Let $X_1$ be the maximum of $N_1$ iid exponential random variables with parameter $\beta_1$. Similar, let $X_2$ be the maximum of $N_2$ iid exponential random variables with parameter $\beta_2$. The PDF and CDF of $X_i$ are \begin{align}\label{} f_{X_i}(x_i)=&N_i\beta_i(1-e^{-x_i\beta_i})^{N_1-1} \\ F_{X_i}(x_i)=&(1-e^{-x_i\beta_i})^{N_1} \end{align}

I use the following representation \begin{align}\label{} f_{X_i}(x_i)=&N_i\beta_i\sum_{n_i=0}^{N_i-1}\binom{N_i-1}{n_i}(-1)^{n_i}e^{-x_i\beta_i(n_i+1)} \\ F_{X_i}(x_i)=&\sum_{n_i=0}^{N_i}\binom{N_i}{n_i}(-1)^{n_i}e^{-x_i\beta_in_i} \end{align} I have the following probabilities \begin{align}\label{} P\{\{X_1+X_2\leq \alpha\}\cap\{X_1\leq X_2\}\}=&\int_{x_1=0}^{\alpha/2}f_{X_1}(x_1)[F_{X_2}(\alpha -x_1)-F_{X_2}(x_1)]dx \\ =&\underbrace{\int_{x_1=0}^{\alpha/2}f_{X_1}(x_1)F_{X_2}(\alpha -x_1)dx}_{I_1}-\int_{x_1=0}^{\alpha/2}f_{X_1}(x_1)F_{X_2}(x_1)dx. \end{align} At this point, the problem is in $I_1$. I have two cases, case one $\beta_1\neq \beta_2$ or case two that $\beta_1=\beta_2=\beta$. Taken case one \begin{align}\label{} I_1=&N_1\beta_1 \sum_{n_1=0}^{N_1-1}\sum_{n_2=0}^{N_2} \binom{N_1-1}{n_1} \binom{N_2}{n_2}(-1)^{n_1+n_2} \int_{x_1=0}^{\alpha/2}e^{-x_1\beta_1(n_1+1)}e^{-(\alpha-x_1)\beta_2n_2}dx_1 \\ = &N_1\beta_1 \sum_{n_1=0}^{N_1-1}\sum_{n_2=0}^{N_2} \binom{N_1-1}{n_1} \binom{N_2}{n_2}(-1)^{n_1+n_2}e^{-\alpha\beta_2n_2} \underbrace{\int_{x_1=0}^{\alpha/2}e^{-x_1(\beta_1(n_1+1)-\beta_2n_2)}dx_1}_{J_1} \end{align} Now for $J_1$

\begin{equation}\label{} \int_{x_1=0}^{\alpha/2}e^{-x_1(\beta_1(n_1+1)-\beta_2n_2)}dx_1=\frac{1}{\beta_1(n_1+1)-\beta_2n_2}\left[e^{-\frac{\alpha}{2}(\beta_1(n_1+1)-\beta_2n_2)}-1\right] \end{equation}

with condition $$\beta_1(n_1+1)-\beta_2n_2\neq0$$. For this case I think we may find some value of $\beta_1$ and $\beta_2$ such that \begin{align}\label{} \beta_1 \neq&\beta_2 \\ \beta_1(n_1+1)-\beta_2n_2\neq&0 \end{align}.

However for the case $\beta_1=\beta_2=\beta$, $J_1$ become \begin{equation}\label{} \int_{x_1=0}^{\alpha/2}e^{-x_1\beta(n_1+1-n_2)}dx_1=\frac{1}{\beta(n_1+1-n_2)}\left[e^{-\frac{\alpha}{2}\beta(n_1+1-n_2)}-1\right] \end{equation} This is problem since for $n_1=0$ and $n_2=1$ we have


so we get division by zero? How we can solve this problem?

For this last reason, I think we need to find an other integration domain rather then $$ X_1\leq X_2\leq \alpha-X_1 $$

to avoid $F_{X_2}(\alpha-x_1)$.



I don't see any problem.

For $\beta_1(n+1)=\beta_2n_2$, you go back to the definition of the integral $J_1$ $$ J_1:=\int_0^{\alpha/2}e^{-x_1(\beta_1(n_1+1)-\beta_2 n_2)}\,\mathrm{d}x_1=\int_0^{\alpha/2}\,\mathrm{d}x_1=\frac\alpha2. $$

What you are claiming is similar to naively saying $\int_1^2 x^n\,\mathrm{d}x=\frac{2^{n+1}-1}{n+1}$ doesn't work for $n=-1$ so you need to choose another domain.

  • $\begingroup$ Hi sir, I did not understand what you are meaning, can you add more detail please. Note that I have $0\leq n_1\leq N_1-1$ and $0\leq n_2\leq N_2$? $\endgroup$ – Monir Jun 15 at 14:32
  • $\begingroup$ So what I can do for solution? $\endgroup$ – Monir Jun 15 at 14:36

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