Question
Suppose that you have two laptops, both of which you begin using at time 0. Each laptop will eventually fail, and we model each one's lifetime as exponentially distributed with the same parameter λ. The lifetimes of the two laptops are independent. One of the laptops will fail first, followed by the other. Define T1 as the time of the first failure and T2 as the time of the second failure.
In parts 1, 2, 4, and 5 below, your answers will be algebraic expressions. Enter 'lambda' for λ and use 'exp()' for exponentials. Follow standard notation.
Determine the PDF of T1.
For t>0,fT1(t)= 0 - incorrect
0
Let X=T2−T1. Determine the conditional PDF fX∣T1(x∣t).
For x,t>0,fX∣T1(x∣t)= 0 - incorrect
0
Is X independent of T1?
Yes, they are independent - correct
Determine the PDF fT2(t).
For t>0,fT2(t)= 0 - incorrect
0
E[T2]=0 - incorrect
0
In parts 1, 2, 4, and 5 below, your answers will be algebraic expressions. Enter 'lambda' for λ and use 'exp()' for exponentials. Follow standard notation.
Determine the PDF of T1.
For t>0,fT1(t)= 0 - incorrect
0
Let X=T2−T1. Determine the conditional PDF fX∣T1(x∣t).
For x,t>0,fX∣T1(x∣t)= 0 - incorrect
0
Is X independent of T1?
Yes, they are independent - correct
Determine the PDF fT2(t).
For t>0,fT2(t)= 0 - incorrect
0
E[T2]=0 - incorrect
0
Answers
1) 2*lambda*exp(-2*lambda*t)
2) lambda*e^(-lambda*x)
3) yes
4.
2*lambda*exp(-lambda*t)*(1-exp(-lambda*t))
5) 3/(2*lambda)
2) lambda*e^(-lambda*x)
3) yes
4.
2*lambda*exp(-lambda*t)*(1-exp(-lambda*t))
5) 3/(2*lambda)
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