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]=
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)
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
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
4: lambda*e^(-lambda*y)
6: 3*lambda*e^(-lambda*t)*(1-e^(-lambda*t))^2
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)=
1) The PDF of T1 can be found by considering that T1 represents the time of the first failure. Since the lifetimes of the engines are exponentially distributed with parameter λ, we can use the exponential distribution function to find the PDF.
For t > 0, fT1(t) = λ * exp(-λ * t)
2) Let X = T2 - T1. To find the conditional PDF fX|T1(x|t), we need to consider that T2 represents the time of the second failure, given that T1 has already occurred at time t. Since the lifetimes are independent, the conditional probability of T2 occurring at x + t, given T1 occurred at time t, is the same as the unconditional probability.
For x, t > 0, fX|T1(x|t) = λ * exp(-λ * (x + t))
3) X is not independent of T1 since the value of X depends on the value of T1.
4) Let Y = T3 - T2. The conditional PDF fY|T2(y|t) represents the probability distribution of Y given that T2 occurred at time t. Similar to the previous question, the conditional probability is the same as the unconditional probability since the lifetimes are independent.
For y, t > 0, fY|T2(y|t) = λ * exp(-λ * (y + t))
5) Y is not independent of T2 since the value of Y depends on the value of T2.
6) To find the PDF of T3, we can consider T3 as the maximum of the three exponential lifetimes. The maximum of a set of exponential random variables with the same parameter λ is also an exponential random variable with parameter λ. Thus, the PDF of T3 is the same as the PDF for each individual engine's lifetime.
For t ≥ 0, fT3(t) = λ * exp(-λ * t)
7) To find E[T3], we can use the fact that the expected value of an exponential random variable with parameter λ is 1/λ.
E[T3] = 1/λ
1) To determine the PDF of T1, we need to find the probability density function for the first failure time. Since the lifetimes of the engines are exponentially distributed with parameter λ, the probability density function for each engine failure is given by:
f(t) = λ * exp(-λt), for t > 0
To find the PDF of T1, we need to find the probability that the first failure occurs at time t. This is equivalent to finding the probability that the first engine fails within the time interval (t, t + dt), given that it hasn't failed before time t. This probability can be calculated as the complement of the probability that the first failure occurs before time t, which is given by the cumulative distribution function (CDF) of the exponential distribution:
F(t) = 1 - exp(-λt), for t > 0
Differentiating the CDF with respect to t gives us the PDF:
fT1(t) = dF(t)/dt = d(1 - exp(-λt))/dt = λ * exp(-λt), for t > 0
Therefore, the PDF of T1 is fT1(t) = λ * exp(-λt), for t > 0.
2) To determine the conditional PDF of X given T1, we need to find the probability density function for the difference between the second and first failure times, conditioned on the first failure occurring at time t. This can be calculated as the probability that the second failure occurs at time (t + x), given that the first failure occurred at time t. Since the lifetimes of the engines are independent, the conditional probability can be calculated as:
fX|T1(x|t) = λ * exp(-λ(t+x)), for x,t > 0
Therefore, the conditional PDF of X given T1 is fX|T1(x|t) = λ * exp(-λ(t+x)), for x,t > 0.
3) To determine whether X is independent of T1, we need to check if the joint PDF of X and T1 can be expressed as the product of their individual PDFs. If X and T1 are independent, then the joint PDF fX,T1(x,t) can be expressed as:
fX,T1(x,t) = fX|T1(x|t) * fT1(t)
Substituting the PDFs we derived earlier:
fX,T1(x,t) = λ * exp(-λ(t+x)) * λ * exp(-λt) = λ^2 * exp(-λ(x+2t)), for x,t > 0
Since this expression cannot be expressed as the product of the individual PDFs, X is not independent of T1.
4) To find the conditional PDF of Y given T2, we need to find the probability density function for the difference between the third and second failure times, conditioned on the second failure occurring at time t. Since the lifetimes of the engines are independent, the conditional probability can be calculated as:
fY|T2(y|t) = λ * exp(-λ(t+y)), for y,t > 0
Therefore, the conditional PDF of Y given T2 is fY|T2(y|t) = λ * exp(-λ(t+y)), for y,t > 0.
5) To determine whether Y is independent of T2, we need to check if the joint PDF of Y and T2 can be expressed as the product of their individual PDFs. If Y and T2 are independent, then the joint PDF fY,T2(y,t) can be expressed as:
fY,T2(y,t) = fY|T2(y|t) * fT2(t)
Since the conditional PDF fY|T2(y|t) and the PDF fT2(t) are both λ * exp(-λ(t+y)), the joint PDF can be expressed as:
fY,T2(y,t) = (λ * exp(-λ(t+y))) * (λ * exp(-λt)) = λ^2 * exp(-λ(2t+y)), for y,t > 0
Therefore, Y is independent of T2.
6) To find the PDF of T3, we can think of T3 as the maximum of three exponential random variables with parameter λ. This means that the time of the third failure is determined by the engine that fails last among the three engines. The maximum of independent exponential random variables follows a gamma distribution.
The gamma distribution has a PDF given by:
f(t) = (λ^k)/(Γ(k)) * t^(k-1) * exp(-λt), for t ≥ 0
where Γ(k) is the gamma function, which is defined as the integral of t^(k-1) * exp(-λt) from 0 to infinity.
In this case, since we have three engines, the time of the third failure T3 follows a gamma distribution with parameters k = 3 and λ.
Therefore, the PDF of T3 is:
fT3(t) = (λ^3)/(Γ(3)) * t^(3-1) * exp(-λt) = (λ^3)/2 * t^2 * exp(-λt), for t ≥ 0.
7) To find E[T3], we need to calculate the expected value of T3 using its PDF. The expected value of a continuous random variable can be calculated by integrating the product of the variable and its PDF over its range.
E[T3] = ∫t * fT3(t) dt, from 0 to infinity
Substituting the PDF of T3 derived earlier:
E[T3] = ∫t * (λ^3)/2 * t^2 * exp(-λt) dt, from 0 to infinity
This integral can be evaluated using integration techniques or using a computer algebra system.
E[T3] = (3!/(2λ^3)) * ∫t^3 * exp(-λt) dt, from 0 to infinity
Simplifying this integral will give us the value of E[T3].