How to Find E(1/(1 + e^Z)) for a Normally Distributed Z?

[Hejdun]

Hi everyone,

I am stuck with this problem. I am looking for E(1/(1 + e^Z)) where Z is a normally distributed random variable.

I know that E(e^Z) and E(1/e^Z) follow lognormal and inverse lognormal distibution and the means of these distributions are standard results. Of course, is also easy to find E(e^Z + 1).

However regarding my problem, does anyone have a suggestion of how to proceed? I tried to use the moment generating function but got stuck...

Thanks in advance!
/Hejdun

 

[Hejdun]

Sorry to bump this.

Still no ideas of how to solve this problem?

Of course, I can approximate it using Taylor expansion, but the
resulting expression isn't very nice.

/Hejdun

 

[bpet]

Maybe Gauss-Hermite quadrature will give you a decent approximation?

 

[Hejdun]

Maybe Gauss-Hermite quadrature will give you a decent approximation?

Yes, the integral may be evaluated numerically,
but I am looking for an analytical answer. I am not sure how the Gauss-Hermite quadrature would help for such a case.

Thanks.

/Hejdun

 

[blue_raver22]

Hello Hejdun,

Thank you for reaching out with your question. In order to find the inverse of the shifted lognormal, we can use the fact that the inverse of a lognormal distribution is a gamma distribution. The shifted lognormal distribution is simply a lognormal distribution with a constant added to it. Therefore, the inverse of the shifted lognormal will also be a gamma distribution with a constant added to it.

The formula for the expected value of a gamma distribution with shape parameter α and scale parameter β is αβ. In your case, we can rewrite the expression as E(1/(1+e^Z)) = E(e^(-Z)/(1+e^Z)). We can then use the fact that the inverse of a lognormal distribution is a gamma distribution to rewrite this as E(e^(-Z)/(1+e^Z)) = E(1/e^Z) - E(1/(1+e^Z)). Since we already know the expected value of 1/e^Z, we can focus on finding the expected value of 1/(1+e^Z).

Using the formula for the expected value of a gamma distribution, we can rewrite this as E(1/(1+e^Z)) = αβ - E(e^Z). We know that the expected value of e^Z is e^(μ+σ^2/2), where μ and σ^2 are the mean and variance of the normal distribution. Therefore, we can further simplify this expression to E(1/(1+e^Z)) = αβ - e^(μ+σ^2/2).

In conclusion, the expected value of 1/(1+e^Z) for a shifted lognormal distribution is αβ - e^(μ+σ^2/2), where α and β are the shape and scale parameters of the corresponding gamma distribution. I hope this helps you with your problem. Best of luck!