# Discrete random variable, joint density question

Discrete random variables $X$ and $Y$ have the joint density shown in the table below.

$\ \ \ \ X \ 1 \ \ \ 2$

$Y$

$1 \ \ \ \ \ \ \frac{1}{10} \ \ \ \frac{2}{10}$

$2 \ \ \ \ \ \ \frac{3}{10} \ \ \ \frac{4}{10}$

What is the conditional density of $Y$ given $X = 1$?

What is the procedure for doing this problem? I have an exam tomorrow and we never covered this in the lectures. I have just found it on an old exam paper which means it could potentially be on tomorrows exam.

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You are being asked for conditional probabilities: If event $A$ is $\{Y = 1\}$ and thus event $A^c$ is $\{Y = 0\}$, what are the _conditional probabilities $P(A\mid B)$ and $P(A^c\mid B)$ where $B = \{X = 1\}$. So, use the _definition of conditional probability to work out the answers. Can you find $P(A \cap B)$ from the given information? How about $P(B)$? – Dilip Sarwate Dec 2 '12 at 22:37

Given $X=1$ we are looking at the first column. The "ratio"in that column is $1$ to $3$.

More formally, we want $\Pr(Y=1|X=1)$ and $\Pr(Y=2|X=1)$. Note that $\Pr (X=1)=\frac{1}{10}+\frac{3}{10}=\frac{4}{10}$.

So $$\Pr(Y=1|X=1)=\frac{\Pr(X=1 \cap Y=1)}{\Pr(X=1)}=\frac{1/10}{4/10}=\frac{1}{4}.$$ By a similar argument, $\Pr(Y=2|X=1)=\dfrac{3}{4}$.

Basically, we confine attention to the column $Y=1$, and scale up the probabilities in that column so they add to $1$.

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The general procedure can be explained as follows: you take the column (or row) that corresponds to the known value of the variable you are conditioning on. In the presented case it's the first column. Then you get probabilties for $Y$ and in order to obtain a distribution - you should normalize them. In you example, you should divide both $\frac{1}{10}$ and $\frac{3}{10}$ by their sum.

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