# Segment 9 (6 February 2013)

## Comment on Segment

With sufficient assumptions, Mathematica is able to compute the characteristic function of the Cauchy distribution:

In[27]:= $Assumptions = (*$Assumptions &&*) (sig > 0) && (mu \[Element] Reals)

Out[27]= sig > 0 && mu \[Element] Reals

In[28]:= p = 1/(Pi sig) (1 + ((x - mu)/sig)^2)^(-1)

Out[28]= 1/(\[Pi] sig (1 + (-mu + x)^2/sig^2))

In[29]:= Integrate[p, {x, -Infinity, Infinity}]

Out[29]= 1

In[30]:= Integrate[p Exp[I t x], {x, -Infinity, Infinity}]

Out[30]= ConditionalExpression[(
E^(I mu t) \[Pi] Cosh[sig t] Sign[
t] - \[Pi] (Cos[mu t] + I Sign[t]^2 Sin[mu t]) Sinh[
sig t])/(\[Pi] Sign[t]), t \[Element] Reals]


-- it is, however, much less pretty.

Maple can do it too:

> assume(sigma > 0, mu::real, t::real);
> p := 1/(Pi*sigma)*(1 + ((x - mu)/sigma)^2)^(-1);
> int(p,x=-infinity..infinity);
1
> int(p*exp(I*t*x),x=-infinity..infinity);
1
- exp(-sigma t + I t mu) signum(t) (signum(t) - exp(2 sigma t)
2

+ exp(2 sigma t) signum(t) + 1)


although if the assumption t::real is ommitted, Maple seems to think the integral is zero, which does not inspire confidence in the answer. (This seems to be true only of Maple 13; Maple 16 simply fails to do the integral without the additional assumption.)

## Computation Problems

1. $\phi_\mathrm{Normal}(t)=e^{i\mu t-\sigma^2t^2/2}$, so if $\phi_X(t)=e^{i\mu_X t-\sigma_X^2t^2/2}$ and $\phi_Y(t)=e^{i\mu_Y t-\sigma_Y^2t^2/2}$, then $\phi_X(t)\phi_Y(t)=e^{i\mu_X t-\sigma_X^2t^2/2+i\mu_Y t-\sigma_Y^2t^2/2}=e^{i(\mu_X+\mu_Y)t-(\sigma_X^2+\sigma_Y^2)t^2/2}$

so the sum of normal variables is normal, with mean $\mu_X+\mu_Y$ and variance $\sigma_X^2+\sigma_Y^2$.

2. $p(x)=\beta e^{-\beta x}$, so $\phi(t)=\int_0^\infty e^{itx}\beta e^{-beta x}\,dx=\beta\int_0^\infty e^{(it-\beta)x}dx$

In[62]:= b*Integrate[Exp[(I t - b) x], {x, 0, Infinity}]
Out[62]= ConditionalExpression[b/(b - I t), Im[t] + Re[b] > 0]


$=\frac{\beta}{\beta-it}$ with $\Im(t)+\Re(b)>0$.

## In-class Problems

Team 2, with User:Kai, User:Tameem, Keerthana Kumar's Solution, User:Jzhang, User:Trettels. Won small diamond.