I am looking at part a) and I have found the marginal p.f for $Y$ to be $e^{-2}2^{y}/y!$. I have set up for the equation for the marginal p.f for $X$ but I have no idea how to start it. Help would be appreciated.
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1$\begingroup$ Hint: If $X$ has value $3$, $Y$ must have value $3$ or more, no? No? It is possible that $(X,Y) = (3,2)$? What is the probability $P(X=3, Y=2)$ as per the formula you gave for the joint probability mass function? For part (b), the eyeball test is easiest to apply: it can be used to get the answer even if you are unable to do part (a) at all. $\endgroup$– Dilip SarwateCommented May 31, 2015 at 17:05
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$\begingroup$ Yes that makes sense so its not possible that (X,Y)=(3,2) because y>=x, so if I sum out the possible values of y to get the marginal, my summation is from y=x to infinity. $\endgroup$– user153009Commented May 31, 2015 at 17:27
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$\begingroup$ Okay, using the substitution z=y-x, I got a similar marginal $\endgroup$– user153009Commented May 31, 2015 at 17:46
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$\begingroup$ @user153009 Similar but not identical though, right? See my answer. $\endgroup$– JohnKCommented Jun 1, 2015 at 0:50
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2 Answers
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You can see right away that these random variables are not independent from their support, as @Dilip Sarwate noted. In general if the joint support is not a product space, you can immediately conclude that the RVs are not independent. This is not a necessary condition though, so be careful.
This is a one-liner then but let's derive the marginal densities as well.
\begin{align} f_X (x)=\sum_{y=x}^{\infty} f_{X,Y} (x,y) & =\sum_{y=x}^{\infty} \frac{e^{-2}}{x! \left(y-x\right)!}\\ &= \frac{e^{-2}}{x!} \sum_{y=x}^{\infty} \frac{1}{\left(y-x\right)!} \\&= \frac{e^{-1}}{x!} \quad (\text{why?}) \end{align}
The marginal support of $X$ consists now of all nonnegative integers, i.e. $S_X=\left\{0,1,\ldots\right\}$. For $Y$ we similarly sum over all possible $X$ values.
\begin{align} f_Y (y)=\sum_{x=0}^{y} f_{X,Y} (x,y) &= \sum_{x=0}^{y}\frac{e^{-2}}{x! \left(y-x\right)!} \\ &= \frac{e^{-2}}{y!} \sum_{x=0}^{y} \binom{y}{x} \\ &=\frac{e^{-2}2^y}{y!} \quad (\text{why?}) \end{align}
Likewise, $S_Y=\left\{0,1,\ldots \right\}$. This pmf is very standard, can you recognize its type? From the marginal distributions now, it is obvious that $f_X(x) \times f_Y (y) \neq f_{X,Y} \left(x,y \right)$, as we had initially suspected.
What you should take out of this is a look at the joint support is a quick way to verify independence and that when summing/integrating over random variables to arrive at a marginal density it is very important to have all restrictions in place. As verification of your work, you can always check whether the resulting mass function/density sums/integrates to $1$.
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1$\begingroup$ The joint distribution given is that of $X$ and $Y = X+Z$ where $Z$ is a Poisson random variable with the same parameter $1$ as $X$ but independent of $X$. $\endgroup$ Commented Jun 2, 2015 at 15:45
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When dealing with bivariate discrete distributions it always helps to write joint distribution as a table. In this case we have (I omit the multiplicative constant $e^{-2}$. Also note the empty cells where distribution is not defined)
\begin{array} {|r|r|r|r|r|r|...} \hline &y=0 & y=1 & y=2 & y=3\\ \hline x=0& \frac{1}{0!0!} & \frac{1}{0!1!} & \frac{1}{0!2!} & \frac{1}{0!3!} & ...\\ \hline x=1& & \frac{1}{1!0!} & \frac{1}{1!1!} & \frac{1}{1!2!} & ...\\ \hline x=2& & & \frac{1}{2!0!} & \frac{1}{0!1!} & ...\\ \hline x=3& & & & \frac{1}{3!0!} & ...\\ \hline ... & & & & & ...\\ \hline \end{array}
Then marginal distributions are simply the sums of columns (for $Y$) or rows (for $X$).
Looking at the rows it should not be hard to spot the pattern and get the marginal distribution for $X$. For $Y$ it is a bit harder, since you need to recall the formula for $n \choose m$ and the fact that $\sum_{m=0}^n {n \choose m}=2^n$.
