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I have a problem with this exercise.

Let $X_1, X_2$ be two independent random variables such that $X_1$ ~ $\text{Exp}(\lambda_1)$ and $X_2$ ~ $\text{Exp}(\lambda_2)$ then with joint density $f_{(X_1,X_2)}(x_1,x_2)=\lambda_1\lambda_2e^{-(\lambda_1x_1+\lambda_2x_2)}$ if $x_1,x_2 > 0$ and $0$ otherwise. Now we define $Z=\min(X_1,X_2)$ and $W$ which is equal to $1$ if $X_1 \le X_2$ and $2$ otherwise.

I have to calculate $\mathbb{P}(Z \ge s, W=1)$ if $s \ge 0$.

I tried to solve the exercise in the following way:

$\mathbb{P}(Z \ge s, W=1) = \mathbb{P}(\min(X_1,X_2) \ge s, X_1 \le X_2) = \mathbb{P}(\min(X_1,X_2) \ge s) + \mathbb{P}(X_1 \le X_2)-\mathbb{P}(\min(X_1,X_2) \ge s \cup X_1 \le X_2)$.

Now, how can I calculate $\mathbb{P}(\min(X_1,X_2) \ge s \cup X_1 \le X_2)$?

Thanks in advance for your help.

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Notice: $\{\min(X_1,X_2)\geqslant s, X_1\leqslant X_2\}=\{s\leqslant X_1\leqslant X_2\}$

So: $\displaystyle \mathsf P(Z\geqslant s, W=1) ~=~ \int_s^\infty \int_x^\infty f_{X_1,X_2}(x,y)\,\mathrm d y\,\mathrm d x$

Take it from there.

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  • $\begingroup$ Thanks for this answer. Could you suggest me a way to prove that W and Z are independent? $\endgroup$ – LJG Aug 31 '17 at 11:58
  • $\begingroup$ Simply show whether it holds that: $\mathsf P(Z\geqslant s, W=1)~=~\mathsf P(Z\geqslant s)\,\mathsf P(W=1)$. $\endgroup$ – Graham Kemp Aug 31 '17 at 21:56

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