# Expectation and variance of quotient of sums of positive, discrete, iid random variables

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?

• Check this answer: math.stackexchange.com/questions/302436/… – Ale Aug 11 '20 at 13:19
• @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 $Y$? – user9343456 Aug 11 '20 at 13:32
• 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 $X_i.$ – whuber Aug 13 '20 at 13:11
• @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. – user9343456 Aug 13 '20 at 13:16
• 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. – whuber Aug 13 '20 at 13:19

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.
• 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 $X_2^2 + \cdots X_n^2.$ – whuber Aug 13 '20 at 17:08
• @whuber How are you getting $E[X_1/(\sum_i X_i)^2] = E[X_1/\sum_i X_i^2]$? – Bertus101 Dec 18 '20 at 10:30