# 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?

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

• Dear Prof Robert Israel, Thank you for this common. In the attached reference, the $X_i$s are IID exponential random variables, and the author claims that the remaining $X_i$s after censoring maximum value preserve IID, but no proof has been given for it. I am eager to know your opinion about this.
– Leon
Sep 9 '20 at 1:52

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

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

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.

• Dear Svenson, The set $Y_1,...Y_{N-1}$ is equal the set $X_1,...X_{N-1}$ after removing the maximum value, and only its name has changed. my question is that the mentioned strategy for calculating the PDF of $Y_1,...Y_{N-1}$ is correct, and the $Y_1,...Y_{N-1}$ samples are IID or not?
– Leon
Sep 8 '20 at 18:32
• I edited my answer and completely rewrote it. Since you mentioned exponential distributions, I tried calculating it directly, and now I think they are not independent. Sorry for the misleading answer earlier. It would seem that there is an error in the paper, unless there are some additional details that weren't mentioned. Sep 9 '20 at 18:35

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