Are the remaining samples still independent, identically distributed (IID) after removing the maximum value of the IID samples? $X_1, X_2, \cdots, X_N$ are independent identically distributed (IID) random variables and $Y_1$, $Y_2, \cdots, Y_{N-1}$ are the remaining after removing the maximum value of $X_k, k=1, \cdots, N .$ Is this assumption true that $Y_1$, $Y_2, \cdots, Y_{N-1}$ are IID?
 A: This is not true if the common distribution of the $X_i$ is discrete (presumably, when there is a tie for maximum, only one of the maximum values is removed).  For example, suppose $N=3$ and $X_i$ has a Bernoulli distribution with $p=1/2$.  Then the possibilities for
$(Y_1, Y_2)$ are $(0,0)$ with probability $1/2$ (namely this happens if there is at most one $X_i=1$), $(1,0)$ and $(0,1)$ with probabilities $3/16$ each,
and $(1,1)$ with probability $1/8$, and it is easy to see that $\mathbb P(Y_1 = Y_2 = 1) \ne \mathbb P(Y_1 = 1) \mathbb P(Y_2=1)$.
A: OK, I tried to redo this with exponential distributions and $N=3$. Let $X,Y,Z$ be independent and exponentially distributed with parameter $\lambda$. I believe that what you want is the joint cdf given by
$$P\left(X>x,Y>y\mid X\leq Z,Y\leq Z\right) = \frac{P\left(x<X\leq Z,y<Y\leq Z\right)}{P\left(X\leq Z,Y\leq Z\right)}.$$
Here all you are told is that $Z$ is greater than both $X$ and $Y$, but otherwise you are not told the value of $Z$ (it is "censored") and you are not told anything about the ordering of $X$ and $Y$ relative to each other.
The denominator of the RHS is $\frac{1}{3}$ since they are i.i.d., so any permutation is equally likely. To get the numerator, we can write
\begin{eqnarray*}
P\left(x<X\leq Z,y<Y\leq Z\right) &=& \mathbb{E}\left(P\left(x<X\leq Z,y<Y\leq Z\,|\,Z\right)\right)\\
&=& \mathbb{E}\left(P\left(x<X\leq Z\mid Z\right)P\left(y<Y\leq Z\mid Z \right)\right)\\
&=& \int^\infty_{\max\left(x,y\right)} \left(e^{-\lambda x}-e^{-\lambda z} \right) \left(e^{-\lambda y}-e^{-\lambda z}\right)\lambda e^{-\lambda z} \, dz.
\end{eqnarray*}
One can work this out, but because the lower limit is $\max\left(x,y\right)$, I don't think there is any way to separate the result into a product of something that depends only on $x$ and something that depends only on $y$. So, the answer would seem to be that $X$ and $Y$ are not conditionally independent given the event $\left\{X\leq Z,Y\leq Z\right\}$. Sorry for the misleading answers earlier. I think this should be much more precise.
A: $Y_1,\ldots, Y_{N-1}$ are not independent. That can be seen by proving that
$$
\Pr(Y_1>3\mid Y_2,\ldots,Y_{N-1}<3) < \Pr(Y_1>3)
$$
under the assumption that $\Pr(X_1<3)>0.$
Proof:
Let $I = \begin{cases} 1 & \text{if } \max\{X_1,\ldots,X_N\}>3, \\ 0 & \text{otherwise.} \end{cases}$
Then
\begin{align}
& \Pr(Y_1>3\mid Y_2,\ldots,Y_{N-1}<3) \\[8pt]
= {} & \operatorname E(\Pr(Y_1>3\mid I) \mid Y_2,\ldots,Y_{N-1} < 3) \\[8pt]
= {} & \phantom{{}+{}} \Pr(Y_1>3 \mid I=0\ \&\ Y_2,\ldots,Y_{N-1} < 3) \Pr(I=0\mid Y_2,\ldots,Y_{N-1}<3) \\
& {} + \Pr(Y_1>3\mid I=1\ \&\ Y_2,\ldots,Y_{N-1} < 3) \Pr(I=1\mid Y_2,\ldots,Y_{N-1} < 3) \\[12pt]
= {} & 0 + \Pr(Y_1>3\mid I=1\ \&\ Y_2,\ldots,Y_{N-1} < 3) \Pr(I=1\mid Y_2,\ldots,Y_{N-1} < 3) \\[8pt]
= {} & \Pr(Y_1>3\mid I=1\ \&\ Y_2,\ldots,Y_{N-1} < 3) \Pr(Y_1 > 3) \\[8pt]
< {} & \Pr(Y_1>3). 
\end{align}
