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(Sorry that I've previously formulated the question in a wrong way, which confused everyone including myself. This is a better version of the question. Thanks!) Here's another order statistics question that I wish to ask.

Question

Consider $n$ random variables $x_1, x_2,\cdots x_n\overset{iid}{\sim} D$. Where $D$ is some unimodal on 0, symmetric, continuous distribution with a finite variance (P.S. This condition might be overly restrictive, suggestions on loosening it would be greatly appreciated! ). The PDF for nth order statistics is

$nF_{D}(x)^{n-1}f(x).$

I'm interested in the properties of the following "expected density" (I'm not sure there's a better way to put it) of the nth order PDF:

$\displaystyle\int_{-\infty}^{+\infty}(n-1)F_{D}(x)^{n-2}f_{D}(x)\times f_{D}(x)\:dx$,

which simplifies to

$\displaystyle\int_{-\infty}^{+\infty}(n-1)F_{D}(x)^{n-2}f_{D}(x)^2\:dx$

What I want to show is that this expression decreases as $n$ increases.

What I have gotten so far:

By an integration by parts trick, we can show that the above can be expressed as:

$\int_{-\infty}^{\infty}f_{D}(x)dF_{D}(x)^{n-1}$

$=f_{D}(x)F_{D}(x)^{n-1}|_{-\infty}^{+\infty}-\int_{-\infty}^{+\infty}F_{D}(x)^{n-1}df_{D}(x)$

$=-\int_{-\infty}^{+\infty}F_{D}(x)^{n-1}f^{\prime}_{D}(x)dx.$

Intuitively, I can see that as $-f^{\prime}_{D}(x)$ is positive on $x>0$ and negative on $x<0$. And as $n$ increases $F^n_{D}(.)$ shifts more mass to extreme values where $f^{\prime}_{D}(x)$ is very close to 0. So, eventually the whole integral will become smaller as $n$ increases. But I'm not sure how to proceed this argument formally. Any help will be greatly appreciated!

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  • $\begingroup$ Can you explain how the first integral (the expected probability) is obtained? $\endgroup$ – Juho Kokkala Aug 5 '14 at 11:21
  • $\begingroup$ @JuhoKokkala I think this is simply the pdf for x being nth order statistics times integrating over all possible values of x (the pdf). $\endgroup$ – Jacob Yerger Aug 5 '14 at 12:27
  • $\begingroup$ The chance that the $k^{\text{th}}$ draw of $n$ variables is the maximum is $1/n$, because each draw has an equal chance and those chances sum to unity. (This conflicts with the integral in the question, which is incorrect.) That leads immediately to an answer to your question whenever $F$ is continuous and it can easily be generalized to discontinuous $F$. $\endgroup$ – whuber Aug 5 '14 at 12:43
  • $\begingroup$ @whuber I'm so sorry that I wasn't clear enough. I was actually interested in the property of the "expected density", rather than probability. The question is revised if you are still interested. $\endgroup$ – Jacob Yerger Aug 5 '14 at 13:22
  • $\begingroup$ Re the edit: what does that integral have to do with the density of the maximum? It has certain traits that suggest it may have been erroneously derived in some kind of analysis. If so, you would be poorly served to have this question answered as it stands, because you would get the correct answer to the wrong question. Where, then, does this integral come from? What is it intended to represent? Why is it of interest? $\endgroup$ – whuber Aug 5 '14 at 13:25
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Expectation of a conditional probability is the total probability, thus the expected probability would be just the probability that $x_k$ is the maximum, which in the case of continuous distribution is $1/n$, from which decreasing in $n$ immediately follows. (This was pointed out in the comments by whuber). In this answer, I also show how to formulate the expectation integral correctly and evaluate it to be $1/n$ - this is not needed for proving the result, but may be useful for learning purposes to see what went wrong in the original approach.

The first integral is incorrect. The conditional probability of $x_k$ being the maximum (conditional on of $x_k$) is simply the probability that all others are less than (or equal, which does not matter in the continuous case) $x_k$: \begin{equation} \mathbb{P}(x_k = \max(x_1,\ldots,x_n) \mid x_k) = \mathbb{P}(x_1\leq x_k,\ldots,x_{k-1}\leq x_k,x_{k+1}\leq x_k,\ldots,x_n \leq x_k) \end{equation} Applying the independence property and the definition of the CDF $F_D$, we obtain \begin{equation} =\mathbb{P}(x_1\leq x_k)\,\ldots\,\mathbb{P}(x_{k-1}\leq x_k)\,\mathbb{P}(x_{k+1}\leq x_k)\ldots\,\mathbb{P}(x_n\leq x_k)=F_D(x_k)^{n-1}. \end{equation} Then, to obtain the expectation of this random variable, we average over different values of $x_k$ weighting by the PDF of $x_k$: \begin{equation} \mathbb{E}(\mathbb{P}(x_k = \max(x_1,\ldots,x_n) \mid x_k)) = \mathbb{E}(F_D(x_k)^{n-1}) = \int_{-\infty}^{\infty}f_D(x)F_D(x)^{n-1}dx. \end{equation} Using the chain rule, the derivative of $\frac{1}{n}F_D(x)^n$ w.r.t. $x$ is $\frac{1}{n}f_D(x)\,n\,F_D(x)^{n-1}$ $= f_D(x)F_D(x)^{n-1}$, and thus our last integral equals \begin{equation} = \left. \frac{1}{n}F_D(x)^n \right|_{x=-\infty}^{x=\infty} = \frac{1}{n}. \end{equation}

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  • $\begingroup$ Thank you so much for this. But I think what I am really looking for is the prove that the specific expression I have in the question decreases with n. I'm so sorry that I've confused everyone including my self. I'd greatly appreciate if you would want to give it another try. Thanks~ $\endgroup$ – Jacob Yerger Aug 5 '14 at 13:19
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    $\begingroup$ Your product-of-probabilities formula is too easy to misread: each event, of the form $x_i\le x_k$, appears to involve two independent random variables $x_i$ and $x_k$. Those probabilities would not be given by $F_D$. It would be very helpful to use a notation that makes it clear that this equation views $x_k$ as a number and the other $x_i, i\ne k$, as random variables. $\endgroup$ – whuber Aug 5 '14 at 13:29
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    $\begingroup$ @whuber Good point, thanks for pointing that out. I'm not sure whether I should correct that or delete this answer completely as the question has been edited to ask something else. $\endgroup$ – Juho Kokkala Aug 5 '14 at 13:48

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