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)= 0 - incorrect
0
Let X=T2−T1. Determine the conditional PDF fX∣T1(x∣t).
For x,t>0,fX∣T1(x∣t)= 0 - incorrect
0
Is X independent of T1?
Yes, they are independent - correct
Determine the PDF fT2(t).
For t>0,fT2(t)= 0 - incorrect
0
E[T2]=0 - incorrect
0
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 determine the PDF of T1, we need to use the fact that the lifetime of each laptop is exponentially distributed with parameter λ. The PDF of an exponential distribution is given by f(x) = λ * exp(-λx), for x > 0.
Since T1 represents the time of the first failure, it follows that the PDF of T1, denoted as fT1(t), is equal to the PDF of the exponential distribution. Therefore, fT1(t) = λ * exp(-λt), for t > 0.
To determine the conditional PDF of X given T1, denoted as fX|T1(x|t), we need to find the probability density function of X when we know the value of T1.
Note that X = T2 - T1, which represents the time between the failure of the first laptop (T1) and the failure of the second laptop (T2). If we know the value of T1, then T2 can be any time greater than T1. This means that X can take any value greater than or equal to 0.
Therefore, for x > 0 (because X is positive), the conditional PDF of X given T1, fX|T1(x|t), is given by the same PDF of an exponential distribution with parameter λ. Thus, fX|T1(x|t) = λ * exp(-λx), for x > 0.
Next, we are asked if X is independent of T1. X is considered independent of T1 if and only if the joint probability density function (PDF) of X and T1, denoted as f(X, T1), is equal to the product of the marginal PDFs of X and T1, i.e., f(X, T1) = fX(x) * fT1(t).
If we substitute the expression of the PDF of X given T1, fX|T1(x|t) = λ * exp(-λx), and the expression of the PDF of T1, fT1(t) = λ * exp(-λt), into the equation f(X, T1) = fX(x) * fT1(t), we can check if the equation holds.
f(X, T1) = λ * exp(-λx) * λ * exp(-λt) = λ^2 * exp(-λ(x+t))
fX(x) = λ * exp(-λx)
fT1(t) = λ * exp(-λt)
Multiplying fX(x) and fT1(t), we have λ * exp(-λx) * λ * exp(-λt) = λ^2 * exp(-λ(x+t))
Comparing this with f(X, T1), we see that they are equal. Therefore, X is indeed independent of T1.
To determine the PDF of T2, we can calculate it by conditioning on the value of T1. Let's denote T1 = t.
Since T2 represents the time of the second failure, it must be greater than T1 (which we have set as t). Therefore, we can express T2 = T1 + X.
To find the PDF of T2, denoted as fT2(t), we need to find the probability density function of T2.
Since X is exponential with parameter λ, we have already determined that the conditional PDF of X given T1 is fX|T1(x|t) = λ * exp(-λx) for x > 0.
Now, we consider the probability density function of T2 given T1 = t. Since T2 = T1 + X, we can write this as T2 = t + X.
For any given value t, we can substitute T2 = t + X into the expression of the conditional PDF of X given T1, fX|T1(x|t) = λ * exp(-λx), to get the joint PDF of T2 and T1, f(T2, T1).
Then, we integrate over all possible values of X (i.e., integrate from x = 0 to x = ∞) to obtain the PDF of T2, fT2(t).
Let's perform the integration:
fT2(t) = ∫[0,∞] λ * exp(-λx) dx
Since the integral of λ * exp(-λx) with respect to x from 0 to ∞ is 1, we have:
fT2(t) = 1
In conclusion, the PDF of T2, fT2(t), is equal to 1 for t > 0.
Finally, to calculate E[T2], we need to find the expected value (mean) of T2.
Since we already determined the PDF of T2, fT2(t) = 1 for t > 0, we can use the formula for calculating the expected value of a continuous random variable:
E[T2] = ∫[0,∞] t * fT2(t) dt
Since fT2(t) = 1, the integral becomes:
E[T2] = ∫[0,∞] t * 1 dt
Integrating t with respect to t from 0 to ∞, we have:
E[T2] = ∞ - 0 = ∞
Therefore, the expected value of T2, E[T2], is infinite.
I hope this explanation helps you understand the concepts and calculations involved in these questions! Let me know if you have any further questions.