# Finding $P(X_3<X_2<X_1)$ for three exponential random variables

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

I tried this: $$P(X_3

Why is the statement I wrote above not correct?

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

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

$$\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}\frac{6}{1000}e^{-(0.1x+0.2y+0.3z)}\,dz\,dy\,dx$$

$$\int_{0}^{\infty}\int_{x}^{\infty}\frac{2}{100}e^{-(0.1x+0.5y)}dy\,dx=\int_{0}^{\infty}\frac{1}{25}e^{-0.6x}\,dx=\frac{1}{15}$$

• Why is the upper bound of the integration for y is not z? and the upper bound for x is not y? Sep 18, 2021 at 12:11
• 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 $\frac{1}{6}$ Sep 18, 2021 at 13:35