Problem source

A random variable $X$ is distributed uniformly on $[\, 0\, , \, a\,]$. 
Show that the variance of $X$ is ${1 \over 12} a^2$.
A sample, $X_1$ and $X_2$, of two independent values of the random variable is drawn, and the variance $V$ of the sample is determined. Show that $V = {1 \over 4} \l X_1 -X_2 \r ^2$, and hence prove that $2 V$ is an unbiased estimator of the variance of X.
Find an exact expression for the probability that the value of $V$ is less than ${1 \over 12} a^2$ and estimate the value of this probability correct to one significant figure.

Solution source

\begin{align*}
&& \E[X] &= \frac{a}{2}\tag{by symmetry} \\
&&\E[X^2] &= \int_0^a \frac{1}{a} x^2 \d x \\
&&&= \frac{a^3}{3a} = \frac{a^2}{3} \\
\Rightarrow && \var[X] &= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12} \\
\end{align*}

\begin{align*}
&& V &=\frac{1}{2} \left (  \left ( X_1 - \frac{X_1+X_2}{2} \right )^2+\left ( X_2- \frac{X_1+X_2}{2} \right )^2 \right ) \\
&&&= \frac{1}{8} ((X_1 - X_2)^2 + (X_2 - X_1)^2 ) \\
&&&= \frac14 (X_1-X_2)^2 \\
\\
&& \E[2V] &= \E \left [ \frac12 (X_1 - X_2)^2 \right] \\
&&&= \frac12 \E[X_1^2] - \E[X_1X_2] + \frac12 \E[X_2^2] \\
&&&= \frac{a^2}{3} - \frac{a^2}{4} = \frac{a^2}{12}
\end{align*}

Therefore $2V$ is an unbiased estimator of the variance of $X$.

\begin{center}
    \begin{tikzpicture}
    \def\functionf(#1){0.5 * exp(-abs(#1))};
    \def\xl{-4}; 
    \def\xu{4};
    \def\yl{-0.2}; \def\yu{.75};
    
    % Calculate scaling factors to make the plot square
    % \pgfmathsetmacro{\xrange}{\xu-\xl}
    % \pgfmathsetmacro{\yrange}{\yu-\yl}
    % \pgfmathsetmacro{\xscale}{10/\xrange}
    % \pgfmathsetmacro{\yscale}{10/\yrange}
    
    % Define the reusable styles to keep code clean
    \tikzset{
        % x=\xscale cm, y=\yscale cm,
        axis/.style={thick, draw=black!80},
        grid/.style={thin, dashed, gray!30},
        curveA/.style={very thick, color=cyan!70!black, smooth},
        curveB/.style={very thick, color=orange!90!black, smooth},
        dot/.style={circle, fill=black, inner sep=1.2pt},
        labelbox/.style={fill=white, inner sep=2pt, rounded corners=2pt} % Protects text from lines
    }

    \filldraw[color=cyan!20] (0,0) -- (0, {4/sqrt(3)}) -- ({4-4/sqrt(3)},4) -- 
                (4,4) -- (4,{4-4/sqrt(3)}) -- ({4/sqrt(3)},0)  -- cycle; 
    \draw[axis] (0,0) rectangle (4,4);

    \node[below] at (2,0) {$X_1$};
    \node[below] at (4,0) {$a$};
    \node[left] at (0,4) {$a$};
    \node[left] at (0, 2) {$X_2$};

    \draw[curveA] (0, {4/sqrt(3)}) -- ({4-4/sqrt(3)},4);
    \draw[curveA] ({4/sqrt(3)},0) -- (4,{4-4/sqrt(3)});

    % \draw[curveA] (0, {4/sqrt(3)}) -- ({1-4/sqrt(3)},1);
        % Set up axes


    \end{tikzpicture}
\end{center}

We need $|X_1 - X_2| < \frac{a}{\sqrt{3}}$

We are interested in the blue area, which is $a^2 - a^2(1- \frac{1}{\sqrt{3}})^2 = a^2 \left ( \frac{2}{\sqrt{3}} - \frac13 \right)$ ie the probability is $\frac{2\sqrt{3}-1}{3} \approx 0.8$