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

  • Thread starter Hejdun
  • Start date
  • #1
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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:

Answers and Replies

  • #2
chiro
Science Advisor
4,797
132
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.
 
  • #3
25
0
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.
 
  • #4
chiro
Science Advisor
4,797
132
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.
 
  • #5
25
0
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
 
  • #6
329
0
Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?
 
  • #7
25
0
Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?

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.
 
  • #8
Hurkyl
Staff Emeritus
Science Advisor
Gold Member
14,950
19
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"?
 
  • #9
25
0
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.
 
  • #10
329
0
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.

[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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