefficients of each term must be equal. Hence, the coefficients of t^4 in the two expressions for f(t) above must be equal. These coefficients are E(Z^4)/4! and 1/8 for () and (**) respectively. Hence: ] E(Z^4) /4! = 1/8 so E(Z^4) = 4!/8 = 3 . [We just multiplied [That is since 4!/8 = 4 3 * 2 * 1 / 8 = 3]