Let $X_i$ for $i=1,2,3$ be independent identically distributed random variables with probability density function $f(x)=e^{-x}$ for $x\in(0,\infty)$ (and $0$ otherwise). Find $$P(X_1<X_2<X_3|X_3<1).$$

By the definition of conditional probability I figured \begin{align} P(X_1<X_2<X_3|X_3<1)&=\frac{P(\left[X_1<X_2<X_3\right]\cap [X_3<1])}{P(X_3<1)}\\ &=\frac{P(X_1<X_2<X_3<1)}{P(X_3<1)} \end{align} I had to figure out a triple integral to represent the numerator and I believe it comes out to $$P(X_1<X_2<X_3<1)=\int_{0}^{1}\int_{0}^{x_3}\int_{0}^{x_3-x_2}e^{-(x_1+x_2+x_3)}\mathrm{d}x_1\mathrm{d}x_2\mathrm{d}x_3$$ and $P(X_3<1)=\int_{0}^{1}e^{-x_3}\mathrm{d}x_3$, is this correct? I am unsure about the triple integral bounds and if they satisfy the inequality given.

  • $\begingroup$ It can't be right if when you integrate for $x_2$ your upper limit also varies with $x_2$. Maybe a typo? $\endgroup$ Dec 26, 2017 at 23:59
  • $\begingroup$ @AlejandroNasifSalum fixed, will you check again? $\endgroup$
    – Teh Rod
    Dec 27, 2017 at 0:08
  • 1
    $\begingroup$ your bounds for $x_1$ are from $0$ to $x_2$ $\endgroup$
    – user365239
    Dec 27, 2017 at 0:10
  • $\begingroup$ do you have independence among $X_i$? there are other ways to do this $\endgroup$
    – user365239
    Dec 27, 2017 at 0:12
  • $\begingroup$ Yes. Like @user365239 said. $\endgroup$ Dec 27, 2017 at 0:12

1 Answer 1


I assume you have independence between the r.v., otherwise your joint density function would be wrong.

On the other hand, if you wanna be sure, just write your conditions $$0<X_1<X_2<X_3<1$$ (I added the $0$ because it's implicit in the distribution).

We could actually write an inequality for every variable as in $$\left\{\begin{matrix} X_2<X_3<1 \\ X_1<X_2<X_3 \\ 0<X_1<X_2 \\ \end{matrix}\right.$$

On the other hand, limits of the iterated integrals you propose are equivalent (via Fubini's Theorem) to a triple integral over the region of $\mathbb{R}^3$ consisting in the points $(X_1,X_2,X_3)$ that satisfy simultaneously $$\left\{\begin{matrix} 0<X_3<1 \\ 0<X_2<X_3 \\ 0<X_1<X_3-X_2\end{matrix}\right.$$

If this region were the same described before you should be able to check that the first group of conditions is true provided that this second group is so, and vice-versa. In this case, for instance, I don't think you can prove starting from the second group, that $X_1<X_2$, which is one of the conditions to prove. So they're not equivalent. (To be sure of this, we should present an element belonging to one of the region and not to the other; for instance, $(0.2,0.4,0.5)$ satisfies the first group of conditions, but not the second one, since $X_1=0.2\ge0.5-0.4=X_3-X_2$.)

To find a right description we need to pick an order of integration: let's maintain the one you used. This means that we need to find $a$, $b$, $\varphi_1$, $\varphi_2$, $\psi_1$ and $\psi_2$ such that $$\left\{\begin{matrix} a<X_3<b \\ \psi_1(X_3)<X_2<\psi_2(X_3) \\ \varphi_1(X_2,X_3)<X_1<\varphi_2(X_2,X_3)\\ \end{matrix}\right.$$ and do it in a way that is equivalent to what you have.

It is clear looking to the first group of conditions that we must choose $a=0$ and $b=1$. Then, for $X_2$ we have that this one have to be between $X_1$ and $X_3$, but the limits $\psi_1$ and $\psi_2$ can only depend on $X_3$, so we pick $\psi_2(X_3)=X_3$ and for the sake of simplicity $\psi_1(X_3)=0$, which is and immediate consequence of $0<X_1$ and $X_1<X_2$.

Finally, for $X_1$ there's no difficulty, since $\varphi_1$ and $\varphi_2$ may depend on $X_2$ and $X_3$, so we can leave this condition as it is in the first group.

Then, we have $$\left\{\begin{matrix} 0<X_3<1 \\ 0<X_2<X_3 \\ 0<X_1<X_2\end{matrix}\right.$$ and it's easy to check that every condition we needed can be immediately deducted from these and vice-versa.


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