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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?

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  • $\begingroup$ There probably is no simpler way to do this exercise than to fill out the table for "probability mass function" for the joint distribution of $(X,Y)$. You seem to be missing the idea that the last row is just the value $x$ achieved by the random variable $X$ (shown already in the top row). $\endgroup$
    – hardmath
    Commented Dec 4, 2016 at 23:42
  • $\begingroup$ Could you show me how you got $(1,1)$ for $x$ then I can most likely do the rest. $\endgroup$
    – Jon
    Commented Dec 5, 2016 at 0:45
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    $\begingroup$ Sure, I will post a completion of your chart with an explanation of the last row. $\endgroup$
    – hardmath
    Commented Dec 5, 2016 at 0:53

2 Answers 2

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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.

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Here is a completed table of the discrete probabilities:

$$ \begin{array}{|c|c|c|c|} \hline (x,y)& (1,1) & (1,2) & (2,1) & (2,2) & (3,1) & (3,2) \\ \hline f(x,y)& 0 & \frac{1}{7} & \frac{1}{7} & 0 & \frac{4}{7} & \frac{1}{7}\\ \hline U=x+y& 2 & 3 &3 & 4 & 4 & 5\\ \hline x & 1 & 1 & 2 & 2 & 3 & 3 \\ \hline \end {array} $$

The Reader will recognize that the last row's entries are simply the $x$ values of the coordinate pairs in the first row. The intent of forming such a row is likely to help answer the second part of the exercise given in the Question, what is the conditional probability distribution of $X$, given that the sum $U = 4$?

The table makes it clear that while the events $X=2,Y=2$ and $X=3,Y=1$ both lead to $U=4$, the former of these events is assigned a zero probability (and the latter is assigned a positive probability). Therefore when the conditional probabilities are assigned, conditioned on $U=4$, we have:

$$ \mathbf{Pr}(X=2 \mid U=4) = 0 \; ; \; \mathbf{Pr}(X=3 \mid U=4) = 1 $$

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