সন্ধানের সহজ উপায়

Consider 3 iid samples drawn from the uniform distribution $u(\theta, 2\theta)$ , where $\theta$ is parameter. I want to find

E [X_{(2)} | X_{(1)}, X_{(3)}]

$\mathbb{E}\left[X_{(2)}| X_{(1)}, X_{(3)}\right]$ where

X_{(i)}

$X_{(i)}$ is order statistic

i

$i$ .

I would expect the result to be

E [X_{(2)} | X_{(1)}, X_{(3)}] = \frac{X_{(1)} + X_{(3)}}{2}

$\mathbb{E}\left[X_{(2)}| X_{(1)}, X_{(3)}\right] = \frac{X_{(1)}+ X_{(3)}}{2}$ 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

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

E [X_{(2)} | X_{(1)}, X_{(3)}] = \int x f (x | x_{(1)}, x_{(3)}) d x

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

বিবরণ:

আমি আদেশের পরিসংখ্যানগুলির ঘনত্বের জন্য সাধারণ সূত্র গ্রহণ করি (সহ) $\mathbb{I}_{\{A\}}$ সেট একটি সূচক $A$ )

f_{x_{(1)}, \dots, x_{(n)}} (x_{1}, \dots, x_{n}) = n! \prod_{i = 1}^{n} f_{x} (x_{i}) I_{{x_{(1)} \leq x_{(2)} \leq \dots \leq x_{(n)}}} (x_{1}, \dots, x_{n})

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

আমার মামলা পেতে

f_{x_{(1)}, x_{(2)}, x_{(3)}} (x_{1}, x_{2}, x_{3}) = 3! \frac{1}{θ^{3}} I_{{x_{1} \leq x_{2} \leq \dots \leq x_{n}}} (x_{1}, \dots, x_{3})

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

প্রান্তিক $f_{x_{(1)}, x_{(3)}}(u, v)$ হয়

f_{x_{(1)}, x_{(3)}} (u, v) = \int f_{x_{(1)}, x_{(2)}, x_{(3)}} (u, x_{2}, v) d x_{2}

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

এটাই

f_{x_{(1)}, x_{(3)}} (u, v) = \int 3! \frac{1}{θ^{3}} I_{{x_{1} = u \leq x_{2} \leq x_{3} = v}} (u, x, v) d x = 3! \frac{1}{θ^{3}} [v - u]

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

তজ্জন্য

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}{θ^{3}} I_{u \leq x_{2} \leq \dots \leq v} (u, x_{2}, v)}{3! \frac{1}{θ^{3}} [v - u]} = [v - u]^{- 1} I_{{u < x_{2} < v}}

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

যা দেয়

E [X_{(2)} | X_{(1)} = u, X_{(3)} = v] = [v - u]^{- 1} \int_{u}^{v} x d x = [v - u]^{- 1} \frac{[v^{2} - u^{2}]}{2} = \frac{u + v}{2}

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

uniform conditional-expectation order-statistics

— them
সূত্র

I didn't look at what you did, but you got an answer of

\frac{u - v}{2}

$\frac{u - v}{2}$ , not

\frac{u + v}{2}

$\frac{u + v}{2}$

— Mark L. Stone

@MarkL.Stone you are right... I fixed that, the last line, integral of

\int x d x

$\int x dx$ was incorrect.

— them

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.

— whuber
সূত্র

whuber, when you write

d F (x_{k})

$dF(x_k)$ you refer to the probability density of X, am I right?

— them

Yes, that is correct. By definition,

d F (x) = \frac{d F}{d x} (x) d x .

$dF(x)=\frac{d F}{d x}(x)\,dx.$ (Technically, I should have called this the "probability element" rather than the "density".)

— whuber