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. (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
fT1(t)=(3*lambda)*e^(-3*lambda*t)
fX|T1(x∣t)=(2*lambda)*e^(-2*lambda*x)
E[T3]=(11)/(6*lambda)
E[T3] should be this....1/(3*lambda)+1/(2*lambda)+1/lambda
Can some one help on this
thanks
1. Determine the PDF of T1. For t>0,
fT1(t) = λ * exp(-λt)
2. Let X = T2 - T1. Determine the conditional PDF fX|T1(x|t). For x, t > 0,
fX|T1(x|t) = λ * exp(-λ(x+t))
3. To find E[T3], we need to find the cumulative distribution function (CDF) of T3, F_T3(t), which is equal to the probability that all three engines have failed by time t. Since the failure times of the engines are independent, we can use the probability density function (PDF) to calculate the CDF.
F_T3(t) = P(T1 ≤ t) * P(T2 ≤ t) * P(T3 ≤ t)
For each engine, the probability of failure by time t is given by the CDF of the exponential distribution:
P(Ti ≤ t) = 1 - exp(-λt)
Plugging in the values, we have:
F_T3(t) = (1 - exp(-λt))^3
Finally, to find the expected value of T3, we integrate the CDF:
E[T3] = ∫(t * dF_T3(t))
E[T3] = ∫(t * 3(1 - exp(-λt))^2 * (-λexp(-λt)) dt
Simplifying and solving the integral, we find:
E[T3] = -3 * exp(-2λt) + 3 * exp(-λt) + t * (exp(-2λt) - 2exp(-λt))
Note: The above expressions assume that the failure times of the engines are exponentially distributed with parameter λ.
To solve these problems, we will use the properties of exponential distribution and the fact that the lifetimes of the engines are independent.
1. The PDF of T1:
Since the lifetime of each engine is exponentially distributed with parameter λ, the probability density function (PDF) of an exponential distribution is given by:
f(t) = λ * exp(-λt) for t ≥ 0.
The first engine failure can occur at any time t > 0. So, the PDF of T1 is given by the same exponential distribution as the individual engine lifetimes:
fT1(t) = λ * exp(-λt) for t > 0.
2. The conditional PDF of X|T1:
To find the conditional PDF of X|T1, we need to consider the distribution of the time difference between the second failure (T2) and the first failure (T1). Since the lifetimes of the engines are independent, the remaining two engines will continue to operate after the first failure. Hence, T2 - T1 follows an exponential distribution with parameter λ.
Therefore, the conditional PDF of X|T1 is the same as the PDF of an exponential distribution with parameter λ:
fX|T1(x|t) = λ * exp(-λx) for x, t > 0.
3. Expected value of T3:
The third failure time, T3, can be thought of as the maximum of the remaining lifetimes of the two remaining engines. The maximum of independent exponentially distributed random variables follows the Erlang distribution.
For the Erlang distribution, the expected value is given by:
E[T3] = (k/λ) where k is the number of independent exponential variables being summed. In this case, k = 2 (since there are two remaining engines after the first failure).
Therefore, E[T3] = (2/λ).