Question

A device containing two key components fails when and only when both components fail. The lifetimes, T1 and T2, of these components are independent with common density function f(t)=e^-t for t > 0. The cost, X, of operating the device until failure is 2T1 + T2.

1. What is the density function of X for x > 0?

Answer 1

`f(x) = (1-e^{-(1)/(2)x})[-e^(-x) +e^0]=(1-e^{-(x)/(2)})[1-e^(-x)]`

Explanation:

If we have two random variables Y1 and Y2 and we have th following limits:

`a_1 \leq Y_1 \leq a_2 , b_1 \leq Y_2 \leq b_2`

We an find the density function with this formula:

`P(a_1 \leq Y_1 \leq a_2 ,b_1 \leq Y_2 \leq b_2)= \int_(b_1)^(b_2) \int_(a_1)^(a_2) f(y_1, y_2) dy_1 dy_2`

Now for our problem we know that for the two times of failure the density function is given by:

`f(t) = e^(-t) t>0`

And we know that the joint density for T1 and T2 is given by:

`f(t_1, t_2) =e^(-t_1)e^(-t_2), t_1 >0, t_2 >0`

And we know that
`X= 2T_1 +T_2`

If we solve for
`T_1[/tex we got:</p><p>[tex] T_1 =(X-T_2)/(2)`

And then we can find the density function like this:

`P(X \leq x) = P(2T_1 +T_2 \leq x)`

`\int_(0)^x \int_(0)^{(1)/(2)(x-t_2)} e^(-t_1)e^(-t_2) dt_1 dt_2`

`=\int_(0)^x e^(-t_2) \int_(0)^{(1)/(2)(x-t_2)}e^(-t_1)dt_1 dt_2`

`=\int_(0)^x e^(-t_2) [-e^(-t_1) \Big|_0^{(1)/(2)(x-t_2)}] dt_2`

`=\int_(0)^x e^(-t_2) [1-e^{-(1)/(2) (x-t_2)}] dt_2`

`=\int_(0)^x e^(-t_2) -e^{-(1)/(2)x}e^(-t_2) dt_2`

`= \int_(0)^x (1- e^{-(1)/(2)x}) e^(-t_2)dt_2`

`= -(1-e^{-(1)/(2)x}) e^(-t_2) \Big|_0^x`

`f(x) = (1-e^{-(1)/(2)x})[-e^(-x) +e^0]=(1-e^{-(x)/(2)})[1-e^(-x)]`