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
\[V(X) \leq \frac{c^2}{4}.\] Click here if solved 3

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

Find a quadratic function $f(x) = ax^2 + bx + c$ such that $f(1) = 3$, $f'(1) = 3$, and $f^{\prime\prime}(1) = 2$.

Here, $f'(x)$ and $f^{\prime\prime}(x)$ denote the first and second derivatives, respectively.

Solution.

Each condition required on $f$ can be turned into an equation involving the constants $a, b, c$.

In particular, $f(1) = 3$ tells us that $a + b + c = 3$.

Because $f'(x) = 2ax + b$, the condition $f'(1) = 3$ gives us $2a + b = 3$.

And finally $f^{\prime\prime}(x) = 2a$, and so $f^{\prime\prime}(1) = 2a = 2$. Thus we have the system of equations
\begin{align*}
a + b + c &= 3 \\
2a + b &= 3\\
2a &= 2
\end{align*}

To solve this system, we could create the augmented matrix and then reduce it.

For this system, though, it is simpler to solve directly. The third equation tells us that $a=1$.

Plugging this value into the second equation, we find $b=1$.

Plugging both of these values into the first equation, we see $c=1$ as well.

Thus the function we want is $f(x) = x^2 + x + 1$.

The post Find a Quadratic Function Satisfying Conditions on Derivatives first appeared on Problems in Mathematics.