Question

Suppose that we have three engines, which we turn on at time 0. Each engine will eventually fail, and we model each engine's lifetime as exponentially distributed with parameter λ. The lifetimes of different engines are independent. One of the engines will fail first, followed by the second, and followed by the last. Let T1 be the time of the first failure, T2 be the time of the second failure, and T3 be the time of the third failure. For answers involving algebraic expressions, enter "lambda" for λ and use "exp()" for exponentials. Follow standard notation.

1)
Determine the PDF of T1. For t>0,
fT1(t)=

2) Let X=T2−T1. Determine the conditional PDF fX|T1(x|t). For x,t>0,

fX|T1(x∣t)=

3) Is X independent of T1?

4) Let Y=T3−T2. Find the PDF of fY|T2(y∣T2).

For y,t>0,

fY|T2(y∣t)=

5) Is Y independent of T2?

6) Find the PDF fT3(t) for t≥0.

For t≥0,

fT3(t)=

Hint: Think of an interpretation of T3 as a maximum of some exponential random variables.

7) Find E[T3].

E[T3]=

Answer 1

3) Is X independent of T1?
Yes

5) Is Y independent of T2?
Yes

7) Find E[T3].

E[T3]=1/(3*lambda)+1/(2*lambda)+1/lambda

Answer 2

2) Let X=T2−T1. Determine the conditional PDF fX|T1(x|t). For x,t>0,

fX|T1(x∣t)=2*lambda*e^(-2*lambda*x)

Answer 3

4) Let Y=T3−T2. Find the PDF of fY|T2(y∣T2).

For y,t>0,

fY|T2(y∣t)=

6) Find the PDF fT3(t) for t≥0.

For t≥0,

fT3(t)=

Answer 4

4: lambda*e^(-lambda*y)
6: 3*lambda*e^(-lambda*t)*(1-e^(-lambda*t))^2

Answer 5

1. (3*lambda)*e^(-3*lambda*t)
2. (2*lambda)*e^(-2*lambda*x)
3. Yes..They are independent
4. lambda*e^(-lambda*y)
5. Yes..They are independent
6. 3*lambda*e^(-lambda*t)*(1-e^(-lambda*t))^2
7. 1/(3*lambda)+1/(2*lambda)+1/lambda