Are the remaining samples still independent, identically distributed (IID) after removing the maximum value of the IID samples?

Question

$X_1, X_2, \cdots, X_N$ 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?

Answer 1

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.

Answer 2

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 $\mathbb P(Y_1 = Y_2 = 1) \ne \mathbb P(Y_1 = 1) \mathbb P(Y_2=1)$.

Answer 3

$Y_1,\ldots, Y_{N-1}$ 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}