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Delta method fails. Any suggestions how to calculate E(Y)? 
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#1
Mar3112, 08:54 PM

P: 25

Hi,
Let Z be a r.v. from N(0, σ^2) and let Y = Z^2*exp(Z)/(1 + exp(Z))^2. What is the exptected value of Y, E(Y)? The delta method (Taylor expansion) doesn't work since the errors seem to accumulate. Any suggestions? This is a nonstandard problem related to my research and I am totally stuck with this one... // H 


#2
Mar3112, 10:21 PM

P: 4,579

If it's hard to get an analytic answer for the distribution, you should do a simulation and then create a simulated distribution for Y in which you can can easily calculate E[Y]. I would use R to do this: it's free and it would take you probably ten minutes to get this result up on screen. The commands you need to use z = rnorm(10000,0,sd=whatever) and then y = z^2*exp(z)/(1 + exp(z))^2 and then mean(y). It should take only ten seconds or less to execute. 


#3
Apr112, 05:31 AM

P: 25

Thanks for your help. However, I am not looking for a numerical solution. I am aware of that no exact analytical solution exists, but maybe there is some approximation that is "good enough". If you plot the function in R, then it is clear that Y=f(Z) is a symmetric bimodal distribution. Perhaps it simplifies if we put a restriction and look at Y when Z>0. But anyway, I am stuck right now. Perhaps not solution exist. 


#4
Apr112, 05:48 AM

P: 4,579

Delta method fails. Any suggestions how to calculate E(Y)?



#5
Apr112, 06:02 AM

P: 25

However, I am not sure how your suggestion in this case would help. What chisquare distribution and how many degrees of freedom? BTW, I tried doing a Padé Approximation instead of Taylor expansion, but that didn't help either. /H 


#6
Apr112, 08:30 AM

P: 327

Since Y(Z) = Y(Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?



#7
Apr112, 08:39 AM

P: 25

That is true. I think the assumption Y(Z)=Y(Z) could be wrong however (even though they are very similar), since when I run simulations to check E(Y) then E(Y) ≠ 0. 


#8
Apr112, 09:09 AM

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#9
Apr112, 09:13 AM

P: 25

But thanks for you comment. 


#10
Apr112, 09:56 AM

P: 327

[edit] Oops... Although it's true that Y(Z) = Y(Z), unfortunately that's not what you would need to get a mean of zero. For that you would need Y(Z) = Y(Z). So please ignore what I wrote. [/edit] 


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