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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.

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  • $\begingroup$ What is stopping you here? $\endgroup$ – Did Feb 2 at 17:00
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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:

enter image description here

We seek the cdf of $Z = X_1 + \max(X_2,X_3)$, namely $P(Z<z)$.

This can be obtained immediately as:

enter image description here

which returns the cdf as:

enter image description here

... 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 $b_3 = 7$:

enter image description here

All looks good.

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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.

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

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