Suppose I have three independent continuous random variables $X$, $Y$ and $Z$. I want to know the probability of $X=a$ given that the minimum of the random variables is $a$.
Attempt:
We want to find $P(X=a |$min$\{X,Y,Z\} = a)$ which implies we have $$\dfrac{P(X=a \cap \text{min}\{X,Y,Z\} = a)}{P(\text{min}\{X,Y,Z\} = a)} = \dfrac{P(X= \text{min}\{X,Y,Z\})}{P(\text{min}\{X,Y,Z\} = a)}$$
For the top part, I think I can solve it by finding $P(X<Y, X<Z)$ which is easy by conditioning on $X$. However for the bottom part, is this even doable given that they are all continuous? Isn't it the fact that the probability of equalities in continuous R.V. is $0$?
I have also thought about assigning probability distribution on the conditional R.V. $(X|\text{min}\{X,Y,Z\} = a)$ but got nothing.
Any help would be appreciated!
By recognizing that $$\{\min(X,Y,Z)=a\}=\{X=a,Y\geq a,Z\geq a\}\cup \{Y=a,X\geq a,Z\geq a\}\cup \{Z=a,X\geq a,Y\geq a\}$$ we gather that the conditional distribution of $(X,Y,Z)$ given $\min(X,Y,Z)=a$ has pdf $$f_{X,Y,Z|E}(x,y,z)=\frac{f_X(x)f_Y(y)f_Z(z) \cdot 1_{E}(x,y,z)}{\iint_{E}f_X(x)f_Y(y)f_Z(z)\mathrm{d}S}$$ where $E$ is the surface $$E=\{x=a,y\geq a,z\geq a\}\cup \{y=a,x\geq a,z\geq a\}\cup \{z=a,x\geq a,y\geq a\}$$ This conditional density can be written as $$f_{X,Y,Z|E}(x,y,z)=\frac{1}{C}f_X(x)f_Y(y)f_Z(z)\cdot 1_{E}(x,y,z)$$ where $$C=\int_a^{\infty}\int_a^{\infty}f_X(a)f_Y(y)f_Z(z)\mathrm{d}y\mathrm{d}z+\int_a^{\infty}\int_a^{\infty}f_X(x)f_Y(a)f_Z(z)\mathrm{d}x\mathrm{d}z+\int_a^{\infty}\int_a^{\infty}f_X(x)f_Y(y)f_Z(a)\mathrm{d}x\mathrm{d}y$$ Integrating $f_{X,Y,Z|E}$ over the surface $\{x=a\}\cap E$ yields your desired probability: $$\frac{1}{C}\int_a^{\infty} \int_a^{\infty}f_X(a)f_Y(y)f_Z(z)\mathrm{d}y\mathrm{d}z$$ Note this above probability equals $1/3$ whenever $X,Y,Z$ are identically distributed i.e. $f_X\equiv f_Y \equiv f_Z$