# Segment 9 Sanmit Narvekar

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## 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:

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

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

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

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

#### To Think About

1. Learn enough about contour integration to be able to make sense of Saul's explanation at the bottom of slide 7. Then draw a picture of the contours, label the pole(s), and show how you calculate their residues.

2. Do you think that characteristic functions are ever useful computationally (that is, not just analytically to prove theorems)?

How would you do numerical procedures with complex numbers...?