P(X

Question

Let's have $X,Y,Z$ independently exponentially distributed with parameters respectively $\lambda_x, \lambda_y, \lambda_z$. I want to calculate $$P(X<Y<Z)$$

My work so far
According to $P(X<Y)$ where X and Y are exponential with means $2$ and $1$

I know that $P(X<Y)=\frac{\lambda_x}{\lambda_x+\lambda_y}$ and $P(Y<Z)=\frac{\lambda_y}{\lambda_y+\lambda_z}$

I want to combine those two to calculate initial probability.

$P(\{X<Y<Z\})=P(\{X<Y\} \cap\{Y<Z\}) \neq P(\{X<Y\})P(\{Y<Z\})$ because indeed $X,Y,Z$ are independent but $X<Y$ and $Y<Z$ are not (due to $Y$ appearance in both inequalities). So I had a problem with this calculation.

Could you please give me a hand with calculation above ? Is there any way how can we smoothly use our knowledge with $P(X<Y)$ to calculate $P(X<Y<Z)$ ?

Answer 1

the links in the comment sure will answer your question but I would like to explain in very poor words how you can attack this kind of problem.

You are requested to calculate

$$\mathbb{P}[X<Y<Z]=\int_{\Omega}f_{XYZ}(x,y,z)dxdydz$$

Where

$$\Omega=\{(x,y,z) \in \mathbb{R}^3: 0<x<y<z<\infty, x,y,z>0\}$$

So first solve the integral in the $X-Plane$

$$\int_{0}^{y}f_{XYZ}(x,y,z)dx$$

then apply the result in the remainig bivariate domain:

$$0<y<z$$

to get the final result.

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Simplyfied Example: Just to understand the procedure avoiding boring calculation, let's solve the same problem starting from

$X,Y,Z$ iid uniform in $(0;1)$

1. First integrate the joint density over the $X-Plane$ obtaining

$$\int_0^y dx=y$$

2. Second integrate the result in the remaining bivariate domain

$$\int_0^1 dz\int_0^z ydy=\frac{1}{6}$$

Answer 2

If $f_Y$ denotes the probability density function of $Y$, then my suggestion would be to solve the intregral $\int_0^\infty f_Y(y) \cdot \mathbb P(X \leq y) \cdot \mathbb P(Z > y) \: \text dy$.