Mixed random variables. Probabilistic models sometimes

QUESTION:

Mixed random variables. Probabilistic models sometimes involve random variables that can be viewed as a mixture of a discrete random variable Y and a continuous random variable Z. By this we mean that the value of X is obtained according to the probability law of Y with a given probability p. and according to the probability law of Z with the complementary probability 1 - p. Then, X is called a mixed random variable and its CDF is given. using the total probability theorem, by Fx (x) = P(X x) = pP(Y x) + (1 - p)P(Z x) = pFy (x) + (1 - p)Fz(x). Its expected value is defined in a way that conforms to the total expectation theorem: E[X] = pE[Y] + (1 - p)E[Z] . The taxi stand and the bus stop near AI's home are in the same location. Al goes there at a given time and if a taxi is waiting (this happens with probability 2/3) he boards it. Otherwise he waits for a taxi or a bus to come, whichever comes first. The next taxi will arrive in a time that is uniformly distributed between 0 and 10 minutes, while the next bus will arrive in exactly 5 minutes. Find the CDF and the expected value of AI's waiting time. Solution. Let A be the event that Al will find a taxi waiting or will be picked up by the bus after 5 minutes. Note that the probability of boarding the next bus, given that Al has to wait, is Pea taxi will take more than 5 minutes to arrive) = . AI's waiting time, call it X, is a mixed random variable. With probability 2 1 1 5 peA) = - + - . - = - 3 3 2 6' it is equal to its discrete component Y (corresponding to either finding a taxi waiting, or boarding the bus), which has PMF py(y) { 3PA) ' if Y = 0, 6PA) ' if Y = 5, { 12 = 5 ) 15 ' if y = 0, if y = 5. [This equation follows from the calculation (0) = P(Y = 0 I A) = P(Y = 0, A) 2 py peA) - 3P(A) ' The calculation for py (5) is similar.] With the complementary probability 1 - peA), the waiting time is equal to its continuous component Z (corresponding to boarding a taxi after having to wait for some time less than 5 minutes), which has PDF fz(z) = { 1/5, if 0 :::; :::; 5, 0, otherwIse. The CDF is given by Fx (x) = P(A)Fy (x) + (1 - P(A))Fz(x), from which { 0, if x < 0, 5 12 1 x Fx (x) = _._ + _.- if O :::; x<5, 6 15 6 5 ' 1, if 5 :::; x. The expected value of the waiting time is 5 3 1 5 15 E[X] = P(A)E[Y] + (1 - P(A))E[Z] = - . - . 5 + - . - = -. 6 15 6 2 12