# What is the variance of multiple indicator random variables?!

Consider the following independent random variables $(V_1,V_2,V_3,\ldots,V_n)$ and a random variable $X$ as a function of these other random variables defined as follow on a set $A=(-\infty,x]$: $$\ X=f(V_{1},V_{2},V_{3},\cdots,V_{n})=\sum_{i=1}^{n}1_{(-\infty,x]}\left( V_{i}\right) \$$

Are the following assertions true $$\ \mathbb{E}\left( X\right) =\sum_{i=1}^{n}\mathbb{E}\left( 1_{(-\infty ,x]}\left( V_{i}\right) \right) =n\mathbb{P}\left( A\right) \$$ and that the variance is

$$\ var\left( X\right) =var\left( \sum_{i=1}^{n}1_{A}(V_{i})\right) =\sum_{j=1}^{n}\sum_{i=1}^{n}cov(1_{A}(V_{i}),1_{A}(V_{j}))=n^{2}% \mathbb{P}\left( A\right) \left( 1-\mathbb{P}\left( A\right) \right) \$$

No, it is not correct. Please note that $A = (-\infty,x]$ is a subset of $\mathbb{R}$ whereas $\mathbb{P}$ is a (probability) measure on the probability space. This means that $\mathbb{P}(A)$ is not even well-defined.

For the first one, note that

$$\mathbb{E} \big( 1_{(-\infty,x]}(V_i) \big) = \int 1_{(-\infty,x]}(V_i(\omega)) \, d\mathbb{P}(\omega) = \mathbb{P}(V_i \leq x).$$

Hence,

$$\mathbb{E}X = \sum_{i=1}^n \mathbb{P}(V_i \leq x).$$

If the random variables are identically distributed, then

$$\mathbb{E}X = n \mathbb{P}(V_1 \leq x).$$

A similar calculation yields the variance of $X$; use that

$$\mathbb{E}(1_A(V_i) 1_A(V_j)) = \begin{cases} \mathbb{P}(V_i \leq x) & i=j \\ \mathbb{P}(V_i \leq x) \mathbb{P}(V_j \leq x) & i \neq j. \end{cases}$$

• Thank you very much Saz, but then again if $V_i$ are all identically distributed $\mathbb{P}\left( V_{i}\leq x\right) \mathbb{P}\left( V_{j}\leq x\right) =\left( \mathbb{P}\left( V_{i}\leq x\right) \right) ^{2}$ even if $i$ is different than $j$ Commented Aug 18, 2014 at 6:44
• @rsc05 Yes, that's correct. But note that for $i=j$, we have $\mathbb{P}(V_i \leq x)$ and not $\mathbb{P}(V_i \leq x)^2$.
– saz
Commented Aug 18, 2014 at 7:01