Conditional Probability Function

Study guide: Exam P-1

A Conditional Probability Function gives the distribution of one random variable given a specific value of another. The conditional distribution integrates (or sums) to 1 over its support given the conditioning value.

$f_{X|Y}(x \mid y) = \frac{f(x, y)}{f_Y(y)}, \quad f_Y(y) > 0$

For discrete random variables:

$P(X = x \mid Y = y) = \frac{P(X = x, Y = y)}{P(Y = y)}$

Conditional PDF from a Joint Density If $f(x,y) = 6(1-y)$ for $0 \leq x \leq y \leq 1$ , what is $f_{X|Y}(x \mid y)$ ?

Answer First find the marginal PDF of $Y$ : $f_Y(y) = \int_0^y 6(1-y)\, dx = 6y(1-y)$ Then apply the conditional PDF formula: $f_{X|Y}(x \mid y) = \frac{6(1-y)}{6y(1-y)} = \frac{1}{y}, \quad 0 \leq x \leq y$ Given $Y = y$ , $X$ is uniformly distributed on $[0, y]$ .