Two individuals, A and B, both require kidney transplants. If she does not receive a new kidney, then A will die after an exponential time with rate µA, and B after an exponential time with rate µB. New kidneys arrive in accordance with a Poisson process having rate λ. It has been decided that the first kidney will go to A (or to B if B is alive and A is not at that time) and the next one to B (if still living). a) What is the probability that A obtains a new kidney? b) What is the probability that B obtains a new kidney?

Question

Mathematics

Diana

27 May, 06:42

Two individuals, A and B, both require kidney transplants. If she does not receive a new kidney, then A will die after an exponential time with rate µA, and B after an exponential time with rate µB. New kidneys arrive in accordance with a Poisson process having rate λ. It has been decided that the first kidney will go to A (or to B if B is alive and A is not at that time) and the next one to B (if still living). a) What is the probability that A obtains a new kidney? b) What is the probability that B obtains a new kidney?

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Edwin Michael · Answer 1

P (A obtains a new kidney) = P (T1 < TA) = λ / (λ + µA)

P (B obtains a new kidney) = [λ / (λ + µB) ] + [ (λ + µA) / (λ + µA + µB) ]

Step-by-step explanation:

P (A obtains a new kidney) = P (T1 < TA) = λ / (λ + µA)

To obtain the probability of B obtains a new kidney, we will condition on the first event (and then unconditioned), i. e., which happens first: a kidney arrives, A dies or B dies. Thus

P (B obtains a new kidney)

=P (B obtains a new kidney|T1 = min {T1, TA, TB}) P (T1 = min {T1, TA, TB}) + P (B obtains a new kidney|TA = min {T1, TA, TB}) P (TA = min {T1, TA, TB}) + P (B obtains a new kidney|TB = min {T1, TA, TB}) P (TB = min {T1, TA, TB})

=P (T2 < TB) P (T1 = min {T1, TA, TB}) + P (T1 < TB) P (TA = min {T1, TA, TB}) + 0

=[λ / (λ + µB) ][λ / (λ + µA + µB) ] + [λ / (λ + µB) ] [µA / (λ + µA + µB) ]

=[λ / (λ + µB) ] + [ (λ + µA) / (λ + µA + µB) ]

Suppose that each person survives a kidney operation (independently) with probability p, 0 < p < 1, and, if so, has an exponentially distributed remaining life with mean 1/µ

We consider the time until either A dies or the kidney arrives, whichever is first. The mean time until that is 1 / (µA + λ). A will of course live until that time. We now consider the expected remaining time A will live after that time. The conditional probability that A will live until the kidney arrives is λ / (λ + µA), independent of the time itself. Let LA be the lifetime of A. Given that the kidney operation is performed, the expected remaining life is p/µ. Thus

E[LA] = 1 / (λ + µA) + (λ / λ + µA) (p/µ) = (µ + λp) / [ (λ + µA) µ]