Question

The lifetimes of switches of a certain brand follow an exponential distribution with parameter 0. Two switches are randomly chosen (assume independence). The first is used until it fails, then the second one is used until it fails. Find the distribution of the sum of the lifetimes using the CDF method.

Answer 1

Given:

X is exponential Distribution with parameter $\theta$ .

Y is exponential Distribution with parameter $\theta$ .

To find the distribution of Z = X + Y using the CDF method:

CDF of Z = X + Y is given by:

P(X+Y < 2) = P(x+y)<z|)P(x)d.

Substituting the given distribution of X, we get:

P(X+Y <z)= | Ply < (2 - x))de-Brdo

So,

we get:

AP-70-90 (–90-9 – 11 ) = (7 5 A+ x)d

Simplifying RHS, we get:

P(X+Y < 2) = [ (@e-ør – 6e-63)dx

i.e.,

                            (1)

P(X+Y53= f*e*do – [*oe**dr

By direct integration, we get:

                                                  (2)

1 0e-ord.x = 1-e- Jo

                                                        (3)

1 0e-o-d.r = Oze-8 Jo

Substituting (2) and (3) in equation (1), we get:

P(X+Y < 2) = 1- e-6:- Aze-8

Thus, CDF of Z = X + Y is given by:

F(x) = 1- e-6:- Oze-8-

Probability Density Function of Z is got by differentiating F(z) with respect to z as follows:

P f(x) = 1 = [1 - e-:- Aze-0-

= 0+ 0e-8:- oſe-6: + (-0--)

– θ2 e-θε-

Thus, the distribution of Z = X + Y using the CDF method is:

f(x) = o-ze-

This is called Erlang Distribution with parameters 2 and $\theta$