$P(X<Y<Z)$ for exponentially distributed independent random variables

Question

Let $X$ $Y$ $Z$ independent, exponentially distributed random variables with parameters $a$ $b$ $c$ respectively. Calculate $P(X<Y<Z)$.

One way is to compute integral of joint density.

Second way which my textbook describe is:

$$P(X<Y<Z)=P(X<\min(Y,Z))P(Y<Z)$$

$\min(Y,Z)$ is exponentially distributed with parameter $b+c$ and the rest is easy integration.

However why is $P(X<Y<Z)=P(X<\min(Y,Z))P(Y<Z)$ true? Why not $P(X<Y<Z)=P(X<Y)P(Y<Z)$? (independent question)

Answer 1

The textbook way can be justified by the following facts for mutually independent exponential random variables $\{W_i\}_{i=1}^{\infty}$ (with rate parameters $\{\lambda_i\}_{i=1}^{\infty})$ that "race" against each other:

1. The winning time is exponentially distributed: $$\min[W_1, ..., W_n] \sim Exp(\lambda_1+...+\lambda_n)$$

2. The probability of winning is proportional to the rate parameter: $$P[W_1<W_2] = \frac{\lambda_1}{\lambda_1+\lambda_2}$$

3. The winning time is independent of who wins: $$\{W_1= min[W_1, ..., W_n]\} \quad \mbox{is independent of } \quad \min[W_1, ..., W_n]$$

So if $X,Y,Z$ are mutually independent exponential random variables with parameters $\lambda_X, \lambda_Y, \lambda_Z$, we can define $T=\min[Y,Z]$ and property 3 implies $T$ is independent of $\{Y<Z\}$. So $(X,T)$ is independent of $\{Y<Z\}$ so \begin{align} P[X<Y<Z] &= P[Y<Z]P[X<\min[Y, Z] | Y<Z] \\ &=P[Y<Z]P[X<T|Y<Z]\\ &=P[Y<Z]P[X<T]\\ &=\frac{\lambda_Y}{\lambda_Y+\lambda_Z}\cdot \frac{\lambda_X}{\lambda_X+\lambda_Y+\lambda_Z} \end{align}

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On the other hand a more direct way of approaching the problem, which would hold for any mutually independent $X,Y,Z$ that have PDFs (even if the random variables are not exponentially distributed), is to condition on all possible values of $Y$: \begin{align} P[X<Y<Z] &= \int_{-\infty}^{\infty} P[X<Y<Z|Y=y]f_Y(y)dy\\ &=\int_{-\infty}^{-\infty} P[X<y<Z|Y=y]f_Y(y)dy\\ &=\int_{-\infty}^{\infty}P[X<y<Z]f_Y(y)dy\\ &=\int_{-\infty}^{\infty} P[X<y]P[y<Z]f_Y(y)dy \end{align} Now if we use the properties of the exponential PDF we get \begin{align} P[X<Y<Z]&=\int_0^{\infty} (1-e^{-\lambda_Xy})e^{-\lambda_Zy}\lambda_Ye^{-\lambda_Yy}dy\\ &=\lambda_Y\int_0^{\infty} e^{-(\lambda_Y+\lambda_Z)y}dy - \lambda_Y\int_0^{\infty}e^{-(\lambda_X+\lambda_Y+\lambda_Z)y}dy\\ &=\frac{\lambda_Y}{\lambda_Y+\lambda_Z} - \frac{\lambda_Y}{\lambda_X+\lambda_Y+\lambda_Z}\\ &=\frac{\lambda_Y}{\lambda_Y+\lambda_Z}\cdot \frac{\lambda_X}{\lambda_X+\lambda_Y+\lambda_Z} \end{align} which is the same answer as before.