Once a row of a joint table is known to have happened, the other variable has a new distribution: the conditional probability idea, applied to every cell of that row. Comparing those conditional distributions is also how independence of two random variables is tested.
Key idea
$$P(X = x \mid Y = y) = \frac{p(x, y)}{p_Y(y)}:$$ divide the row $Y = y$ by its total. $X$ and $Y$ are independent when $p(x, y) = p_X(x)\,p_Y(y)$ for every cell; equivalently, every row's conditional distribution is the same.
Why it works
Given $Y = y$, the sample space shrinks to that row, and dividing by the row total makes the new probabilities add to 1. If every cell factors as $p_X(x)p_Y(y)$, then dividing row $y$ by $p_Y(y)$ leaves $p_X(x)$: knowing $Y$ changes nothing about $X$. If even one cell fails to factor, the conditional distribution in that row differs from the marginal, so $Y$ carries information about $X$.
| $X = 0$ | $X = 1$ | $X = 2$ | $P(Y = y)$ | |
|---|---|---|---|---|
| $Y = 0$ | $\frac{1}{6}$ | $\frac{1}{6}$ | $0$ | $\frac{1}{3}$ |
| $Y = 1$ | $\frac{1}{12}$ | $\frac{1}{4}$ | $\frac{1}{3}$ | $\frac{2}{3}$ |
| $P(X = x)$ | $\frac{1}{4}$ | $\frac{5}{12}$ | $\frac{1}{3}$ | $1$ |
Worked examples
Worked example: A conditional distribution
From the table, find the distribution of $X$ given $Y = 1$.
- Row $Y = 1$: $\frac{1}{12}, \frac{1}{4}, \frac{1}{3}$, with total $\frac{2}{3}$.
- Divide: $P(X = 0 \mid Y = 1) = \frac{1}{8}$, $P(X = 1 \mid Y = 1) = \frac{3}{8}$, $P(X = 2 \mid Y = 1) = \frac{1}{2}$.
- They add to 1. Given $Y = 0$ instead, the distribution is $\frac{1}{2}, \frac{1}{2}, 0$: very different, so $X$ and $Y$ are dependent.
Answer: $\left(\frac{1}{8}, \frac{3}{8}, \frac{1}{2}\right)$
Worked example: One matching cell is not enough
A joint table has rows $Y = 0$: $\frac{3}{16}, \frac{3}{16}, \frac{1}{8}$ and $Y = 1$: $\frac{1}{16}, \frac{5}{16}, \frac{1}{8}$ (for $X = 0, 1, 2$). Are $X$ and $Y$ independent?
- Marginals: $p_X = \left(\frac{1}{4}, \frac{1}{2}, \frac{1}{4}\right)$ and $p_Y = \left(\frac{1}{2}, \frac{1}{2}\right)$.
- Column $X = 2$ factors: $\frac{1}{8} = \frac{1}{4} \cdot \frac{1}{2}$ in both rows.
- But $p(0, 0) = \frac{3}{16}$ while $p_X(0)p_Y(0) = \frac{1}{8}$.
Answer: No. Two cells factor, but independence needs all six.
Conditional distributions can also be the starting data. If $P(Y = 0) = \frac{1}{4}$ and you know the distribution of $X$ in each row, then each cell is $$p(x, y) = P(X = x \mid Y = y)\,P(Y = y)$$ adding a column gives $P(X = x)$, and dividing a cell by its column total reverses the conditioning, as in Bayes' theorem.
Common mistake
Checking one cell, finding that it factors, and declaring $X$ and $Y$ independent. Independence is a statement about every cell. Check: compare whole conditional distributions, or test each cell against the product of its two marginals.