Q. 8.6 - A First Course in Probability

Question

$\left(a\right)$Let $X$be a discrete random variable whose possible values are$1, 2, ...$. If $P[X=k]$is nonincreasing$k = 1, 2, ...$, prove that

$P(X=k)\le 2\frac{E\left[X\right]}{k^2}$

$\left(b\right)$Let $X$be a non-negative continuous random variable having a nonincreasing density function. Show that$f\left(x\right)\le 2\frac{E\left[X\right]}{x^2}$ for all$x>0$.

Step-by-Step Solution

Answer

We get,

$$\begin{gathered} \left(a\right)P(X=k)\le 2\frac{E\left[X\right]}{k^2} \\ \left(b\right)f\left(x\right)\le 2\frac{E\left[X\right]}{x^2} \end{gathered}$$

Step 1 Given Information.

Let $X$be a discrete random variable whose possible values are$1, 2, ...$ and $P[X=k]$are nonincreasing in$k = 1, 2, ...$.

Step 2 Part (a) Explanation.

Let $\mathrm{X}$be a discrete RV whose possible values are $1,2,\dots$If $P(X=k)$is nonincreasing in $k=1,2, . .$we want to show

$P(X=k)\le 2\frac{E\left[X\right]}{k^2}$

We will start with the definition of expectation of discrete RV and draw an inequality from there.

$$\begin{gathered} E\left[X\right]=\sum _1^{\infty }xP(X=x) \\ \ge \sum _1^{k}xP(X=x) \end{gathered}$$

And since $P(X=k)$is nonincreasing$k$, it follows$P(X=i)\ge P(X=k)$ for$i\le k$.

$$\begin{gathered} \Rightarrow E\left[X\right]\ge \sum _1^{k}xP(X=k) \\ \ge P(X=k)\sum _1^{k}x \\ \ge P(X=k)\frac{k(k+1)}{2} \\ \ge P(X=k)\frac{k^2}{2} \end{gathered}$$

By rearranging this inequality we get the desired result.

Step 3 Part (b)Explanation.

The continuous case is very similar. To avoid confusion we will write the density as$f_X\left(t\right)$.

$$\begin{gathered} \Rightarrow E\left[X\right]={\int }_0^{\infty }t{f}_X\left(t\right)dt \\ \ge {\int }_0^{x}t{f}_X\left(t\right)dt \end{gathered}$$

We know the density $f_X\left(t\right)$is non-increasing, so $f_X\left(t\right)\ge f_X\left(x\right)$for$t\le x$.

$$\begin{gathered} \Rightarrow E\left[X\right]\ge {\int }_0^{x}t{f}_X\left(x\right)dt \\ \ge f_X\left(x\right){\int }_0^{x}tdt \\ \ge {\left.f_X\left(x\right)\frac{t^2}{2}\right|}_0^x \\ \ge f_X\left(x\right)\frac{x^2}{2} \end{gathered}$$

By rearranging this inequality we get the desired result.