Asked by Fay
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 Ti be the time of the first failure·T, be the time of the second failure, and 13 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. Find E[T3]=
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. Find E[T3]=
Answers
Answered by
Anonymous
Can some one help on this
thanks
thanks
Answered by
Anonymous
fT1(t)=(3*lambda)*e^(-3*lambda*t)
fX|T1(x∣t)=(2*lambda)*e^(-2*lambda*x)
E[T3]=(11)/(6*lambda)
fX|T1(x∣t)=(2*lambda)*e^(-2*lambda*x)
E[T3]=(11)/(6*lambda)
Answered by
me
E[T3] should be this....1/(3*lambda)+1/(2*lambda)+1/lambda
Answered by
Anonymous
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
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
There are no AI answers yet. The ability to request AI answers is coming soon!
Submit Your Answer
We prioritize human answers over AI answers.
If you are human, and you can answer this question, please submit your answer.