Conditional distribution of one of the two exponential random variables, given one is smaller than the other

Question

Let $X$ be a random variable with exponential distribution with parameter $a$, i.e. $X\sim Exp(a)$. See https://en.wikipedia.org/wiki/Exponential_distribution for the definition of exponential distribution and its pdf. Let $Y$ be a random variable with $Y\sim Exp(b)$. Suppose $X$ and $Y$ are independent. Let $A$ denote the event $X<Y$. Given $A$, the conditional distribution of $X$ is $Exp(a+b)$, if I compute it correctly. This happens to be the (unconditional) distribution of the random variable $Z=\min\{X,Y\}$. This observation leads me to the conjecture that the conditional distribution of $Y$ given $A$ should be the same as the (unconditional) distribution of $W=\max\{X,Y\}$. But I found that the cdf of $W$ is $1-e^{-at}-e^{-bt}+e^{-(a+b)t}$, whereas the cdf of the $Y$ given $A$ is $1+\frac{b}{a}e^{-(a+b)t}-\frac{a+b}{a}e^{-bt}$. Is my conjecture wrong? Or did I make some mistakes in calculations?

Answer 1

Given $A$, the conditional distribution of $X$ is $\mathcal{Exp}(a+b)$, if I compute it correctly.

Yes, that is the case.

$$\begin{align}f_{X\mid A}(x) &= \dfrac{f_X(x)\,\mathsf P(Y\geq x)}{\mathsf P(X<Y)}\\&=\dfrac{a\,\mathrm e^{-ax}\cdot\mathrm e^{-b x}}{a/(a+b)}\,\mathbf 1_{0\leqslant x}\\&=(a+b)\,\mathrm e^{-(a+b) x}\,\mathbf 1_{0\leqslant x}\end{align}$$

This happens to be the (unconditional) distribution of the random variable Z=min{X,Y}.

No. Well, yes, but no. This is the conditional distribution of $X$ when given that it is the minimum of $\{X,Y\}$. That is not the same thing.

Although, yes, the Law of Total Probability does add up to the coincidental:

$$\begin{align}f_Z(z) &= f_{\small X\mid A}(z)\mathrm P(X<Y)+f_{\small Y\mid A^{\tiny\complement}}(z)\mathrm P(Y<X)\\ &= (a+b)\mathrm e^{-(a+b) z}\,\mathbf 1_{0\leqslant z}\end{align}$$

---

It just does not work out that neatly for the maximum.

To evaluate the conditional probability density of $Y$ given $A$ in the same manner as your first calculation:

$$\begin{align}f_{Y\mid A}(y) &= \dfrac{f_Y(y)\,\mathsf P(X\leqslant y)}{\mathsf P(X<Y)}\\&=\dfrac{b (a+b)}{a}\,\left(\mathrm e^{-by}-\mathrm e^{-(a+b) y}\right)\,\mathbf 1_{0\leqslant y}\end{align}$$

$$\begin{align}f_W(w) &= f_{X\mid A^{\complement}}(w)\mathsf P(Y<X)+f_{Y\mid A}(w)\mathsf P(X<Y)\\&= (a(\mathrm e^{-aw}-\mathrm e^{-(a+b)w})+b(\mathrm e^{-bw}-\mathrm e^{-(a+b)w}))\,\mathbf 1_{0\leqslant w}\\&=(a\mathrm e^{-aw}+b\mathrm e^{-bw}-(a+b)\mathrm e^{-(a+b)w})\,\mathbf 1_{0\leqslant w}\end{align} $$

---

Is my conjecture wrong? Or did I make some mistakes in calculations?

It was a nice observation, however your conjecture has proven wrong.

The distribution of $Y$ when it is maximum, and the distribution of $X$ when it was the minimum, did not lend to an argument for symmetry.