2
$\begingroup$

I'm struggling to see how the distribution function of an order statistic is obtained. So far, from here, I understand that the distribution function $F_k$ of $X_{(k)}$ is given by:

\begin{equation} \sum_{j=k}^n \binom{n}{j} \left[F(x)\right]^j \left[1 - F(x)\right]^{n - j} \end{equation}

Now, I want to find the CDF of $F_1(x)$, that is, I choose $k$ to be equal to 1. I'm striving to obtain the same solution shown in the link above, which is

\begin{equation} F_1(x) = 1 - \left[1 - F(x)\right]^n. \end{equation}

I tried to use the transformations and change of variables that are used here. Mainly this one: $\dbinom r l = \dfrac r l \dbinom {r - 1} {l - 1}$. The problem is that, in our case, $l$ depends on the summation so I'm not able to take it out to do the change of variables.

Could anyone please shed some light on this? Is there a better way to solve this? Is there a closed expression for any $k$?

Thanks in advance for the time you take in reviewing this question.

$\endgroup$
  • $\begingroup$ Hint: raise $1=F(x)+(1-F(x))$ to the $n$ power and apply the Binomial Theorem to obtain $$1^n=\sum_{j=0}^n\binom{n}{j}F(x)^j(1-F(x))^{n-j}=(1-F(x))^n + F_1(x),$$then solve. $\endgroup$ – whuber Jun 8 '17 at 18:40
  • $\begingroup$ I know, but I think the problem is that, in the Binomial Theorem, the summation is from $j = 0$ to $j = n$. Conversely, in this case the summation is from $j = k$. If I try to apply $\dbinom r l = \dfrac r l \dbinom {r - 1} {l - 1}$ to make the summation go from $j = 0$ to $j = n$ via change of variable, a new term appears and I cannot apply the Binomial Theorem. Is there something I am missing? Maybe I am completely wrong. $\endgroup$ – Gabriel Jun 9 '17 at 10:18
  • $\begingroup$ All I can imagine is that you might be misinterpreting the summation notation. This identity is much simpler than you might think. Why not write out the sum for a small $n$, such as $n=2$, so you can see what it means? $\endgroup$ – whuber Jun 9 '17 at 14:43
  • $\begingroup$ Thanks for your response whuber . I still don't see it but I will try to do it. Thanks again. $\endgroup$ – Gabriel Jun 12 '17 at 12:31
1
$\begingroup$

Ok, so finally I think I have solved it. I leave here the solution for anyone who is curious about it:

Let $F(x)$ be the CDF fo the random variable from which we sample points:

We know that the CDF of the $k$-th order statistic is

\begin{equation}\label{eq:original} F_k(x) = \sum_{j=k}^n \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} \end{equation}

Additionally, note that, by the Binomial Theorem:

\begin{equation}\label{eq:binomialTh} \sum_{j=0}^n \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} = [ F(x) + (1 - F(x))]^n = 1^n \end{equation}

By using eq.~\ref{eq:original}, we may expand eq.~\ref{eq:binomialTh} as follows:

\begin{equation}\label{eq:combination} 1^n = \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} + \sum_{j=k}^{n} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} \end{equation}

Note, in eq.~\ref{eq:combination}, that the second summation is the expression of $F_k(x)$. Therefore, we may rewrite~\ref{eq:combination} as follows and derive the expression for $F_k(x)$:

\begin{equation} 1 = \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} + F_k(x) \end{equation}

\begin{equation}\label{eq:final} F_k(x) = 1 - \sum_{j=0}^{k-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n-j} \end{equation}

Now, particularising eq.~\ref{eq:final} for both $k = 1$ and $k = n$, we obtain the following expressions for the $1$st and $n$-th order statistics.

\begin{equation} F_1(x) = 1 - \sum_{j=0}^{0} \binom{n}{0} F(x)^0 [1 - F(x)]^{n} = 1 - [1 - F(x)]^{n} \end{equation}

\begin{equation} \begin{aligned} F_n(x) &=1 - \sum_{j=0}^{n-1} \binom{n}{j} F(x)^j [1 - F(x)]^{n} = \\ &= 1 - [1 - \sum_{j=n}^{n} \binom{n}{n} F(x)^n [1 - F(x)]^{n-n}] = \\ &= 1 - [1 - F(x)^n] = F(x)^n \end{aligned} \end{equation}

I would appreciate if anyone could help me with labeling equations in this forum, since it does not appear to work as in standard latex.

Regards.

$\endgroup$
  • $\begingroup$ This post seems overly elaborate for what it conveys. Isn't it just trying to say that $\sum_{j=0}^n x_j = x_n + \sum_{j=0}^{n-1}x_j$ for any sequence $x_j$? $\endgroup$ – whuber Jun 15 '17 at 18:20
  • $\begingroup$ That is the case for $F_n(x)$. In a general case I would say the main point is that : $\sum_{j=0}^{n} \binom{n}{j} x_j = \sum_{j=0}^{k-1} \binom{n}{j} x_j + \sum_{j=k}^{n} \binom{n}{j} x_j$ for any $k$. Anyway, shouldn't I answer the questions just because it is easy? No offence, I'm just asking, as you can see I am a noob here. $\endgroup$ – Gabriel Jun 16 '17 at 14:15
  • $\begingroup$ The issue I'm bringing up has nothing to do with a question or answer being easy or hard. The point is that there is a single key idea and I think your elaborate answer has obscured it. Indeed, even your comment fails to expose the idea fully: this has nothing to do with binomial coefficients. It simply comes down to the fact that when you sum a sequence of $n+1$ numbers, you may sum the first $k$ of them, sum the remaining $n+1-k$ of them, and add those two sums. That's it. $\endgroup$ – whuber Jun 16 '17 at 14:31
  • $\begingroup$ I still don't get your point. My goal is to obtain $F_n(x)$, not showing the key idea itself. If anyone comes here and is able to understand how to get to the $F_n(x)$ expression, which is the ultimate goal, he/she will implicitly understand any key ideas in between. I don't get what is your suggestion for improving my answer. Is it about erasing everything about $F_n(x)$ stuff and leaving only the key idea to solve it? $\endgroup$ – Gabriel Jun 16 '17 at 18:03
1
$\begingroup$

You ask if there's a "better way to solve this" so I assume you're open to the idea of solutions which don't involve messing with binomials, and that you're only looking to solve this for the first order statistic. If I've misinterpreted this I apologize.

We have:

$$ Pr(X_{(1)} \leq x) = 1 - Pr(X_{(1)} \geq x) $$

$$ Pr(X_{(1)} \geq x) = Pr(X_{1} \geq x, X_{2} \geq x, ... , X_n \geq x) $$

$$ = Pr(X_1 \geq x) \times Pr(X_2 \geq x) \times ... \times Pr(X_n \geq x) $$

$$ = [Pr(X_1 \geq x)]^n = [1-Pr(X_1 \leq x)]^n = [1-F_x(x)]^n $$

$$ \therefore Pr(X_{(1)} \leq x) = 1 - [1-F_x(x)]^n $$

$\endgroup$
  • $\begingroup$ Thanks klumbard, I appreciate this alternative solution. Thanks for taking the time to answer :) $\endgroup$ – Gabriel Jun 16 '17 at 18:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.