Originally Posted by

**Robb** 1.

It is clear from the definition of a marginal pmf that they would be the same, given you are summing either from 0 to infinity of i, or j to calculate either marginal.

$\displaystyle P_X (X=x)=\sum_{y=0}^\infty P(X=x,Y=y) = \sum_{x=0}^\infty P(X=x,Y=y) = P_Y (Y=y)$

2.

Letting $\displaystyle S=X+Y$, to calculate $\displaystyle P(S=k)$ we note;

For:

$\displaystyle k=0 \implies X=0,Y=0 \text{ so } P(S=0)=P(X=0,Y=0)=\frac{\alpha}{(1)!}$

Where by the definition of factorials $\displaystyle 0! = 1!$

$\displaystyle k=1 \implies X=0,Y=1;X=1,Y=0 \text{ so } P(S=1)=2 \cdot P(X=1,Y=0)=2\cdot \frac{\alpha}{(1+1+0)!} = \frac{2\alpha}{1\cdot 2}=\frac{\alpha}{1!}

$

Since $\displaystyle P(X=1,Y=0)=P(X=0,Y=1) $

$\displaystyle k=2 \implies X=0,Y=2;X=2,Y=0;X=1,Y=1 \text{ so } P(S=2)= 2 \cdot \frac{\alpha}{(1+2+0)!}+\frac{\alpha}{(1+1+1)!}=\f rac{3\cdot \alpha}{3!}=\frac{\alpha}{2!}$

So iterating for all k, we end up with the given formula.

3.

For any pmf, we know that the sum over all possible values equals 1. So we can solve for $\displaystyle \alpha$

$\displaystyle 1=\sum_{k=0}^\infty \frac{\alpha}{k!}=\alpha \cdot e^1$

Thefore, $\displaystyle \alpha=e^{-1}$, so S is poison distributed with paramater $\displaystyle \lambda=1$

4.

$\displaystyle P(X=0)=\sum_{y=0}^\infty \frac{e^{-1}}{(1+y)!}=e^{-1}\sum_{y=1}^\infty \frac{1}{(y)!}=e^{-1}(e^1-1)=1-e^{-1}

$

By the definition of independence;

$\displaystyle P(X=0,Y=0)=P(X=0)P(Y=0)$

LHS: $\displaystyle P(X=0,Y=0)=P(S=0)=e^{-1}=0.3679$

RHS: $\displaystyle P(X=0)P(Y=0)=(1-e^{-1})^2=0.399 \neq \text{LHS}$ as X and Y have the same distribution.

Hence they are not independent

5.

Since $\displaystyle S \sim \text{Poi}(1)$ its expected value is $\displaystyle \lambda=1$

Using the fact from (a), X and Y have the same distribution and hence same expected value;

$\displaystyle E(S)=1=E(X)+E(Y)=2\cdot E(X)$

Therefore, $\displaystyle E(X)=0.5$