Conditional distribution $\mathbb{P}(X_1 | \max \{X_1,...,X_n\})$

Question

few days ago i asked similar thing about conditional expectation and at first leonbloy's and Eupraxis's solutions seemed logical to me. Right now i can't explain myself following problem:

Let $X_i$ be i.i.d $U([0,1])$. And $Z:=\max\{X_1,...,X_n\}$.

Then i understand that $\mathbb{P}(X_1=z|Z=z)=\frac{1}{n}$. Since $\mathbb{P}(X_1\leq z|Z=z)=1$ it follows that \begin{align*} \mathbb{P}(X_1 < z|Z=z)=\frac{n-1}{n}. \end{align*}

But i was told before $X_1$ conditioned being less than $Z=z$ has $U([0,z))$ distribution. Does it not imply that $\mathbb{P}(X_1 <a |Z=z)=\frac{a}{z-0}$ i.e $\mathbb{P}(X_1 <z|Z=z)=\frac{z}{z-0}=1$?

Something is definetly wrong, but what did i misunderstand?

The cdf in my case is probably: \begin{align*} \mathbb{P}(X_1\leq t|Z=z)= \begin{cases} 0 & \text{ if } t\leq0 \\ \frac{t(n-1)}{zn} & \text{ if } 0 < t <z \\ 1 & \text{ elsewhere} \end{cases} \end{align*}

Answer 1

Your $\mathbb{P}(X_1 \lt z|Z=z)=\frac{n-1}{n}$ looks correct.

Your $\mathbb{P}(X_1 \lt a |Z=z)=\frac{a}{z-0}$ and $\mathbb{P}(X_1 \lt z|Z=z)=\frac{z}{z-0}=1$ are in fact conditional probabilities and might be rewritten as: $$\mathbb{P}(X_1 \lt a |Z=z \land X_1 \lt z )=\frac{a}{z}$$ $$\mathbb{P}(X_1 \lt z|Z=z \land X_1 \lt z)=1$$ or, taking account of $\mathbb{P}(X_1 \lt z|Z=z)=\frac{n-1}{n}$, as: $$\mathbb{P}(X_1 \lt a \le z |Z=z )=\frac{(n-1)a}{nz}$$ $$\mathbb{P}(X_1 \lt z|Z=z )=\frac{n-1}{n}.$$

Your cumulative distribution function looks correct. Note that it is discontinuous at $t=z$.

Answer 2

This is basically equivalent to Henry's answer.

Let $E$ be the event $X_1=Z$.

$$\mathbb{P}(X_1 \lt a \mid Z=z)=\mathbb{P}(X_1 <a \land E \mid Z=z)+\mathbb{P}(X_1 <a \land E^c \mid Z=z)$$

For $0<a \le z$ the first term is zero, so

$$\begin{array}{rcl} \mathbb{P}(X_1 \lt a \mid Z=z)&=&\mathbb{P}(X_1 <a \land E^c \mid Z=z)\\ &=&\mathbb{P}(X_1 <a \mid Z=z \land E^c ) \, \mathbb{P}(E^c\mid Z=z) \\ &=&\frac{a}{z} \frac{n-1}{n} \end{array} $$

You are wrongly assuming that $\mathbb{P}(X_1 <a \mid Z=z)$, when $a<z$, should equal $\mathbb{P}(X_1 <a \mid Z=z \land E^c )$, but this is not true.