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

  1. Mar 31, 2012 #1
    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: Mar 31, 2012
  2. jcsd
  3. Mar 31, 2012 #2

    chiro

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    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.
     
  4. Apr 1, 2012 #3

    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.
     
  5. Apr 1, 2012 #4

    chiro

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    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.
     
  6. Apr 1, 2012 #5
    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
     
  7. Apr 1, 2012 #6
    Since Y(Z) = Y(-Z) and Z's distribution is symmetric about 0, doesn't this imply that E(Y) = 0?
     
  8. Apr 1, 2012 #7
    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.
     
  9. Apr 1, 2012 #8

    Hurkyl

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    I don't understand this. Why can't a numerical solution be "good enough"?
     
  10. Apr 1, 2012 #9
    Because this expectation is a part of a larger derivation of a proof and needs to be more general.

    But thanks for you comment.
     
  11. Apr 1, 2012 #10
    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: Apr 1, 2012
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