Upper Bound of the Variance When a Random Variable is Bounded
Problem 753
Let $c$ be a fixed positive number. Let $X$ be a random variable that takes values only between $0$ and $c$. This implies the probability $P(0 \leq X \leq c) = 1$. Then prove the next inequality about the variance $V(X)$.
\[V(X) \leq \frac{c^2}{4}.\]
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Proof.
Recall that the variance $V(X)$ of a random variable $X$ can be computed using expected values as
\[V(X) = E[X^2] – \left(E[X]\right)^2.\]
We try to find the upper bound $c^2/4$ of the right-hand side.
As we know $0\leq X \leq c$, we get
\begin{align*}
E[X^2] &= E[XX]\\
&\leq E[cX]\\
&= cE[X],
\end{align*}
where the last step follows since $c$ is a constant and by the linearity of the expected values.
It follows that
\begin{align*}
V(X) &= E[X^2] – \left(E[X]\right)^2\\
& \leq cE[X] – \left(E[X]\right)^2.
\end{align*}
For the sake of simplicity, let us put $z = E[X]$. Then the last expression is a quadratic function
\[-z^2 + cz\]
with variable $z$. Note that the graph of the equation
\[-z^2 + cz = -z(z-c)\]
is a parabola that opens downward. This attains the maximal value at its vertex, whose $z$-coordinate is the middle point of the $z$-intercept $z=0, c$.
Thus, the maximal value is obtained when $z = \frac{0 + c}{2} = \frac{c}{2}$ and it is
\begin{align*}
-\left(\frac{c}{2}\right)^2 + c \left(\frac{c}{2}\right) = \frac{c^2}{4}.
\end{align*}
Therefore, we have obtained the desired inequality
\[V(X) \leq \frac{c^2}{4}.\]
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