0
$\begingroup$

Let $X_1,X_2,X_3$ be independent $Exp(\lambda)$ distributed random variables. I need to find $P(X_1<X_2<X_3)$.

I think we need to use independence, then we can rewrite $P(X_1<X_2<X_3)=P(X_1<X_2)P(X_2<X_3)$, but I don't know how to proceed next. Maybe considering $Y=X_1-X_2$, but then we need to find the distribution of Y, which is hard for me.

Any help would be appreciated, thanks.

$\endgroup$
2
$\begingroup$

By symmetry,

we have

\begin{align}P(X_1<X_2<X_3)&=P(X_1<X_3<X_2)\\&=P(X_2<X_1<X_3)\\ &=P(X_2<X_3<X_1)\\ &=P(X_3<X_1<X_2)\\ &=P(X_3<X_2<X_1)\\ \end{align}

Since they have to sum up to $1$, $$P(X_1<X_2<X_3)=\frac16$$

$\endgroup$

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.