EXAMPLE 5.12

Reconsider the situation of EXAMPLE 5.3 and EXAMPLE 5.4 (EX 5.3 continued) involving

• $X=$ the proportion of time that a bank’s drive-up facility is busy
• $Y=$ the analogous proportion for the walk-up window.

The conditional pdf of $Y$ given that $X=.8$ is

$$f_{Y\mid X}\left(y\mid .8\right)=\frac{f\left(.8,y\right)}{f_X\left(.8\right)}=\frac{1.2\left(.8+y^2\right)}{1.2\left(.8\right)+.4}=\frac{1}{34}\left(24+30y^2\right)0<y<1$$

The probability that the walk-up facility is busy at most half the time given that $X=.8$ is then

$$P\left(Y\le .5\mid X=.8\right)={\int }_{-\infty }^{.5}{f}_{Y\mid X}\left(y\mid .8\right)dy={\int }_0^{.5}\frac{1}{34}\left(24+30y^2\right)dy=.390$$

Using the marginal pdf of $Y$ gives $P\left(Y\le .5\right)=.350$. Also $E\left(Y\right)=.6$, whereas the expected proportion of time that the walk-up facility is busy given that $X=.8$ (a conditional expectation) is

$$E\left(Y\mid X=.8\right)={\int }_{-\infty }^{\infty }y\cdot f_{Y\mid X}\left(y\mid .8\right)dy=\frac{1}{34}{\int }_0^{1}y\left(24+30y^2\right)dy=.574$$