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

by Hejdun
Tags: delta, fails, method, suggestions
 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 non-standard problem related to my research and I am totally stuck with this one... // H
P: 4,542
 Quote by Hejdun 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 non-standard problem related to my research and I am totally stuck with this one... // H
Hey Hejdun and welcome to the forums.

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.
P: 25
 Quote by chiro Hey Hejdun and welcome to the forums. 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.

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.

P: 4,542

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

 Quote by Hejdun 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.
Instead of using a normal, why don't you use the square root of a chi-square. This distribution will act like a positive normal distribution and is only defined from 0 onwards.
P: 25
 Quote by chiro Instead of using a normal, why don't you use the square root of a chi-square. This distribution will act like a positive normal distribution and is only defined from 0 onwards.
Thanks again for you interest in my problem and your suggestions.
However, I am not sure how your suggestion in this case would help. What chi-square 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
 P: 325 Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?
P: 25
 Quote by awkward Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?

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.
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P: 16,094
 Quote by Hejdun 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".
I don't understand this. Why can't a numerical solution be "good enough"?
P: 25
 Quote by Hurkyl I don't understand this. Why can't a numerical solution be "good enough"?
Because this expectation is a part of a larger derivation of a proof and needs to be more general.

But thanks for you comment.
P: 325
 Quote by Hejdun Thanks for your answer. 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.
It's not an assumption, it's an identity. See if you can prove it.

 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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