# Segment 9 Sanmit Narvekar

## Segment 9

#### To Calculate

1. Use characteristic functions to show that the sum of two independent Gaussian random variables is itself a Gaussian random variable. What is its mean and variance?

From the slides, we know that the characteristic function of the sum of two independent random variables S = X + Y with distributions p_X(t) and p_Y(t) respectively is:

$\displaystyle \phi_S(t) = \phi_X(t) \phi_Y(t)$

We also know the characteristic function of a Gaussian random variable is:

$\displaystyle \phi_X(t) = e^{i \mu_X t - \frac{1}{2} \sigma_X^2t^2}$

So we plug in and simplify:

$\displaystyle \phi_{X+Y}(t)$

$\displaystyle = \phi_X(t) \phi_Y(t)$

$\displaystyle = e^{i \mu_X t - \frac{1}{2} \sigma_X^2t^2} e^{i \mu_Y t - \frac{1}{2} \sigma_Y^2t^2}$

$\displaystyle = e^{i t ( \mu_X + \mu_Y) - \frac{1}{2} t^2 (\sigma_X^2 + \sigma_Y^2)}$

By comparing the form of this equation to the characteristic function of a Gaussian random variable, its easy to see the result is another Gaussian random variable with mean $\displaystyle \mu_X + \mu_Y$ and variance $\displaystyle \sigma^2_X + \sigma^2_Y$ .

2. Calculate (don't just look up) the characteristic function of the Exponential distribution.

Since the exponential distribution's domain is the set of non-negative reals, we integrate from 0 to infinity:

$\displaystyle \int_0^\infty e^{itx} \beta e^{-\beta x} dx$

$\displaystyle = \beta \int_0^\infty e^{x(it - \beta)} dx$

$\displaystyle = \frac{\beta}{it - \beta} e^{x(it - \beta)} \Big|_0^\infty$

$\displaystyle = - \frac{\beta}{it - \beta}$