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

In summary, the conversation revolved around finding the expected value of Y, where Y = Z^2*exp(Z)/(1 + exp(Z))^2 and Z is a random variable from a normal distribution. The use of the delta method and simulation were suggested, but the person seeking help was looking for an analytical solution. The discussion also touched on using a chi-square distribution and the assumption that Y(Z) = Y(-Z), which would imply E(Y) = 0. However, it was later realized that this assumption was incorrect and the search for a solution continues.
  • #1
Hejdun
25
0
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:
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  • #2
Hejdun said:
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
chiro said:
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
Hejdun said:
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
chiro said:
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
Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?
 
  • #7
awkward said:
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
Hejdun said:
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
Hurkyl said:
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
Hejdun said:
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]
 
Last edited:

What is the Delta Method and why does it fail?

The Delta Method is a mathematical approach used to approximate the distribution of a function of random variables. It fails when the function is highly non-linear or when the random variables are not independent.

What are some common situations when the Delta Method fails?

The Delta Method often fails in situations where the function involves division, multiplication, or exponentiation of random variables. It also fails when there is a high degree of correlation between the random variables.

Can the Delta Method be used to calculate the expected value of a function?

Yes, the Delta Method can be used to calculate the expected value of a function. However, it is important to note that the results may not be accurate if the method fails.

Are there alternative methods for calculating the expected value of a function?

Yes, there are alternative methods such as Monte Carlo simulation or numerical integration that can be used to calculate the expected value of a function when the Delta Method fails.

What are some suggestions for calculating the expected value when the Delta Method fails?

One suggestion is to use a more accurate approximation method, such as the Taylor expansion. Another suggestion is to use simulation techniques, such as bootstrapping or Markov Chain Monte Carlo, to estimate the expected value. Additionally, it may be helpful to assess the sensitivity of the results to any assumptions made in the calculation.

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