Given three independent random variables:
$X_1\sim \exp(0.1), X_2\sim \exp(0.2),X_3\sim \exp(0.3)$
I want to calculate $P(X_3<X_2<X_1)$

I tried this: $P(X_3<X_2<X_1) = P(X_2\in[X_3,X_1]) = P(X_2<X_1)-P(X_2<X_3)$

Why is the statement I wrote above not correct?

  • 2
    $\begingroup$ $P(A) -P(B)=P(A\setminus B)$ when $B \subset A$, not in general. $\endgroup$ Sep 18, 2021 at 9:41
  • $\begingroup$ Can I get a hint on how to approach this? I tried integrating but it seems a little complicated here $\endgroup$
    – Oriya
    Sep 18, 2021 at 9:43
  • $\begingroup$ To my knowledge, "lightbulb" is not a probability theory term. What is meant by that? $\endgroup$ Sep 18, 2021 at 9:58
  • $\begingroup$ The question does not seem complete. What does random variable $X_i$ represent? Also you need to state that they are independent random variables. $\endgroup$
    – Math Lover
    Sep 18, 2021 at 10:12
  • $\begingroup$ You're correct, I forgot to write that, fixed it $\endgroup$
    – Oriya
    Sep 18, 2021 at 10:30

1 Answer 1


Since the three rv's are independent. For ease I will write the rv as $x,y,z$

The joint density function for the three variables is given by:-

$$f(x,y,z)=0.1\cdot 0.2\cdot 0.3\cdot e^{-0.1x}e^{-0.2y}e^{-0.3z}=\frac{6}{1000}e^{-(0.1x+0.2y+0.3z)}\,\,\,x,y,z \in [0,\infty)^{3}$$.

So the probability is given by $(x<y<z)$



  • $\begingroup$ Why is the upper bound of the integration for y is not z? and the upper bound for x is not y? $\endgroup$
    – Oriya
    Sep 18, 2021 at 12:11
  • 1
    $\begingroup$ Because it is already taken care of when you lower bounded z by y. It is standard for multivariable calculus. Not convinced?. Convince yourself by letting the 3 exponentials to be of the same parameter. After that compute the integrals using the limits I gave. You will get $\frac{1}{6}$.Which is the logical answer if you consider that there are $3!=6$ ways to arrange $X_{1},X_{2},X_{3}$ in ascending order. so each has probability $\frac{1}{6}$ $\endgroup$ Sep 18, 2021 at 13:35

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