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Let $X_1, X_2 \stackrel{\text{iid}}{\sim}\mathrm{Uniform}(0,1)$ and then sort $X_1,X_2$ to get $X_{(1)} < X_{(2)}$.

Based on the pdfs of $X_{(i)}$, we know $X_{(1)} \sim \mathrm{Beta}(1,2)$ and $X_{(2)} \sim \mathrm{Beta}(2,1)$, with $\mathbb{E}(X_{(1)}) = \frac{1}{3}$ and $\mathbb{E}(X_{(2)}) = \frac{2}{3}$.

Consider the constant approximation for the function $f(x) = x$.

  1. Sample two points $x_1, x_2$ uniformly on $(0, 1)$ and then sort them to get $x_{(1)}$ and $x_{(2)}$, $x_{(1)} < x_{(2)}$. Denote the expectation as $\mathbb{E}_1(\|f - c\|_2^2) = \mathbb{E}_1(\displaystyle\int_{x_{(1)}}^{x_{(2)}} (f(t)-c)^2 \; dt)$, where $c = \frac{1}{x_{(2)}-x_{(1)}}\displaystyle\int_{x_{(1)}}^{x_{(2)}} f(t)\; dt$ is a constant.

  2. Sample $y_{(1)}$ from $\mathrm{Beta}(1,2)$ and $y_{(2)}$ from $\mathrm{Beta}(2,1)$. Denote the Expectation as $\mathbb{E}_2(\|f - c\|_2^2) = \mathbb{E}_2(\displaystyle\int_{y_{(1)}}^{y_{(2)}} (f(t)-c)^2 \; dt)$, where $c = \frac{1}{y_{(2)}-y_{(1)}}\displaystyle\int_{y_{(1)}}^{y_{(2)}} f(t)\; dt$ is a constant.

(Note: the definitions of $\mathbb{E}_1$ and $\mathbb{E}_2$ are identical; the only difference is the method of obtaining the points $x_{(i)}, y_{(i)}$.)

Compare $\mathbb{E}_1(\|f - c\|_2^2)$ and $\mathbb{E}_2(\|f - c\|_2^2)$. It's observed from the numerical experiments that $\mathbb{E}_1(\|f - c\|_2^2) < \mathbb{E}_2(\|f - c\|_2^2)$. This result confuses me.

I expect them to be equal because both $x_{(i)}$ and $y_{(i)}$ from the same Beta distribution as discussed above.

Is this because $x_{(1)} \;\mathrm{and}\; x_{(2)}$ are not independent? If so, what is the joint probability density of $x_{(1)}$ and $x_{(2)}$?


Is the following the correct joint density?

$f_{x_{(1)}, x_{(2)}}(u, v) = n! f_{x_{(1)}}(u) f_{x_{(2)}}(v) \mathbf{1}_{u<v}$, where $f_{x_{(1)}}$ and $f_{x_{(2)}}$ are pdfs of $\mathrm{Beta}(1,2)$ and $\mathrm{Beta}(2,1)$?

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    $\begingroup$ I don't see any "piecewise constant approximation" to the identity function $f.$ Perhaps you omitted something? $\endgroup$
    – whuber
    2 days ago
  • 1
    $\begingroup$ Even though the marginals are the same, $X_{(1)}$ and $X_{(2)}$ are indeed negatively correlated, while there is a positive probability that $Y_{(2)}<Y_{(1)}$. $\endgroup$
    – Xi'an
    2 days ago
  • $\begingroup$ @whuber Yes, I forgot to change the description. In fact, it's just a constant approximation to the function $f$ on an interval. $\endgroup$
    – learner
    2 days ago
  • $\begingroup$ Because both those expectations are easily computed in closed form and have simple formulas, I wonder whether you are asking the question you intend to ask. Regardless, the procedure differ because the $y_{(i)}$ are independent whereas the $x_{(i)}$ are not. Finally, you have already concluded the $x_{(i)}$ have Beta distributions, so why is your ultimate question about their "pdfs or cdfs"? Just look them up! $\endgroup$
    – whuber
    2 days ago
  • $\begingroup$ @whuber Thank you for your comments. I am not asking for the pdfs or cdfs of the Beta distribution, which are "trivial". My question was if $x_(1)$ and $x_(2)$ are not independent, what is the joint density? Through my search today, I think the joint density should be $f_{x_{(1)}, x_{(2)}}(u, v) = n! f_{x_{(1)}}(u) f_{x_{(2)}}(v) \mathbf{1}_{u<v}$, where $f_{x_{(1)}}$ and $f_{x_{(2)}}$ are pdfs of $\mathrm{Beta}(1,2)$ and $\mathrm{Beta}(2,1)$, respectively. But I am not sure do I need the indicator function $\mathbf{1}_{u<v}$. $\endgroup$
    – learner
    yesterday

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