Question

We have three light bulbs with lifetimes T1,T2,T3 distributed according to Exponential(λ1), Exponential(λ2), Exponential(λ3). In other word, for example bulb #1 will break at a random time T1, where the distribution of this time T1 is Exponential(λ1). The three bulbs break independently of each other. The three light bulbs are arranged in series, one after the other, along a circuit—this means that as soon as one or more light bulbs fail, the circuit will break. Let T be the lifetime of the circuit—that is, the time until the circuit breaks.

(a) What is the CDF of T, the lifetime of the circuit?

(b) Next, suppose that we only check on the circuit once every second (assume the times T1,T2,T3,T are measured in seconds). Let S be the first time we check the circuit and see that it’s broken. For example, if the circuit breaks after 3.55 seconds, we will only observe this when 4 seconds have passed, and so S = 4. Calculate the PMF of S.

(c) Finally, suppose that instead of checking on the circuit every second, we instead do the following: after each second, we randomly decide whether to check on the circuit or not. With probability p we check, and with probability 1−p we do not check. This decision is made independently at each time. Now let N be the number of times we check and see the circuit working. For example, if the circuit breaks at time 3.55, and our choices were to check at time 1 second, not to check at times 2 or 3 or 4, and to check at time 5, then N = 1, since the circuit was broken the 2nd time we checked. What is the PMF of N? (Hint: start by finding the joint PMF of N and S. It’s fine if your answer is in summation form.)

Answer 1

(a)

  

The lifetime T of circuit = min{ T1,T2,T3}

P(T<x) = 1 - P(T>x) =1 - P( min{T1,T2,T3} >x) = 1 - P(T1>x,T2>x,T3>x)

= 1- P(T1>x).P(T2>x).P(T3>x) since T1,T2,T3 are independent

= 1 - e^-₁x.e^-₂x.e^-₃x

= 1 - e^{-(₁+₂+3)x}

= 1 - e^-x where =₁+₂+₃

(b)

For S=1,2,3,....

P(S=n) = P(circuit is brocken for T<=n )- P(circuit is broken for T<=n-1)

= (1 - e^-n) - (1 - e^-(n-1))

= e^-(n-1) - e^-n

^(c)

^{For S=m, N = number of times we checked and see circuit working before m, which follows Binomial(m-1,p).}

^{The joint PMF of N and S is given in following table}

N	0	1	2	3	4	5	6
1	1	0	0	0	0	0	0
2	1-p	p	0	0	0	0	0
3	(1-p)2	2p(1-p)	p2	0	0	0	0
4	(1-p)3	3p(1-p)2	3p2(1-p)	p3	0	0	0
5	(1-p)4	4p(1-p)3	6p2(1-p)2	4p3(1-p)	p4	0	0
6	(1-p)5	5p(1-p)4	10p2(1-p)3	10p3(1-p)2	5p4(1-p)	p5	0
7	(1-p)6	6p(1-p)5	15p2(1-p)4	20p3(1-p)3	15p4(1-p)2	6p5(1-p)	p6

( where n<m )