$P(X \le 2Y \le 3Z)= ?$ Suppose X,Y,Z are iid expo($\lambda$)
$P(X \le 2Y \le 3Z)= ?$
i see that : $0\le X\le 2Y$;$\qquad \frac{X}{2}\le Y\le \ \frac{3}{2}Z \qquad \frac{2}{3}Y \le Z \le \infty$
But i cannot evaluate their limits for integration $\int \int \int dxdydz$.
I am not used to dealing with 3 random variables.Any other method which can be used in these kind of situations
Thank you.
 A: Because of independece:$$P(X\leq 2Y\leq 3Z)=\iint_DP\left(\frac12x\leq Y\leq \frac32z\right)e^{-\lambda(x+z)}\ dz \ dx$$
for some domain $D$.
And because of the distribution given
$$P\left(\frac12x\leq Y\leq \frac32z\right)=\begin{cases}e^{-\lambda\frac12x}-e^{-\lambda\frac32z}&\text{ if }\ \ 0\leq x\leq3z\\0&\text{ otherwise }. \end{cases}$$
Now we have an idea about $D$:

So, our integral to be evaluated is
$$\lambda^2\int_0^{\infty}\int_{\frac13x}^{\infty}\left[e^{\lambda\frac12z}-e^{-\lambda\frac32x}\right]e^{-\lambda(x+z)}\ dz \ dx.$$
A: In general you´re on  the right track. We have just to ensure that 
$$\frac{2}{3}Y \le Z < \infty , \ \frac{X}{2}\le Y< \infty \  \textrm{and} \ 0\leq X<\infty$$
The random variable $X$ must not be further constrained, since $Y$ is already constrained by $X$. Therefore the integral is
$$P(X\leq 2Y\leq 3Z)=\int_0^{\infty} \int_{0.5\cdot x}^{\infty} \int_{2/3\cdot y}^{\infty} e^{-z\cdot \lambda} e^{-y \cdot \lambda} e^{-x \cdot \lambda}  dz \, dy \, dx$$
For $\lambda=1$ we get $P(X\leq 2Y\leq 3Z)=\frac{18}{55}$
