Conditional Probability Distribution of Max$(X, Y)$ given $X$

Question

Say $X$ and $Y$ are I.I.D uniform variables in $[0,1]$. Conditional Probability Distribution of Max$(X, Y)$ given $X$. I think it would be simply $P(X)$ when $X>=Y$ and $P(Y)$ the other way around. Is that so?

Answer 1

Let $Z := \max(X, Y).$ I assume that you are actually interested in something like $\mathbb{P}(Z \geq t \vert X \geq s)$ for some $s, t \in [0, 1],$ but please do correct me if I'm wrong. Note that if $t \leq s,$ then obviously $\mathbb{P}(Z \geq t \vert X \geq s) = 1.$ If $t \geq s,$ then $\mathbb{P}(Z \geq t \vert X \geq s) = \frac{\mathbb{P}((Z \geq t) \cap (X \geq s))}{\mathbb{P}(X \geq s)} = \frac{\mathbb{P}(X \geq t, X \geq s) + \mathbb{P}(X < t, Y \geq t, X \geq s)}{1 - s} = \frac{t + (t - s)(1 - t)}{1 - s} = \frac{- t^2 + (2 + s)t - s}{1 - s}.$

Also note that the CDF of $Z$ can be computed in the following way ($t \in [0, 1]$ is fixed). $\mathbb{P}(Z \leq t) = \mathbb{P}(X \leq t, Y \leq t) = \mathbb{P}(X \leq t) \mathbb{P}(Y \leq t) = t^2.$

I hope this helps. :)

Answer 2

Let $Z=\max\{X,Y\}$ with X, Y iid uniform on [0,1]. Then $f(\max\{X, Y\}|X=\epsilon)=f(\max\{\epsilon, Y\})$ has a mixed distribution. $P(Z=\epsilon|X=\epsilon)=\epsilon$. For $\epsilon <Z\le1$, $f(Z|X=\epsilon)=1$, and $f(Z|X=\epsilon)=0$ otherwise.