Suppose 3 people, A,B,C, decide to go to LA for vacation. They use independent transportation means to go to LA. Suppose the time it takes A to arrive in LA is exponentially distributed with mean 2 hrs, the time it takes B to arrive in LA is exponentially distributed with mean 4 hrs, and the time it takes C to arrive in LA is exponentially distributed with mean 8 hrs.
Questions:
(a) What is the probability that B arrives in LA before A and C
(b) Given that B arrives in LA first, what is the expected amount of time it takes B to arrive in LA?
(c) Given that A arrives in LA first, what is the expected amount of time it takes B to arrive in LA?
For (a), I used property from exponential random variables, and get the probability to be (1/4)/(1/4 + 1/2 + 1/8) = 2/7
The problems are really with (b) and (c). Let the amount of time it takes B to arrive in LA to be T, then E[T] = ∫ E[T|B arrives first] * P(B arrives first), then I do not know how to proceed from here, especially how to find E[T|B arrives first]?
Thanks very much for help!
In general for a jointly continuous random variable $(X,Y)$ you can compute $E[X \mid X>Y]$ through the joint density:
$$E[X \mid X>Y]=\frac{\int_{-\infty}^\infty \int_y^\infty x f_{X,Y}(x,y) dx dy}{\int_{-\infty}^\infty \int_y^\infty f_{X,Y}(x,y) dx dy}.$$
In your particular problem, since you have independence, you can "marginalize" these integrals by writing
$$E[X \mid X>Y]=\int_{-\infty}^\infty E[X \mid X>y] f_Y(y) dy.$$
This is useful to do because now that middle expectation (involving conditioning on $X>y$ for a fixed real number $y$) can be computed using the form of the memoryless property that you know.