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Consider 3 iid samples drawn from the uniform distribution $u(\theta, 2\theta)$, where $\theta$ is parameter. I want to find $$ \mathbb{E}\left[X_{(2)}| X_{(1)}, X_{(3)}\right] $$ where $X_{(i)}$ is order statistic $i$.

I would expect the result to be $$ \mathbb{E}\left[X_{(2)}| X_{(1)}, X_{(3)}\right] = \frac{X_{(1)}+ X_{(3)}}{2} $$ But the only way I can show this result seems to be too lengthy, I cannot come up with simple solution, am I missing something, is there some shortcut?


What I do is the following:

  • I find the conditional density

    $$ f(x_{(2)}| x_{(1)}, x_{(3)}) = \frac{ f(x_{(1)}, x_{(2)}, x_{(3)})}{f(x_{(1)}, x_{(3)})} $$

  • I integrate

$$ \mathbb{E}\left[X_{(2)}| X_{(1)}, X_{(3)}\right] = \int x f(x| x_{(1)}, x_{(3)}) dx $$


Details:

I adopt general formula for density of order statistic (with $\mathbb{I}_{\{A\}}$ an indicator of set $A$)

$$ f_{x_{(1)},\ldots , x_{(n)}}(x_1,\cdots, x_n) = n! \prod_{i=1}^n f_{x}(x_i)\mathbb{I}_{\{x_{(1)} \leq x_{(2)} \leq \cdots \leq x_{(n)}\}}(x_1,\cdots, x_n) $$

to obtain for my case

$$ f_{x_{(1)}, x_{(2)}, x_{(3)}}(x_1, x_2, x_3) = 3!\frac{1}{\theta^3}\mathbb{I}_{\{x_1 \leq x_2 \leq \cdots \leq x_n\}}(x_1,\cdots, x_3) $$

marginal of $f_{x_{(1)}, x_{(3)}}(u, v)$ is

$$f_{x_{(1)}, x_{(3)}}(u, v) = \int f_{x_{(1)}, x_{(2)}, x_{(3)}}(u, x_2, v) dx_2$$

that is

$$ f_{x_{(1)}, x_{(3)}}(u, v) = \int 3!\frac{1}{\theta^3}\mathbb{I}_{\{x_1 = u \leq x_2 \leq x_3 = v\}}(u, x, v) dx = 3!\frac{1}{\theta^3} [v-u] $$

therefor

$$ f(x_{(2)}| x_{(2)} = u, x_{(3)} = v) = \frac{ f(x_{(1)} = u, x_{(2)}, x_{(3)} = v)}{f(x_{(1)}= u, x_{(3)} = v)} = \frac{3!\frac{1}{\theta^3}\mathbb{I}_{u \leq x_2 \leq \cdots \leq v}(u,x_2, v) }{3!\frac{1}{\theta^3} [v-u]}= [v-u]^{-1}\mathbb{I}_{\{u<x_2<v\}} $$

which gives

$$ \mathbb{E}\left[X_{(2)}| X_{(1)} = u, X_{(3)} = v\right] = [v-u]^{-1}\int_{u}^{v} x dx = [v-u]^{-1}\frac{ [v^2 - u^2]}{2} = \frac{u+v}{2} $$

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  • $\begingroup$ I didn't look at what you did, but you got an answer of $\frac{u - v}{2}$, not $\frac{u + v}{2}$ $\endgroup$ Aug 18, 2016 at 16:13
  • $\begingroup$ @MarkL.Stone you are right... I fixed that, the last line, integral of $\int x dx$ was incorrect. $\endgroup$
    – them
    Aug 18, 2016 at 16:18

1 Answer 1

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Because the $X_i$ all have a uniform distribution, all (unordered) variables are assumed independent, and no other order statistic lies between $X_{(1)}$ and $X_{(3)}$, $X_{(2)}$ has a truncated uniform distribution supported on the interval $[X_{(1)}, X_{(3)}]$. Its mean obviously is $(X_{(1)}+X_{(3)})/2$, QED.


If you would like a formal demonstration, note that when the $X_i$ are iid with an absolutely continuous distribution $F$, the conditional density of $X_{(k)}$ (conditional on all the other order statistics) is $dF(x_k)/(F(x_{(k+1)}) - F(x_{(k-1)}))$, which is the truncated distribution. (When $k=1$, $F(x_{0})$ is taken to be $0$; and when $k=n$, $F(x_{n+1})$ is taken to be $1$.) This follows from Joint pdf of functions of order statistics, for instance, together with the definition of conditional densities.

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  • $\begingroup$ whuber, when you write $dF(x_k)$ you refer to the probability density of X, am I right? $\endgroup$
    – them
    Aug 24, 2016 at 14:34
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    $\begingroup$ Yes, that is correct. By definition, $$dF(x)=\frac{d F}{d x}(x)\,dx.$$ (Technically, I should have called this the "probability element" rather than the "density".) $\endgroup$
    – whuber
    Aug 24, 2016 at 14:41

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