and I'm still messed up on this problem XD
http://math.stackexchange.com/questions/1055016/inverse-fourier-transform-for-the-heat-equation

$E[ace]=13 \frac{1}{13}=1$ and $E[spade]=13\frac{1}{4}=\frac{13}{4}$

still working out $E[XY]$

Show that if $f(x)$ is absolutely integrable on $(-\infty, \infty)$ then \
1b. $[f(\frac{x+b}{a})]^{\xi}=ae^{ib \xi}\bar {f}(a \xi),a,b$ real, $a \neq 0$\\
@anon

1. Show that if $f(x)$ is absolutely integrable on $(-\infty, \infty)$ then \\


1b. $[f(\frac{x+b}{a})]^{\xi}=ae^{ib \xi}\bar {f}(a \xi),a,b$ real, $a \neq 0$\\

A real of complex-valued function defined on $(-\infty, \infty)$ is said to be absolutely integrable on $(-\infty, \infty)$ if $\int_{-R}^{R} \mid f(x) \mid dx$ exists for all $R>0$ and\\

$\int_{-\infty}^{\infty} \mid f(x) \mid \equiv $$\lim_{R \to\infty} $$ \int_{-R}^{R} \mid f(x) \mid dx < \infty$\\

$E[XY]=\frac{13}{52}$ chances that you get the ace of spades at all, plus the chance you get an ace or a spade at all, but not both cause you did that first, so the ace is already done and that spade is gone as well: $\frac{13*12}{52}$
so we get $E[XY]=\frac{169}{52}=\frac{13}{4}$ and $E[X]E[Y]=1*\frac{13}{4}$ so we $\sigma(X,Y)=0$?? woops typo

using

Fourier Transform

$f(\frac{x+b}{a})e^{-i \xi x}$

Fourier Transform

$\int^{-\infty}_{\infty}f(x)e^{i \xi x} \frac{1}{\sqrt{2 \pi}}$

$\frac{1}{\sqrt{2 \pi}} \int^{-\infty}_{\infty}f(\frac{x+b}{a})e^{i \xi x} dx$
Let u = x+b

$\frac{1}{\sqrt{2 \pi}} \int^{-\infty}_{\infty}f(\frac{u}{a})e^{i \xi x} dx$.

Now if we let $ v = u/a$ then .

$\frac{1}{\sqrt{2 \pi}} \int^{-\infty}_{\infty}f(v)e^{i \xi x} dx$.