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

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

Last edited:

chiro
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.

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.

chiro
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.

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

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

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.

Hurkyl
Staff Emeritus
Gold Member
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"?

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.