I am computing the following probability: $\Pr \left[ \sum_{j} (P G_j R_j) \le T\right]$, where $P$ is a fixed number, $G_j$ is a random variable i.i.d. $\forall j$ and $R_j$ is a term that depends on index $j$.

Knowing that $P$ is fixed I can write: $\Pr \left[ \sum_{j} (P G_j R_j) \le T\right] = \Pr \left[P \sum_{j} ( G_j R_j) \le T\right]$, I am wondering if I can do the same with $G_j$ considering it is independent from index $j$ and so bringing it outside the summation. In this manner, knowing the CDF of $G_j$ I can conclude the computation. I.e., if $G_j \sim Exp(\mu)$ I have:

$\Pr \left[P \sum_{j} ( G_j R_j) \le T\right] = \Pr \left[P G_j \sum_{j} ( R_j) \le T\right] = \Pr \left[ G_j \le \frac{T}{P \sum_{j} R_j }\right]=$

$= 1 - e^{-\frac{\mu T}{P\sum_{j} R_j }}$

Can I move the r.v. $G_j$ outside the summation in this manner? Or I should do differently?

  • 3
    $\begingroup$ No you can't do that. You basically have a weighted sum of iid exponentials. If the $R_j$ were all equal, your sum would be Gamma distributed. I'm not sure how to deal with the case where the $R_j$ aren't equal (hopefully someone else will answer) but your approach is not legal as it stands. See this question: math.stackexchange.com/questions/375295/… $\endgroup$
    – Will
    Jun 5, 2017 at 21:01
  • 4
    $\begingroup$ @Will one approach is outlined at stats.stackexchange.com/questions/72479/…. The case for a sum of two terms is solved explicitly at stats.stackexchange.com/questions/252191. If the $R_j$ don't vary much and the sum has very many terms, the CLT will apply. It can be corrected in various ways to give approximations for moderate numbers of terms. When the $R_j$ vary a lot, all bets are off. $\endgroup$
    – whuber
    Jun 5, 2017 at 21:06

1 Answer 1


If you have iid random variables $G_1, \dots, G_n \sim F$ then $\sum_j r_j G_j$ is a very different thing from $G \sum_j r_j$ where $G \sim F$ too, and $G_j \sum_j r_j$ is not meaningful notation since $j$ is not defined outside of the summation. You can only factor out $G$ if $G_1 = \dots = G_n = G$, i.e. they're all actually the same RV. But that is very different from being iid.

As an example, suppose that the $G_i$ are iid $\mathcal N(0,1)$ and let $r_i = \frac 1n$ for all $i$. Then $\sum G_i r_i = \bar G \sim \mathcal N(0, \frac 1n)$ while $G \sum r_i \sim \mathcal N(0, 1)$. There's a reduction in variance from averaging $n$ iid RVs, but not from averaging the exact same RV $n$ times.


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