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I´m working on trying to approach the value of $E\left[ \dfrac{e^x}{x+1} \right]$. Where $x$ is an exponential random variable. All that data I have to work with is a gamma random variable with parameters $3/2$ and $1$. I also know that $x$ has mean $1$.

I think that I could get some relations by knowing that if I sum k exponential random variables I would get a gamma random variable. But I´m a bit confused on how I could apply this to the computation of the Expected Value. Can anyone suggest any valid process to compute the above value using the fact that I know the Gamma Random Variable??

EDIT:

POSSIBLE ANSWER: I have received several comments about the divergence of the integral, however I think I can use this estimator to approach the value:

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Where $T$ is the expected value of the gamma random variable and $Si$ the i-th value generated by the gamma distribution

Thank you

Pablo Estrada
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I'm not sure whether you mean $E\left[ \dfrac{e^x}{x+1} \right]$ or $E\left[ \dfrac{e^x}{x} + 1\right]$. But neither of these exists (the improper integrals diverge). Your gamma random variable is irrelevant to this question.

Robert Israel
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  • I´ve edited the question to correct the expression. Thanks for your answer, I don´t think the expected value does not exist for both since you can approximate it with simulation or other methods. My question is about how can I use this Gamma 'black box' to approximate the expression, even though they diverge. Thanks for your help – Pablo Estrada Sep 17 '14 at 03:50
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    No!!! You can not use simulation or any other method to "approximate" something that does not exist. – Robert Israel Sep 17 '14 at 04:00
  • I think it does exist, since x is an exponential random variable with mean 1. But maybe I´m not being able to see something that you do. – Pablo Estrada Sep 17 '14 at 04:03
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    The density for an exponential random variable of mean $1$ is $\exp(-t)$. So $$ E\left[\dfrac{e^x}{x+1}\right] = \int_0^\infty \dfrac{e^t}{t+1} e^{-t}; dt = \int_0^\infty \dfrac{dt}{t+1} = \infty$$ – Robert Israel Sep 17 '14 at 04:05
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    Now, the divergence is rather slow, so in a simulation it may not be very easy to see that it diverges, but nevertheless it really does diverge. – Robert Israel Sep 17 '14 at 04:13
  • Thanks for your answer. So then it is possible to simulate it using the gamma variable? If that´s true what´s the best way to proceed? – Pablo Estrada Sep 17 '14 at 04:14
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    You've repeatedly been told that the expectation is infinite, and despite this, you somehow continue to insist on "simulating" something. You refused to believe you are wrong and had to have it proven to you. I don't think any further assistance on your questions is justified because you have demonstrated incredibly obstinate behavior. – heropup Sep 17 '14 at 04:57
  • Hello I´m not being obstinate. I do understand what Robert said, but there must be another way around then. I´m reading a simulation book and got interested by this particular exercise I don´t think a simulation book has exercises with no solutions. – Pablo Estrada Oct 01 '14 at 22:12
  • This would not be the first time a textbook exercise had a mistake. Maybe try writing to the author? – Robert Israel Oct 02 '14 at 00:48
  • Ok thank you very much. I´m sorry if I was obstinate, I was just trying to get an appropriate answer. Thanks for your time and knowledge – Pablo Estrada Oct 03 '14 at 17:06