5.4.1 problem 26

a. Let $T_w\sim N(w,{\sigma }^2)$ be the time it takes Walter to arrive, $T_c\sim N(c,4{\sigma }^2)$ be the time it takes Carl to arrive. We have $-T_w\sim N(-c,{\sigma }^2)$ since flipping the sign of the rv flips the sign of the expectation but does not change the variance. Then $T_c-T_w\sim N(c-w,5{\sigma }^2)$ per the important fact given in the problem. If $Z\sim N(0,1)$, then $T_c-T_w=\sqrt{5}\mathrm{\sigma Z}+c-w$.

For Carl to arrive first, we require $T_c-T_w<0$ (the time it takes Carl is less than the time it takes Walter). Let us find this probability:

b. If Carl has a greater than 1/2 chance of arriving first, then $\Phi \left(\frac{w-c}{\sigma \sqrt{5}}\right)>1/2$. Since $\Phi$ is an increasing function and equals 1/2 when its input is 0, this implies we need $\frac{w-c}{\sigma \sqrt{5}}>0$, which in turn implies $c>w$. So, as long as Carl’s car lets him be faster on average than Walter’s walking, Carl has a better than 1/2 chance of arriving first.

c. To make it to the meeting at time, either individual needs to make sure the amount of time they take to arrive is less than $w+10$.

Since $\Phi$ is an increasing function, if we want Carl to have a greater chance than Walter to make it on time, then we require $\frac{w+10-c}{2\sigma }>\frac{10}{\sigma }$. This then implies that we need $w>c+10$.