# Find $P\{\{X_1+X_2\leq \alpha\}\cap\{X_1\leq X_2\}\}$ for $X_1$ and $X_2$ have the same parameter or different parameter

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 $$J_1$$

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

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 $$$$\label{} \int_{x_1=0}^{\alpha/2}e^{-x_1\beta(n_1+1-n_2)}dx_1=\frac{1}{\beta(n_1+1-n_2)}\left[e^{-\frac{\alpha}{2}\beta(n_1+1-n_2)}-1\right]$$$$ This is problem since for $$n_1=0$$ and $$n_2=1$$ we have

$$n_1+1-n_2=0+1-1=0$$

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

Thanks.

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.

• 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 $0\leq n_2\leq N_2$? – Monir Jun 15 at 14:32
• So what I can do for solution? – Monir Jun 15 at 14:36