# Segment 9

## Calculation Problems

1. Show that the sum of two independent Gaussian random variables is itself a Gaussian random variable. What is the mean and variance?

Let ${\displaystyle X}$ and ${\displaystyle Y}$ be two independent Gaussian random variables. with mean ${\displaystyle \mu _{X}}$and ${\displaystyle \mu _{Y}}$ and variance ${\displaystyle \sigma _{X}^{2}}$ and ${\displaystyle \sigma _{Y}^{2}}$
${\displaystyle \Phi _{X}(t)=exp\left(it\mu _{X}-{\frac {\sigma _{X}^{2}t^{2}}{2}}\right)}$

Similarly for ${\displaystyle \Phi _{Y}(t)}$.

Since the sum of two independent random variable X and Y is the product of the two seperate characteristic function.

{\displaystyle {\begin{aligned}\Phi _{X+Y}(t)&=\Phi _{X}(t)\Phi _{Y}(t)\\&=exp\left(it\mu _{X}-{\frac {\sigma _{X}^{2}t^{2}}{2}}\right)exp\left(it\mu _{Y}-{\frac {\sigma _{Y}^{2}t^{2}}{2}}\right)\\&=exp\left(it(\mu _{X}+\mu _{Y})-{\frac {(\sigma _{X}^{2}+\sigma _{Y}^{2})t^{2}}{2}}\right)\\\Phi _{Z}(t)&=exp\left(it\mu _{Z}-{\frac {\sigma _{Z}^{2}t^{2}}{2}}\right)\end{aligned}}}

The new random variable ${\displaystyle Z=X+Y}$ has mean ${\displaystyle \mu _{X}+\mu _{Y}}$ and variance ${\displaystyle \sigma _{X}^{2}+\sigma _{Y}^{2}}$.

2. Calculate the characteristic function of the Exponential distribution.

{\displaystyle {\begin{aligned}p(x)&=\beta e^{-\beta x}&x\geq 0\\\Phi (t)&=\int _{0}^{\infty }e^{itx}p(x)\\&=\int _{0}^{\infty }e&{itx}\beta e^{-\beta x}\\&=\left|{\frac {\beta }{it-\beta }}e^{(it-\beta )x}\right|_{0}^{\infty }\\&={\frac {\beta }{\beta -it}}\end{aligned}}}

## Class Activity

Group : Noah, Kai, Tameen, Jin, Trettels