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?
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 $(x<y<z)$
$$\int_{0}^{\infty}\int_{x}^{\infty}\int_{y}^{\infty}\frac{6}{1000}e^{-(0.1x+0.2y+0.3z)}\,dz\,dy\,dx$$
$$\int_{0}^{\infty}\int_{x}^{\infty}\frac{2}{100}e^{-(0.1x+0.5y)}dy\,dx=\int_{0}^{\infty}\frac{1}{25}e^{-0.6x}\,dx=\frac{1}{15}$$