Communication, Networking, Signal and Image Processing (CS)

Question 1: Probability and Random Processes

January 2002

5. (20 pts)

Let $\mathbf{X}$ be a random variable with absolutely continuous probability distribution function. Show that for any $\alpha>0$ and any real number $s$ :$P\left(e^{s\mathbf{X}}\geq\alpha\right)\leq\frac{\phi\left(s\right)}{\alpha}$ where $\phi\left(s\right)$ is the moment generating function, $\phi\left(s\right)=E\left[e^{s\mathbf{X}}\right]$ . Note: $\phi\left(s\right)$ can be related to the Laplace Transform of $f_{\mathbf{X}}\left(x\right)$ .

## Solution 1

Note

This is similar to the proof of Chebyshev Inequality.

$g_{1}\left(x\right)=1_{\left(x\right)_{\left\{ r:e^{sx}\geq\alpha\right\} }},\; g_{2}\left(x\right)=\frac{e^{sx}}{\alpha}.$

$E\left[g_{2}\left(\mathbf{X}\right)-g_{1}\left(\mathbf{X}\right)\right]=E\left[g_{2}\left(\mathbf{X}\right)\right]-E\left[g_{1}\left(\mathbf{X}\right)\right]=\frac{\phi\left(s\right)}{\alpha}-P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\geq0.$

$\therefore\; P\left(\left\{ e^{s\mathbf{X}}\geq\alpha\right\} \right)\leq\frac{\phi\left(s\right)}{\alpha}.$