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I am really stuck on the following question during my exam preparation.

Let $X_1$, $X_2$ and $X_3$ be independent exponential random variables with respective rates $\mu_1$, $\mu_2$ and $\mu_3$. Find:
a) $P(X_1 < X_2 < X_3)$
b) $E(X_2 | X_1 < X_2 < X_3) $

Question a

For question a) I've tried two approaches but they do not give me the same answer. Why the discrepancy and is one of the two even correct?

First approach

$P(X_1 < X_2 < X_3) =$

[using the law of total probability]
$ = P(X_1 < X_2 < X_3 | min X_i = X_1) P(min X_i = X_1) + P(X_1 < X_2 < X_3 | min X_i = X_2) P(min X_i = X_2) + P(X_1 < X_2 < X_3 | min X_i = X_3) P(min X_i = X_3)$

[The second and third term lead to contradictions and therefor are 0]
$= P(X_1 < X_2 < X_3 | min X_i = X_1) P(Xi = min X1)$ $= P(X_1 < X_2 < X_3, min X_i) = P(X_1 < X_2 < X_3, X_1 < X_2, X_1 < X_3)$
$= P(X_2 < X_3, X_1 < X_2, X_1 < X_3)$ [first entry contains redundant information]
$= P(X_2 < X_3)P(X_1 < X_2)P(X_1 < X_3)$ [They are independent]
$= u_2/(u_2+u_3) \cdot u_1/(u_1+u_2) \cdot u_1/(u_1+u_3)$

Second approach

The joint distribution is given by $f(x_1,x_2,x_3) = f_1(x_1)f_2(x_2)f_3(x_3)$ due to independence

$P(X_1 < X_2 < X_3) = \int_{x_2}^{\infty}\int_0^{\infty}\int_0^{x_2}f_1(x_1)f_2(x_2)f_3(x_3)dx_1 dx_2 dx_3 $ $= \int_0^{\infty}\int_0^{x_2}f_1(x_1)dx_1 \int_{x_2}^{\infty}f_3(x_3)dx_3 f_2(x_2)dx_2$

$= u_2/(u_2+u_3) - u_2/(u_1 + u_2 + u_3) $

Question b

For question b) I want to do the following:

$\int_{x_2}^{\infty}\int_0^{\infty}\int_0^{x_2}x_2*f_1(x_1)f_2(x_2)f_3(x_3)dx_1 dx_2 dx_3 $ am I on the right track?

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  • $\begingroup$ I've fixed the formatting of the subscripts. Is your answer still 1/6? $\endgroup$ – Henk Apr 17 '16 at 15:54
  • $\begingroup$ Not the same, I did not see the $\mu_i$. My fault, I have trouble reading non-TeX stuff. $\endgroup$ – André Nicolas Apr 17 '16 at 16:08
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Your first approach has a flaw, because:

$$\mathsf P(X_2 <X_3 ,X_1 <X_2 ,X_1 <X_3 )\neq \mathsf P(X_2 <X_3 )~\mathsf P(X_1 <X_2 )~\mathsf P(X_1 <X_3 )$$

The random variables are independent, but the events of their pairwise ordering are not.   When given that $X_3$ is somewhat larger than $X_2$ (whatever that is) then you would be more likely to anticipate that it is also larger than $X_1$ than without that information. $$\mathsf P(X_1<X_3\mid X_2<X_3)\gt \mathsf P(X_1<X_3)$$

Your second approach holds up:

$$\begin{align} \mathsf P(X_1<X_2<X_3) =&~ \int_0^\infty F_{X_1}(x) f_{X_2}(x)(1-F_{X_3}(x))\operatorname d x \\[1ex] =&~ \int_0^\infty (1-e^{-\mu_1 x})\mu_2 e^{-\mu_2 x} e^{-\mu_3x}\operatorname d x \\[1ex] =&~ \mu_2\int_0^\infty e^{-(\mu_2+\mu_3) x}-e^{-(\mu_1+\mu_2+\mu_3) x} \operatorname d x \\[1ex] =&~ \dfrac{\mu_2}{\mu_1+\mu_2}-\dfrac{\mu_2}{\mu_1+\mu_2+\mu_3} \end{align}$$


For question (b) you are on the right track, but you need to normalise: $$\begin{align}\mathsf E(X_2\mid X_1<X_2<X_3) =&~ \dfrac{\int_0^\infty x f_{X_2}(x) F_{X_1}(x) (1-F_{X_2}(x))\operatorname d x}{\int_0^\infty f_{X_2}(x) F_{X_1}(x) (1-F_{X_2}(x))\operatorname d x} \end{align}$$

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HINT:

a) $P(X_1 < X_2 < X_3) = \int_{0}^{\infty} P(X_1 < X_2 < X_3 \ | X_2 = t)\ dP(X_2=t) = \int_{0}^{\infty} P(X_1 < t < X_3 ) \ f_{X_2}(t)\ dt$

where $P(X_1 < t < X_3) = P(X_1<t) \cdot (1 - P(X_3 \leq t)) = (1-e^{-\mu_{1} \ t}) \cdot e^{-\mu_{3} \ t}$

b) $ E(X_2 | X_1 < X_2 < X_3) = E(X_2 \cdot \mathbb{1}_{\{X_1 < X_2 < X_3\}} \ ) \ / \ P(X_1 < X_2 < X_3)$

where $E(X_2 \cdot \mathbb{1}_{\{X_1 < X_2 < X_3\}} \ ) = \int_{x_2}^{\infty} \int_{0}^{\infty} \int_{0}^{x_2} \ x_2 \ f_{X_1}(x_1) f_{X_2}(x_2) f_{X_3}(x_3) \ dx_1 dx_2 dx_3 $

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