Joint generating function

(This problem is from A First Course in Stochastic Processes by Karlin and Taylor.) Let {X} and {Y} be jointly distributed nonnegative integer-valued random variables. For {|s|<1}, {|t|<1} define the joint generating function

\displaystyle  P_{X,Y}(s,t) = \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^it^j

and the marginal generating functions

\displaystyle  \begin{aligned} P_X(s)&=\sum_{i=0}^\infty \mathbb P(X=i)s^i\\ P_Y(t)&=\sum_{j=0}^\infty \mathbb P(Y=j)t^j. \end{aligned}

Theorem 1 {X} and {Y} are independent if and only if the joint generating function is equal to the product of the marginal generating functions.

Proof: For the “only if” implication, we have

\displaystyle  \begin{aligned} P_{X,Y}(s,t) &= \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^it^j\\ &= \sum_{i,j=0}^\infty \mathbb P(X=i)\mathbb P(Y=j)s^it^j\\ &= \left(\sum_{i=0}^\infty \mathbb P(X=i)s^i \right)\left(\sum_{j=0}^\infty \mathbb P(Y=j)t^j \right)\\ &= P_X(s)P_Y(t). \end{aligned}

For the “if” implication, from

\displaystyle  \sum_{i,j=0}^\infty \mathbb P(X=i,Y=j)s^it^j = \sum_{i,j=0}^\infty \mathbb P(X=i)\mathbb P(Y=j)s^it^j

for all {s} and {t} we see that {\mathbb P(X=i,Y=j)=\mathbb P(X=i)\mathbb P(Y=j)} for all {i} and {j}, and hence {X} and {Y} are independent. \Box

Theorem 2 Independence is sufficient, but not necessary for the generating function of {X+Y} to be the product of the marginal generating functions of {X} and {Y}.

Proof: Let {X} and {Y} each be uniformly distributed over {\{0,1,2\}}, with joint distribution

\displaystyle  \mathbb P(X=i,Y=j) = \begin{cases} \frac19,& (i,j)\in\{(0,0), (1,1), (2,2)\}\\ \frac29,& (i,j)\in\{(0,2), (1,0), (2,1)\}\\ 0,&\text{otherwise} \end{cases}

Then

\displaystyle \phi_X(s)\phi_Y(s)=\frac13(1+s+s^2)\frac13(1+s+s^2) = \frac1{9}(1+s+s^2)^2

and

\displaystyle  \begin{aligned} \phi_{X,Y}(s,s) &= \frac19 + \frac29 s + \frac39 s^2 + \frac29 s^3 +\frac19 s^4\\ &= \frac19(1 + 2s + 3s^2 + 2s^3 + s^4)\\ &= \frac19(1+s+s^2)^2\\ &= \phi_X(s)\phi_Y(s). \end{aligned}

But {X} and {Y} are not independent, as for example

\displaystyle \mathbb P(X=0)\mathbb P(Y=1)=\frac19\ne0=\mathbb P(X=0,Y=1).

\Box

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