Consider the similar question as this problem, where I wanna calculate the expectation of a sum of random variables, it's safe to assume they're independent, but no identical distribution holds.

Is the following equation generally holds? What kind of assumption can we make it hold if not always equal?

$$E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right)\ \underset{=}{?}\ \ \frac{\sum_{j=1}^k E[X_j]}{\sum_{i=1}^n E[X_i]}.$$

Tried to use the linearity of expectation but seems only workable for numerator: $$E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right)\ = \ \ \sum_{j=1}^k E\left[\frac{ X_j}{ \sum_{i=1}^n X_i}\right]$$ but then don't know how to deal with the remains


1 Answer 1


This is almost never true, only in the most trivial cases, and conditions to make it true will be contrived.

Here is a very simple example: suppose that $k = 1$, $n = 2$, that $X_1 = 1$ with probability 1 and that $X_2$ is a standard Bernoulli random variable. (That is, takes values 0 and 1 with probability $\tfrac12$ each.) Then $$ \Bbb E\left(\frac{\sum_{j=1}^k X_j}{\sum_{i=1}^{n} X_i}\right)= \Bbb E\left(\frac1{1+X_2}\right) = \frac34 \neq \frac23 = \frac1{\Bbb E\left(1+X_2\right)} = \frac{\sum_{j=1}^k E[X_j]}{\sum_{i=1}^n E[X_i]}. $$

Edit: Below is an earlier version of this answer, where I incorrectly read the variables in numerator and denominator as distinct; it shows more broadly that the key problem is that taking the inverse and $\mathbb E$ just do not commute.

Note that your observation for the numerator is correct, and if random variables $X, Y$ are independent then $\mathbb E[XY] = \mathbb E[X]\mathbb E[Y]$. This means that in the (already special) case of independent variables, we furthermore have $$ \sum_{j=1}^k \Bbb E\left[\frac{ X_j}{ \sum_{i=1}^n X_i}\right] = \sum_{j=1}^k \Bbb E[X_j] \Bbb E\left[\frac1{ \sum_{i=1}^n X_i}\right], $$ and because you already know that you can distribute sums over $\Bbb E$ the interesting question, really, is whether or not $$ \Bbb E\left[\frac1X\right] \overset{?}{=} \frac1{\Bbb E[X]}, $$ and it is easy to verify in simple cases that this does not hold. E.g., $X$ takes values $1, 2$ both with probability $\tfrac12$: then $\Bbb E[1/X] = \tfrac34$ while $1/\Bbb E[X] = \tfrac23$.

  • $\begingroup$ Even for independence case, the first equality is not true since $X_j$ ($j\leq n$) dependes on $\sum_{i=1}^n X_i$. $\endgroup$
    – NCh
    Commented Feb 27, 2020 at 14:01
  • $\begingroup$ Ah, thanks. I hadn't even realized that it was the same random variables in the numerator as in the denominator. $\endgroup$ Commented Feb 27, 2020 at 14:05

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