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Given that $X$ is random variable that takes values:

$0,\dots,H-1$

The PMF of $X$ is unknown, but I can tell what is the expected value which is $\bar{X}$

There is event $Y$ when calculated it gives the value:

$P(Y)=E[\frac{1}{X+1}]$

Is there a way to find expected value $\bar{Y}$?

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Random variables have expected values, but an event is a non-numeric quantity and cannot be said to have an expected value. Probability, yes; expected value no. So are you using expected value of an event? to mean its probability? – Dilip Sarwate Nov 6 '11 at 22:16
The question is indeed poorly phrased: $P(Y)$ is meaningless if $Y$ is a random variable. Furthermore, $\bar X$ and $\bar Y$ usually denote empirical averages of samples, rather than the expectations $E[X]$ and $E[Y]$. In any case, unless $H=2$, you cannot find $E[1/(X+1)]$ from $E[X]$. – Xi'an Nov 7 '11 at 6:33

3 Answers

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$$\begin{align*} E\left[\frac{1}{X+1}\right] &= 1\cdot p(0) + \frac{1}{2}\cdot p(1) + \frac{1}{3}\cdot p(2) + \cdots + \frac{1}{H}\cdot p(H-1)\\ &\leq p(0) + 1\cdot p(1) + 2\cdot p(2) + \cdots + (H-1)\cdot p(H-1)\\ &= p(0) + E[X]\\ &\leq 1 + E[X]. \end{align*} $$ The weak upper bound $E[\frac{1}{X+1}] \leq p(0) + E[X]$ requires some knowledge of the pmf of $X$, though the even weaker upper bound $1 + E[X]$ requires knowing only the expected value of $X$ which the OP claims he knows, or does he mean that he has the sample mean $\bar{X}$ available to him?

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+1 For showing how to compute the expectation. But why did you choose to replace $1/i$ by $i$ and not by the tighter values of $1$, which would give the much stronger bound $E[1/(X+1)]\le 1$? This is still trivial, since $0\lt 1/(X+1)\le 1$, but it can't be improved upon without more information. – whuber♦ Aug 15 '12 at 14:31
@whuber I was trying to show that knowing $E[X]$ is not enough to make any assertions about $E[1/(X+1)]$. One possible reason might have been that the bound $E[1/(X+1)] \leq 1$ would likely have induced an additional question of the form "But I also know the value of $E[X]$. Doesn't that help in getting a better bound?" – Dilip Sarwate Aug 16 '12 at 3:23
Right: and the answer clearly is no! You could go further if you want; e.g., suppose we know the value of the mgf at $1$; call it $\phi$. Then we easily obtain $\phi-1$ as a lower bound for $E[1/(X+1)]$. This hints that various bounds will be available in terms of moments of $X$. – whuber♦ Aug 16 '12 at 14:10
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No, you cannot. You would need to the know the pmf for X to be able to compute the expectation of a non-linear transformation of X.

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I think this paper would be helpful http://www.sciencedirect.com/science/article/pii/S0167715201000086?np=y

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