This has been bothering me for a while now:

Q: If $\displaystyle X_i,\,i=1,2,3$ are independent exponential random variables with rates $\displaystyle \lambda_i,\,i=1,2,3$, find

(a) $\displaystyle \mathbb{P}(X_1<X_2<X_3)$

Here's what I figured out so far:

Let $\displaystyle Y=\min(X_2,X_3)$ Then $\displaystyle Y$ is $\displaystyle \mathcal{E}(\lambda_2+\lambda_3)$ and is independent of $\displaystyle X_1$. Now observe that $\displaystyle \mathbb{P}(X_1<Y)=\frac{\lambda_1}{\lambda_1+\lamb da_2+\lambda_3}$. We now see that

$\displaystyle \begin{aligned}\mathbb{P}(X_1<X_2<X_3) &= \mathbb{P}(X_1=\min(X_1,X_2,X_3))\mathbb{P}(X_2<X_ 3\mid X_1=\min(X_1,X_2,X_3))\\&=\frac{\lambda_1}{\lambda _1+\lambda_2+\lambda_3}\mathbb{P}(X_2<X_3\mid X_1=\min(X_1,X_2,X_3))\end{aligned}$

Now, take note that the distribution of $\displaystyle X_2-x$ and $\displaystyle X_3-x$ given that they are larger than $\displaystyle X_1 = x$ is the same as that of $\displaystyle X_2$ and $\displaystyle X_3$, so

$\displaystyle \begin{aligned}\mathbb{P}(X_2<X_3\mid\min(X_2,X_3) \geq x) &= \mathbb{P}((X_2-x<X_3-x)\mid\min(X_2,X_3)\geq x)\\ &=\mathbb{P}(X_2<X_3)\\ &=\frac{\lambda_2}{\lambda_2+\lambda_3}\end{aligne d}$

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At this stage, I'm not sure how to tie it all together to get $\displaystyle \mathbb{P}(X_1<X_2<X_3)$. Is it valid for me to state that we should simply get $\displaystyle \mathbb{P}(X_1<X_2<X_3)=\left(\frac{\lambda_1}{\la mbda_1+\lambda_2+\lambda_3}\right)\left(\frac{\lam bda_2}{\lambda_2+\lambda_3}\right)$?

I would appreciate any help with this!