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?
Conditional distribution of one of the two exponential random variables, given one is smaller than the other
conditional probabilityexponential distributionprobabilityprobability distributionsprobability theory
Best Answer
Yes, that is the case.
$$\begin{align}f_{X\mid A}(x) &= \dfrac{f_X(x)\,\mathsf P(Y\geq x)}{\mathsf P(X<Y)}\\&=\dfrac{a\,\mathrm e^{-ax}\cdot\mathrm e^{-b x}}{a/(a+b)}\,\mathbf 1_{0\leqslant x}\\&=(a+b)\,\mathrm e^{-(a+b) x}\,\mathbf 1_{0\leqslant x}\end{align}$$
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:
$$\begin{align}f_{Y\mid A}(y) &= \dfrac{f_Y(y)\,\mathsf P(X\leqslant y)}{\mathsf P(X<Y)}\\&=\dfrac{b (a+b)}{a}\,\left(\mathrm e^{-by}-\mathrm e^{-(a+b) y}\right)\,\mathbf 1_{0\leqslant y}\end{align}$$
$$\begin{align}f_W(w) &= f_{X\mid A^{\complement}}(w)\mathsf P(Y<X)+f_{Y\mid A}(w)\mathsf P(X<Y)\\&= (a(\mathrm e^{-aw}-\mathrm e^{-(a+b)w})+b(\mathrm e^{-bw}-\mathrm e^{-(a+b)w}))\,\mathbf 1_{0\leqslant w}\\&=(a\mathrm e^{-aw}+b\mathrm e^{-bw}-(a+b)\mathrm e^{-(a+b)w})\,\mathbf 1_{0\leqslant w}\end{align} $$
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.