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If $X_1,...,X_n$ are iid random variables with $r$th order statistic $X_{(r)}$ I'm trying to prove it's pdf is


Is is true for $r=1$. I then assume true for $r$. The only step in this proof is I don't understand why this assumption implies that the cdf for $X_{(r)}$ is:

$$F_{(r)}(x)=\sum_{j=r}^n {n\choose j} F(x)^{j}[1-F(x)]^{n-j}$$

Is someone could explain this I would be very grateful

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up vote 3 down vote accepted

The formula $$F_{(r)}(x)=\sum_{j=r}^n {n\choose j} F(x)^{j}[1-F(x)]^{n-j}\qquad\qquad(\ast)$$ comes from basic probability considerations.

Fix $x$. If $X_i \le x$, call the result on trial $i$ a "success." By the definition of $F(x)$, the probability of success on the $i$-th trial is $p=F(x)$, and the probability of failure is $1-p$.

The $r$-th order statistic $X_{(r)}$ is $\le x$ precisely if there are at least $r$ successes. The number of successes has binomial distribution. The probability of exactly $j$ successes is $\binom{n}{j}p^j(1-p)^{n-j}$. So the probability of $r$ or more successes is $$\sum_{j=r}^n \binom{n}{j}p^j(1-p)^{n-j}.$$ Since $p=F(x)$, this is exactly your formula $(\ast)$.

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Thank you! I knew it must be something like that.. my brain doesn't get along with the probabilistic/statistic side of maths. – Freeman Mar 14 '12 at 16:11

Since it seems it's a proof by induction, we can also show this intermediate result by induction. We have \begin{align*} F_{(r)}(x)&=\int_{-\infty}^xf_{(r)}(t)dt\\ &=\frac{n!}{(r-1)!(n-r)!}\left[-\frac{(1-F(t))^{n-r+1}F(t)^{r-1}}{n-r+1} \right]_{-\infty}^x\\ &+\frac{n!}{(r-1)!(n-r)!}\cdot\int_a^x\frac{(1-F(t))^{n-r+1}}{n-r+1}F(t)^{r-2}(r-1)f(t)dt\\ &=-\frac{n!}{(r-1)!(n-(r-1))!}(1-F(x))^{n-r+1}F(x)^{r-1}+\sum_{j=r-1}^n\binom njF(x)^j(1-F(x))^{n-j}\\ &=\sum_{j=r}^n\binom njF(x)^j(1-F(x))^{n-j} \end{align*} and applying the result for $r-1$ we get the wanted result.

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At the third line of the align I add to put $a$ instead of $–\infty$ since I didn't work. – Davide Giraudo Mar 14 '12 at 15:38
In your second line, you use integration by parts am I correct? you integrate $[1-F(x)]^{n-r}f(x)$, why does the power of $F(t)$ change? And the second part don't you need to alter your constant term multiplying it? – Freeman Mar 14 '12 at 16:00
@LHS I've edited my post, since as you noticed there were some typos. I hope it's correct now. Thanks for pointing it out. – Davide Giraudo Mar 14 '12 at 16:05
That looks better now, I thought I was going insane there for a moment. It's comforting it can be done directly, but if you see the other post I believe that that is a good intuitive way to remember this in an exam. Thank you very much for your help though. – Freeman Mar 14 '12 at 16:13

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