If the joint probability distribution of X and Y is given by
$$f(x,y)= \frac{(x-y)^2}{7}, \text{for }x=1,2,3;y=1,2 $$
$(1)$ Find the probability distribution of $U = X + Y; $
$(2)$ Find the conditional probability distribution of $X$ given $U =4.$
In order to solve this problem one must draw a chart.
$$\begin{array}{|c|c|c|c|}
\hline
(x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline
f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline
U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline
x & & &\\ \hline
\end {array}$$
How does one fill up the rest of the table and answer questions one and two.
EDIT
In order to find solve $(1)$ one must add all the related $f(x,y)$ relations. Thus
$(1)$ $$ \quad P(U=2) =0, \\ P(U=3) = \frac{1}{7} + \frac{1}{7} = \frac27, \\ P(U=4)= 0+\frac{4}{7} = \frac47, \\ P(U=5)=\frac{1}{7}$$
One must use this notation to solve.
$$P(x=1|U=4)= \frac{P(x=1,U=4)}{P(U=4)} = ? \\P(x=2|U=4) = ? \\ P(x=3|U=4) = ? $$
Knowing this does anyone know how to solve $(2)$ using this notation? What does one substitute for this question to derive the answer?
Best Answer
Ok so the first thing you notice is that so far your attempt has $$ \begin{align} P(U = 2) &= 0, \\ P(U = 3) &= \frac{1}{7}, \\ P(U = 4) &= \frac{4}{7}, \\ P(U = 5) &= \frac{1}{7} \end{align} $$ and zero elsewhere, but summing over all possible situations only takes us to $\frac{6}{7}$ so something has clearly gone wrong! So what you have missed is that $$ P(U=3) = P(x=1,y=2) + P(x=2,y=1) = \frac{2}{7}. $$ For the second part of your question look at your table and study the different combinations of $x,y$ that will make $U=4$ and then look at the joint probability of these combinations, and you should see clearly what the distribution of $x$ must be.