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Let $\{X_i\}_{i=1}^n$ be $n$ positive, discrete (so positive integers) and IID random variables. Let $\{c_i\}_{i=1}^n$ be constants and $$Y=\frac{\sum c_iX_i}{\big(\sum X_i\big)^2}\ \ \ ;\ \ \ Z=\frac{1}{\sum X_i}$$

I'm trying to calculate $\mathbb{E}[Y]$ and $\text{var}(Y)$ in terms of $\mathbb{E}[X_i]$'s. Similarly for expectation and variance of $Z$. I've looked at other answers related to calculating the expectation of inverses and quotients, but they deal with more general cases and involve integrals and all.

Given the assumptions about $X_i$'s that I listed out, how can $\mathbb{E}[Y]$ and $\text{var}(Y)$ be calculated?

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  • $\begingroup$ Check this answer: math.stackexchange.com/questions/302436/… $\endgroup$ – Ale Aug 11 '20 at 13:19
  • $\begingroup$ @Ale: I already did, but that one involves integration, which I don't think is needed for my case. And the second answer there provides a lower bound. There must be a simpler way to compute expectation and variance of $Y$? $\endgroup$ – user9343456 Aug 11 '20 at 13:32
  • $\begingroup$ There is no universal formula that improves on the definitions. For instance, to find $E[Z]$ first find the distribution of $X=\sum X_i.$ Then, by definition, $E[Z] = \sum_{x} \Pr(X=x)/x.$ If you are hoping for anything simpler, then you will need to be more specific about the distribution of the $X_i.$ $\endgroup$ – whuber Aug 13 '20 at 13:11
  • $\begingroup$ @whuber: completely fair, but what you listed as the formula for $E[Z]$ is for case when it's a function of only one RV, i.e. $X$. What about the case when it's a function of multiple RV's $X_1,\ldots,X_n$? As for the distribution, assume that we can calculate, for any $X_i$, the probability $P(X_i=k)$, where $k$ is a positive integer. $X_i$'s don't have a standard probability distribution like uniform or normal though. $\endgroup$ – user9343456 Aug 13 '20 at 13:16
  • $\begingroup$ That's such a general situation there's nothing to add. I hope the generalization from a univariate discrete variable to a multivariate discrete variable is obvious: a single sum becomes a sum over $N$ variables, that's all. $\endgroup$ – whuber Aug 13 '20 at 13:19
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I don't think you'll get a tractable expression in terms of $E(X_i)$. If you have a concrete expression for $P(X_j=n)$, and denote $S_j\equiv X_j/(\sum X_i)^2$, then in this case you can calculate: $$ \begin{align} E(S_j)&=\sum_{a_1,\ldots,a_n}P(X_1=a_1, \ldots,X_n=a_n)\frac{a_j}{(\sum_ia_i)^2} \\&=\sum_{a_1,\ldots,a_n}P(X_1=a_1)\ldots P(X_n=a_n)\frac{a_j}{(\sum_ia_i)^2} \end{align} $$ where the second equality follows from independence of $X_i$. In addition, since $X_i$ are identically distributed, so are $S_i$, which means the above expression holds for all $j=1,\ldots,n$. Finally, note that $Y=\sum_j c_jS_j$, so by linearity of expectation, $$E(Y)=\sum_jc_jE(S_j)=\mu_S\sum_jc_j$$ since all $E(S_j)$ are the same and we denote the common value by $\mu_S$.

Similarly you can calculate $E(Z)$. To get the variance, you can calculate $E(Y^2)$ and the variance calculation follows.

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    $\begingroup$ Since the $X_i$ are assumed iid, your formula is unnecessarily cumbersome: it will suffice to compute the expectation of $X_1/\sum_i(X_i^2).$ Even that is potentially amenable to simplification, because it can be expressed in terms of the distributions of $X_1$ and $X_2^2 + \cdots X_n^2.$ $\endgroup$ – whuber Aug 13 '20 at 17:08
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    $\begingroup$ @whuber: Ah yes you're correct! I forgot about the identically distributed part. I've at least partially simplified it. $\endgroup$ – Shirish Kulhari Aug 13 '20 at 17:28
  • $\begingroup$ @whuber How are you getting $E[X_1/(\sum_i X_i)^2] = E[X_1/\sum_i X_i^2]$? $\endgroup$ – Bertus101 Dec 18 '20 at 10:30

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