5.1.5 Conditional Distributions

Suppose

• $X=$ the number of major defects in a randomly selected new automobile
• $Y=$ the number of minor defects in that same auto.

Question

• If we learn that the selected car has one major defect, what now is the probability that the car has at most three minor defects-that is, what is $P\left(Y\le 3\mid X=1\right)$?
• Similarly, if $X$ and $Y$ denote the lifetimes of the front and rear tires on a motorcycle, and it happens that $X=10,000$ miles, what now is the probability that $Y$ is at most 15,000 miles, and what is the expected lifetime of the rear tire “conditional on” this value of $X$ ?

Questions of this sort can be answered by studying conditional probability distributions.

conditional probability density function

Let $X$ and $Y$ be two continuous rv’s with joint pdf $f\left(x,y\right)$ and marginal $X$ pdf $f_X\left(x\right)$. Then for any $X$ value $x$ for which $f_X\left(x\right)>0$, the conditional probability density function of $Y$ given that $X=x$ is

$$f_{Y\mid X}\left(y\mid x\right)=\frac{f\left(x,y\right)}{f_X\left(x\right)}\text{ for }-\infty <y<\infty$$

If $X$ and $Y$ are discrete, replacing pdf’s by pmf’s in this definition gives the conditional probability mass function of $Y$ when $X=x$.

Notice that the definition of $f_{Y\mid X}\left(y\mid x\right)$ parallels that of $P\left(B\mid A\right)$, the conditional probability that $B$ will occur, given that $A$ has occurred. Once the conditional pdf or pmf has been determined, questions of the type posed at the outset of this subsection can be answered by integrating or summing over an appropriate set of $Y$ values.

EXAMPLE 5.12

If the two variables are independent, the marginal pmf or pdf in the denominator will cancel the corresponding factor in the numerator. The conditional distribution is then identical to the corresponding marginal distribution.