## 4 AM Math Post Anyone?

Suppose two random variables $X$ and $Y$ have a discrete joint distribution $p(x,y) = \frac{{C }}{{x!(y - x)!}}$, for some real constant C. Suppose we want to find the moment-generating function for this distribution when $y = 0,1,2,...$ and $x = 0,1,2,...,y$.

By definition, the MGF of this distribution is $M(t_1 ,t_2 ) = E(e^{t_1 x} e^{t_2 y})$, where the function $E$ represents the expected value. Since its a discrete distribution, we write $M$ as:

$C \cdot \sum\limits_{y = 0}^\infty {\sum\limits_{x = 0}^y {\frac{{e^{t_1 x} e^{t_2 y}}}{{x!(y - x)!}}} }$ $= C \cdot \sum\limits_{y = 0}^\infty {\frac{1}{{y!}}\sum\limits_{x = 0}^y {\frac{{y!e^{t_1 x} e^{t_2 y} }}{{x!(y - x)!}}} }$

$= C \cdot \sum\limits_{y = 0}^\infty {\frac{1}{{y!}}\sum\limits_{x = 0}^y {\left( {\begin{array}{*{20}c} y \\ x \\ \end{array} } \right)e^{t_1 x} e^{t_2 y} } }$ … by the definition of the binomial coefficient.

Multiply through by $\frac{{e^{t_2 x} }}{{e^{t_2 x} }}$ and rearrange terms to get:

$C \cdot \sum\limits_{y = 0}^\infty {\frac{1}{{y!}}\sum\limits_{x = 0}^y {\left( {\begin{array}{*{20}c} y \\ x \\ \end{array} } \right)\left( {e^{t_1 + t_2 } } \right)^x \left( {e^{t_2 } } \right)^{y - x}}}$

The inner summation reduces nicely by the binomial formula and we get:

$C \cdot \sum\limits_{y = 0}^\infty {\frac{{\left( {e^{t_1 }e^{t_2 } + e^{t_2}}\right)^y}}{{y!}}}$

By the Taylor series expansion for the exponential function, we can rewrite this as:

$C \cdot \exp (e^{t_1}e^{t_2} + e^{t_2 } )$

Isn’t multiplying by one such an awesome little trick?