# Difference between revisions of "Eleisha's Segment 9: Characteristic Functions"

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?

The characteristic function of the sum of independent random variables is the product of their characteristic functions.

$\displaystyle \text{The characteristic function of S is:} \phi_S(t)$

$\displaystyle X \sim Normal(\mu_1, \sigma_1)$

$\displaystyle Y \sim Normal(\mu_2, \sigma_2)$

So let S = X + Y (the sum of two independent Gaussian random variables)

$\displaystyle \phi_S(t) = \phi_X(t) \phi_Y(t) \text{ since X and Y are independent}$

$\displaystyle \phi_normal(t) = e^{i \mu t - \frac{1}{2} \sigma^2 t^2 }$

$\displaystyle \phi_S(t) = e^{i \mu_1 t - \frac{1}{2} \sigma_1^2 t^2 } \cdot e^{i \mu_2 t - \frac{1}{2} \sigma_2^2 t^2 }$

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

$\displaystyle \phi_X(t) = \int_0^\infty e^{itx} \cdot \lambda e^{\lambda x} dx$

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

$\displaystyle \phi_X(t) = \lambda\lim_{u \to \infty} \left[ \int_0^u e^{-(\lambda - it)x} dx \right]$

$\displaystyle \phi_X(t) = \lambda \lim_{u \to \infty} \left[ \frac{-1}{\lambda - it}\right] e^{-(\lambda - it)x}\Big|_0^u$

$\displaystyle \phi_X(t) = \lambda \left[ \lim_{u \to \infty} e^{-(\lambda - it)u} + \frac{1}{\lambda - it} \right]$

$\displaystyle \phi_X(t) = \lambda \left[ \frac{1}{\lambda - it}\right]$

$\displaystyle \phi_X(t) = \frac{\lambda}{\lambda - it}$