Upper Bound of the Variance When a Random Variable is Bounded

Yu — Sun, 02 Feb 2020 20:21:50 +0000

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 $V(X)$.
\[V(X) \leq \frac{c^2}{4}.\]

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
\[V(X) \leq \frac{c^2}{4}.\]

The post Upper Bound of the Variance When a Random Variable is Bounded first appeared on Problems in Mathematics.

Lower and Upper Bounds of the Probability of the Intersection of Two Events

Yu — Sat, 18 Jan 2020 03:42:21 +0000

Problem 741

Let $A, B$ be events with probabilities $P(A)=2/5$, $P(B)=5/6$, respectively. Find the best lower and upper bound of the probability $P(A \cap B)$ of the intersection $A \cap B$. Namely, find real numbers $a, b$ such that
\[a \leq P(A \cap B) \leq b\] and $P(A \cap B)$ could take any values between $a$ and $b$.

Solution.

For the lower bound, we use the basic equality of the probability theory that
\[P(A \cap B) = P(A) + P(B) – P(A \cup B).\] Note also that we always have the inequality $P(A \cup B) \leq 1$. Altogether we obtain the following inequality.
\begin{align*}
P(A \cap B) &= P(A) + P(B) – P(A \cup B)\\
&\geq P(A) + P(B) – 1\\[5pt] &= \frac{2}{5} + \frac{5}{6} – 1\\[5pt] &= \frac{12 + 25 -30}{30}\\[5pt] &= \frac{7}{30}.
\end{align*}

This gives the lower bound $a = 7/30$. Note that $P(A \cap B)$ could take this lower bound when $P(A \cup B) = 1$ and this happens if $A\cup B$ is the whole sample space.

Next, we’ll obtain the upper bound. As the intersection $A \cap B$ is contained in the set $A$ and in the set $B$, we have
\begin{align*}
P(A \cap B) &\leq \min(P(A), P(B))\\
&= \min \left(\frac{2}{5}, \frac{5}{6}\right)\\
& = \frac{2}{5}.
\end{align*}

This yields the upper bound $b = 2/5$. The probability $P(A \cap B)$ could take this upper bound when $A \cap B = A$ (this happens when $A \subset B$).

In conclusion, we obtain the following bounds
\[\frac{7}{30} \leq P(A \cap B) \leq \frac{2}{5}.\]

We remark that as a probability we clearly have bounds $0 \leq P(A \cap B) \leq 1$. However, these bounds are not optimal in a sense that $P(A \cap B)$ never takes values less than $7/30$ or above $2/5$ as we determined above.