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. The PDF of T1 is given by the exponential distribution with parameter λ:
fT1(t) = λ * exp(-λt)
2. To determine the conditional PDF fX|T1(x|t), we need to consider the relationship between T2 and T1. Since T2 is the time of the second failure after the first failure at T1, we can say that T2 = T1 + X. From this, we can calculate X = T2 - T1.
The conditional PDF can then be calculated using the fact that X follows the exponential distribution with parameter λ:
fX|T1(x|t) = λ * exp(-λ(t + x))
3. To find E[T3], we can use the fact that the sum of independent exponential random variables with the same rate parameter is a gamma random variable. Since T3 is the time of the third failure after the first two failures, we can say that T3 = T1 + T2 + Y, where Y follows the exponential distribution with parameter λ.
The expectation of T3 can then be calculated as:
E[T3] = E[T1] + E[T2] + E[Y]
Since T1 and T2 are independent exponential random variables, their expectations are given by:
E[T1] = 1/λ
E[T2] = 1/λ
Since Y follows the exponential distribution with parameter λ, its expectation is:
E[Y] = 1/λ
Therefore, we can find E[T3] as:
E[T3] = 1/λ + 1/λ + 1/λ = 3/λ
1. To determine the PDF of T1, we can use the fact that the lifetime of each engine is exponentially distributed with parameter λ.
Since T1 represents the time of the first failure, it can be calculated as the minimum of the three exponential random variables. The minimum of independent exponential random variables follows the exponential distribution with the sum of their individual parameters.
Let's assume the exponential random variables representing the lifetimes of the three engines are E1, E2, and E3, with parameters λ1, λ2, and λ3, respectively. Since the lifetimes are independent, the parameter for T1 will be the sum of the individual parameters:
λ1 + λ2 + λ3 = λ
Thus, the PDF of T1 for t > 0 is:
fT1(t) = λ * exp(-λ * t)
2. To determine the conditional PDF of X|T1, we need to consider the relationship between T1 and T2. Since T2 represents the time of the second failure, it will occur after the first failure, which is already known to be T1.
Therefore, X = T2 - T1 represents the time between the first and second failures. Given T1 = t and assuming independence between the failures, we can express T2 - T1 as T2 - t.
Since the exponential distribution is memoryless, the time between the first and second failures, given T1 = t, will also follow the exponential distribution with parameter λ. Thus, the conditional PDF of X|T1, for x, t > 0, is:
fX|T1(x|t) = λ * exp(-λ * x)
3. E[T3] represents the expected time of the third failure. Since the lifetime of each engine is exponentially distributed with parameter λ, the expected lifetime of each engine is equal to 1/λ.
Since the lifetimes are independent, the expected time of the third failure (T3) will be the sum of the expected lifetimes of the three engines:
E[T3] = 1/λ + 1/λ + 1/λ = 3/λ
To solve these questions, we need to understand the exponential distribution and how to derive the PDF (Probability Density Function) and expectation for exponential random variables.
1. Determine the PDF of T1:
The exponential distribution has the probability density function f(t) = λ * exp(-λt) for t > 0.
Since T1 represents the time of the first failure, we are interested in finding the probability density function fT1(t). However, to find this, we need to know the cumulative distribution function (CDF) of T1. The CDF is the probability that the failure occurs before time t.
The CDF of T1 is given by F(t) = P(T1 ≤ t) = 1 - P(T1 > t).
Since the failures of different engines are independent, we have:
P(T1 > t) = P(no engine fails before time t).
The probability of no engine failing before time t is the product of the probabilities of each engine surviving until time t. The survival probability for each engine is exp(-λt). Therefore:
P(T1 > t) = exp(-λt) * exp(-λt) * exp(-λt) = exp(-3λt).
Finally, we can find the PDF of T1 by differentiating the CDF:
fT1(t) = d/dt (1 - P(T1 > t))
= 3λ * exp(-3λt)
= 3λ * exp(-λ * 3t)
2. Let X = T2 - T1. Determine the conditional PDF fX|T1(x|t):
We want to find the conditional probability density function of X given T1 = t, denoted as fX|T1(x|t).
Since the lifetimes of the engines are independent, the time of the second failure (T2) does not depend on T1. Therefore, X = T2 - T1 has the same distribution as T2.
Using the PDF of T2 (derived in step 1):
fT2(t) = 3λ * exp(-λ * 3t)
Since X = T2 - T1, we need to consider the case when T2 > T1. This means X can only take positive values.
Therefore, for x, t > 0, the conditional PDF of X given T1 = t is:
fX|T1(x|t) = fT2(t + x) / P(T2 > t)
= (3λ * exp(-λ * 3(t + x))) / P(T2 > t)
To find P(T2 > t), we need to calculate the survival probability for T2:
P(T2 > t) = 1 - P(T2 ≤ t)
= 1 - F(T2 ≤ t)
= 1 - (1 - exp(-λ * 3t))
= exp(-λ * 3t)
Substituting this back into the conditional PDF, we have:
fX|T1(x|t) = (3λ * exp(-λ * 3(t + x))) / exp(-λ * 3t)
= 3λ * exp(-λ * 3x)
3. Find E[T3]:
The expectation of T3 can be found by integrating the product of t and the PDF of T3 from 0 to infinity.
Using the same distribution derived in step 1, we have:
fT3(t) = 3λ * exp(-λ * 3t)
The expectation of T3, denoted as E[T3], is given by:
E[T3] = ∫ t * fT3(t) dt from 0 to infinity
= ∫ t * (3λ * exp(-λ * 3t)) dt from 0 to infinity
Integrating this expression will give you the expected value of T3.
Note: In all these calculations, ensure you substitute the appropriate values for λ as given in the problem statement.