# CDF of $Z=X_1+\max\{X_2,\,X_3\}$

Let $$X_1, X_2$$ and $$X_3$$ be expontial random varibles with paramtre $$\beta_1, \beta_2$$ and $$\beta_3$$ respectivly.

The PDF and CDF of $$X_i$$ for $$x\geq 0$$ are $$f_{X_i}(x)$$ and $$F_{X_i}(x)$$ given by

\begin{align} f_{X_i}(x)&=b_ie^{-x \beta_i} \\ F_{X_i}(x)&=1-e^{-x \beta_i}. \end{align}

I would like to find the CDF of random varible $$Z$$ define by

$$Z=X_1+\max\{X2,X3\}$$ for difrent parametre.

What is the CDF if $$\beta_i=\beta$$ for all $$i\in\{1,2,3\}$$?.

Thanks.

• What is stopping you here? – Did Feb 2 at 17:00

This is the type of problem that is not well-suited to manual solution, and which can benefit considerably from the assistance of a computer algebra system.

If $$X_1$$, $$X_2$$ and $$X_3$$ are independent $$\text{Exponential}(b_i)$$ random variables, then the joint pdf $$f(x_1,x_2,x_3)$$ is: We seek the cdf of $$Z = X_1 + \max(X_2,X_3)$$, namely $$P(Z.

This can be obtained immediately as: which returns the cdf as: ... where I have used the Prob function from the mathStatica add-on to Mathematica to help automate the calculation (and as disclosure, of which I am one of the authors).

Identical parameters

In the case where the $$b_i$$ are identical, the set-up is identical: simply replace each $$b_i$$ with $$b$$, which yields a much more elegant solution for the cdf:

$$F(z) = 2 e^{-b z} (\sinh (b z)-b z) \quad \quad \text{ for } z > 0$$

Monte Carlo check

It is always a good idea to test symbolic work by alternative methods. Here is a quick comparison of the theoretical pdf (red dashed curve) with the empirical pdf (squiggly blue curve - generated by Monte Carlo) when $$b_1 = 10$$, $$b_2 = 0.4$$, and $$b_3 = 7$$: All looks good.

Let $$Y:=\max\{X_2,\,X_3\}$$. Since for $$y\ge 0$$ $$P(Y\le y)=P(X_2\le y\land X_3\le y)=1-\exp -y\beta_2-\exp -y\beta_3+\exp -y(\beta_2+\beta_3),$$ the pdf of $$Y$$ is $$f_Y(y):=\beta_2\exp -y\beta_2+\beta_3\exp -y\beta_3-(\beta_2+\beta_3)\exp -y(\beta_2+\beta_3).$$Since $$X_1$$ has pdf $$f_X(x):=\beta_1\exp -x\beta_1$$ for $$x\ge 0$$, $$Z$$ has pdf$$\int_0^z f_X(z-y)f_Y(y)dy,$$and this you can easily compute and integrate.

• Ok but I am loking for CDF not PDF?, thank – Mokh Tar Bou Feb 2 at 13:58
• @MokhTarBou I know; I told you to (i) compute the integral, which is a PDF, then (ii) integrate that function to get the CDF. – J.G. Feb 2 at 14:09
• ok I wil try but how I cheek if it the correct anser? – Mokh Tar Bou Feb 2 at 14:12
• I am not sure I would agree that the integral is easy to compute, or that the derivation of the pdf of the max is correct. – wolfies Feb 2 at 15:07