I am tossing a fair coin 6 times. I define $X$ to be the number of heads in first 3 tosses and $Y$ to be the number of heads in all 6 tosses. $X$ and $Y$ are random variables.
I want to create the joint and marginal distribution tables for $p(X,Y)$ and $p(X)$ and $p(Y)$. For creating the joint distribution table, I have the following (drawing from this post):
$X = B(3, 0.5)$ and $Y = B(6, 0.5)$
Then are the table columns and rows something like this $P(X = i, Y=j)$ with all the values filled in?
\begin{array}{|c|c|c|c|}
\hline
& i=1 & i=2 & i=3 \\ \hline
j=1& & &\\ \hline
j=2& & &\\ \hline
j=3& & &\\ \hline
j=4& & &\\ \hline
j=5& & &\\ \hline
j=6& & &\\ \hline
\end{array}
For example, if I have $P(X = 2, Y= 5)$, I'm interpreting the statement as probability of getting at "exactly 2 heads in the first three tosses and exactly 5 heads from the 6 tosses." So then I calculated that $P(X=2) = 0.375$ and $P(X=6) = 0.09375$. So would I multiply these two values and enter them as such:
\begin{array}{|c|c|c|c|}
\hline
& i=1 & i=2 & i=3 \\ \hline
j=1& & &\\ \hline
j=2& & &\\ \hline
j=3& & &\\ \hline
j=4& & &\\ \hline
j=5& & 0.375\times 0.09375 = 0.035 &\\ \hline
j=6& & &\\ \hline
\end{array}
I'm not sure if this the proper way to represent this and if assumptions are correct or if I'm erring in my concept.
Best Answer
First point to note is that $X$ and $Y$ can be $0$ with positive probability, so you might want to extend your table
Second point to note is that $Y \ge X$, so $P(Y < X)=0$
Third point to note is that $Y$ is not independent of $X$, and the higher $X$ is, the higher $Y$ can be. But if $Z$ is the number of heads in the last three tosses then you can say
(a) $Z \sim B(3,0.5)$,
(b) $X$ and $Z$ are independent,
(c) $Y=X+Z$ so $Z=Y-X$, and
(d) $P(X=i,Y=j) =P(X=i, Z=j-i)= P(X=i) \times P(Z=j-i)$ so for example $P(X=2,Y=5)=P(X=2)\times P(Z=3)$
So the start of your table could be
\begin{array}{|c|c|c|c|c|} \hline & i=0 & i=1 & i=2 & i=3 \\ \hline j=0& & 0& 0&0\\ \hline j=1& & & 0&0\\ \hline j=2& & & &0\\ \hline j=3& & & &\\ \hline j=4& & & &\\ \hline j=5& & & 0.375\times 0.125 = 0.046875 &\\ \hline j=6& & & &\\ \hline \end{array}