Density function of waiting time

Question

    ◮ A post office has two clerks.
    ◮ Service time for each clerk is 
      exponentially distributed by parameters λ1 , λ2. 
    ◮ When you arrive, 
     both clerks are busy but no one else waiting.
    You will enter service when either clerk becomes free.
    ◮ Calculate the density function
 of the time the customer will have to wait until his turn.

The answer is:

Can someone explain me why? I also have the full explanation of the answer but I can't really understand it. They set 2 expansional variables that indicate about the time of each clerk (T1 and T2), and found the minimum between them and few more things and got to the answer.

I did read about density functions and did see here: https://en.wikipedia.org/wiki/Exponential_distribution

that the PDF is : ${λe}^{λx}$ and

but I can't figure it out, how should I use it it order to solve the question. Should I take it as granted? that every PDF is in that form, without the need to prove anything?

Thanks.

Answer 1

The answer is very easy: the time you have to wait before being served is the mininum between the two waiting times.

The only problem is to understand which is the CDF of the minimum: here is an easy proof

Set $Z=min(X,Y)$

$$P(Z>z)=P(X>z;Y>z)=P(X>z)P(Y>z)=e^{-\lambda_1 z}e^{-\lambda_2 z}=e^{-(\lambda_1+\lambda_2)z}$$

Thus the CDF of Z is

$$F_Z(z)=P(Z\leq z)=1-e^{-(\lambda_1+\lambda_2)z}$$

take the derivative of F to get the desired density finding (setting $\lambda=\lambda_1+\lambda_2$) exactly the solution you posted.

$$f_Z(z)=\frac{d}{dz}=(\lambda_1+\lambda_2)e^{-(\lambda_1+\lambda_2)z}\cdot\mathbb{1}_{[0;+\infty)}(z)$$