Question

Let $X$ be a random variable with pdf $f(x)=\dfrac{1}{2\sqrt{3}}$ for $-\sqrt{3}<x<\sqrt{3}$, and $0$ otherwise. Then, the variance of $X$ is _____(in integer)}.

Hint:

Uniform $U(a,b)$ shortcut: mean $=\dfrac{a+b}{2}$ and variance $=\dfrac{(b-a)^2}{12}$. Symmetry about $0$ immediately gives $E[X]=0$.

Answer 1

Correct Answer: 1

This is a uniform distribution on $(-\sqrt{3},\sqrt{3})$.
Method 1 (known formula). For $X\sim U(a,b)$, $\displaystyle \mathrm{Var}(X)=\frac{(b-a)^2}{12}$. Here $a=-\sqrt{3}$, $b=\sqrt{3}$, so
$(b-a)=2\sqrt{3}\Rightarrow \mathrm{Var}(X)=\dfrac{(2\sqrt{3})^2}{12}=\dfrac{12}{12}=\boxed{1}$.
Method 2 (direct integration). Since the pdf is symmetric, $E[X]=0$.
\[ E[X^2]=\int_{-\sqrt{3}}^{\sqrt{3}} x^2\frac{1}{2\sqrt{3}}\,dx =\frac{1}{2\sqrt{3}}\left[\frac{x^3}{3}\right]_{-\sqrt{3}}^{\sqrt{3}} =\frac{1}{2\sqrt{3}} . \frac{6\sqrt{3}}{3}=1. \] Thus $\mathrm{Var}(X)=E[X^2]-E[X]^2=1-0=\boxed{1}$.