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

Question

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.

Answer 1

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 $P(Z<z)$.

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.

Answer 2

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.