Respuesta :
Answer:
[tex]f(x) = (1-e^{-\frac{1}{2}x})[-e^{-x} +e^0]=(1-e^{-\frac{x}{2}})[1-e^{-x}][/tex]
Step-by-step explanation:
If we have two random variables Y1 and Y2 and we have th following limits:
[tex]a_1 \leq Y_1 \leq a_2 , b_1 \leq Y_2 \leq b_2[/tex]
We an find the density function with this formula:
[tex] 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 [/tex]
Now for our problem we know that for the two times of failure the density function is given by:
[tex]f(t) = e^{-t} t>0[/tex]
And we know that the joint density for T1 and T2 is given by:
[tex]f(t_1, t_2) =e^{-t_1}e^{-t_2}, t_1 >0, t_2 >0[/tex]
And we know that [tex] X= 2T_1 +T_2[/tex]
If we solve for [tex]T_1[/tex we got:
[tex] T_1 =\frac{X-T_2}{2}[/tex]
And then we can find the density function like this:
[tex] P(X \leq x) = P(2T_1 +T_2 \leq x)[/tex]
[tex]\int_{0}^x \int_{0}^{\frac{1}{2}(x-t_2)} e^{-t_1}e^{-t_2} dt_1 dt_2 [/tex]
[tex] =\int_{0}^x e^{-t_2} \int_{0}^{\frac{1}{2}(x-t_2)}e^{-t_1}dt_1 dt_2[/tex]
[tex]=\int_{0}^x e^{-t_2} [-e^{-t_1} \Big|_0^{\frac{1}{2}(x-t_2)}] dt_2[/tex]
[tex]=\int_{0}^x e^{-t_2} [1-e^{-\frac{1}{2} (x-t_2)}] dt_2[/tex]
[tex]=\int_{0}^x e^{-t_2} -e^{-\frac{1}{2}x}e^{-t_2} dt_2[/tex]
[tex]= \int_{0}^x (1- e^{-\frac{1}{2}x}) e^{-t_2}dt_2[/tex]
[tex]= -(1-e^{-\frac{1}{2}x}) e^{-t_2} \Big|_0^x [/tex]
[tex]f(x) = (1-e^{-\frac{1}{2}x})[-e^{-x} +e^0]=(1-e^{-\frac{x}{2}})[1-e^{-x}][/tex]
