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)=
- unanswered
Let X=T2−T1. Determine the conditional PDF fX∣T1(x∣t).
For x,t>0,fX∣T1(x∣t)=
- unanswered
Is X independent of T1?
Yes, they are independent - unanswered
Determine the PDF fT2(t).
For t>0,fT2(t)=
- unanswered
E[T2]=
- unanswered
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)
To find the PDF (Probability Density Function) of T1, we know that both laptops have exponentially distributed lifetimes with the same parameter λ. The exponential distribution has the formula f(t) = λ * exp(-λ * t) for t>0.
Let's calculate the PDF of T1:
For t>0, fT1(t) = λ * exp(-λ * t)
Now, let's move on to the next part.
To determine the conditional PDF fX∣T1(x|t), we need to use the fact that X = T2 - T1. Given that T1 has occurred at time t, we want to find the probability distribution of X.
Since T2 follows T1, we know that T2 ≥ T1. Therefore, X = T2 - T1 will always be greater than or equal to 0.
To find the conditional PDF, we can use the property that the conditional PDF of X given T1 is equal to the PDF of the difference T2 - T1.
The difference of two independent exponential random variables with rate parameters λ will result in a Gamma distribution with parameters (scale=k, shape=n), where k is the difference between the rate parameters and n is the number of independent exponential variables being subtracted.
In this case, since T1 and T2 are both exponentially distributed with rate parameter λ, we have k = λ and n = 1.
Therefore, the conditional PDF of X given T1 is:
For x, t > 0, fX|T1(x|t) = (λ * exp(-λ * x)) / λ = exp(-λ * x)
Moving on to the next point.
To determine if X is independent of T1, we need to check if the joint PDF of X and T1 is equal to the product of their individual PDFs.
The joint PDF of X and T1 would be obtained by multiplying the PDF of T1 and the conditional PDF of X given T1.
The joint PDF f(X, T1) = fT1(t) * fX|T1(x|t)
Substituting the PDFs we found earlier, we have:
f(X, T1) = (λ * exp(-λ * t)) * exp(-λ * x) = λ * exp(-λ * t) * exp(-λ * x) = λ * exp(-λ * (t + x))
Now, if the joint PDF factorizes into the product of their individual PDFs, then X and T1 are independent. Let's check if this is the case:
fX(x) * fT1(t) = (λ * exp(-λ * x)) * (λ * exp(-λ * t)) = λ^2 * exp(-λ * x) * exp(-λ * t) = λ^2 * exp(-λ * (t + x))
Comparing this with the joint PDF we found (λ * exp(-λ * (t + x))), we can see that they are equal. Therefore, X and T1 are independent.
Moving on to the next point.
To find the PDF of T2, we can use the fact that T2 = T1 + X. Since X is independent of T1, we can find the PDF of T2 by convolution.
The convolution of two independent exponential random variables with rate parameters λ will result in a gamma distribution with parameters (scale = k, shape = n), where k is the sum of the rate parameters and n is the number of independent exponential variables being added.
In this case, since T1 and X are both exponentially distributed with rate parameter λ, we have k = 2λ and n = 1.
Therefore, the PDF of T2 is:
For t > 0, fT2(t) = (2λ * exp(-2λ * t))
Finally, let's calculate E[T2], which represents the expected value of T2.
The expected value of a random variable T, denoted as E[T], can be obtained by integrating T times its PDF over its entire range.
For T2, we have:
E[T2] = ∫ t * fT2(t) dt
Substituting the PDF we found earlier, we have:
E[T2] = ∫ t * (2λ * exp(-2λ * t)) dt
Integrating this expression will give us the expected value of T2.
Please note that due to the complexity of the integration, the exact value of E[T2] cannot be provided without specifying the limits of integration.
I hope this explanation helps you understand how to find the PDF of T1, the conditional PDF of X given T1, the independence of X and T1, the PDF of T2, and how to calculate E[T2]!