Variance of Sum of Indicator RV's

You randomly select $$N$$ points independently and uniformly within the unit square $[0,1]\times[0,1]$ . Let $$I_i$$ be the indicator variable equal to 1 if the $$i$$ -th point lies inside the quarter-circle $x^2 + y^2 \le 1$ in the first quadrant, and 0 otherwise. What is the variance of the random variable $\frac{I_1 + I_2 + \cdots + I_N}{N}$ in terms of $$N$$ ?

Answer

Since the area of the quarter-circle $x^2 + y^2 \le 1$ in the first quadrant is $\frac{\pi}{4}$ , each $$I_i$$ has

\mathbb{E}[I_i] = p = \frac{\pi}{4}, \quad \mathrm{Var}(I_i) = p\,(1-p).

Because the $$I_i$$ are i.i.d., their sum $S = I_1 + I_2 + \cdots + I_N$ follows a Binomial $\bigl(N,\,\tfrac{\pi}{4}\bigr)$ . Then the variance of the average $\frac{S}{N}$ is

\mathrm{Var}\!\Bigl(\tfrac{S}{N}\Bigr) = \frac{1}{N^2}\,\mathrm{Var}(S) = \frac{1}{N^2} \cdot N\,p\,(1-p) = \frac{p\,(1-p)}{N} = \frac{\frac{\pi}{4}\Bigl(1-\tfrac{\pi}{4}\Bigr)}{N}.

Notes and comments

Comment 1: For an indicator random variable $$I$$ taking values 0 or 1 with probability $$p$$ of being 1, its variance is

\mathrm{Var}(I) = p - p^2 = p\,(1 - p).

This is due to the fact that $$I^2 = I$$ .