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I am reading this paper "Continuous Diffusion Analysis" and in page 7 there are a next formula

$\mathbb{E}_{x,y}[d_S(x,y)]=\dfrac{1}{4}\sum_{i=0}^{n-1}\int^{1}_{-1}\int^{1}_{-1}|sgn(x_i)-sgn(y_i)|dx_idy_i$.

I am trying to obtain this formula using the formula $\mathbb{E}[z]=\int_{\mathbb{R}}zf(z)dz$ presented in "Absolutely continuous case" of wikipedia, but I do not understand how the authors of "Continuous Diffusion Analysis" reach that formula. For now, considering $z=d_S{(x,y)}$ follows the uniform distribution, I think that the probability density function of $d_S{(x,y)}$ is $\dfrac{1}{2n}$. Then according to wikipedia

$\mathbb{E}[z]=\int_{\mathbb{R}}zf(z)dz=\mathbb{E}[z]=\int_{\mathbb{R}}\dfrac{1}{2n}d_S(x,y)dz$.

Could you help to obtain that formula with an explanation, please? For example, how they go from $\mathbb{R}$ to $-1, 1$ limits?

juaninf
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    Why the downvote? Clearly, the OP has made some effort trying to figure out the answer. – Clement C. Aug 28 '21 at 07:35
  • @OP They assume each $x_i,y_i$ independent and uniformly distributed over $[-1,1]$ (that is, $x,y$ uniformly distributed over $[-1,1]^n$). Then, you can use linearity of expectation and the definition of $ d_S(x,y) = \sum_{i=0}^{n-1} |\operatorname{sgn}(x_i)-\operatorname{sgn}(y_i)| $ to get what they claim. – Clement C. Aug 28 '21 at 07:40
  • thanks @ClementC. Following your advice $\int_{\mathbb{R}}zf(z)dz=\int_{\mathbb{R}}d_S(x,y)dz =\sum_{i=0}^{n-1}\int^{1}{-1}\int^{1}{-1}|sgn(x_i)-sgn(y_i)|dx_idy_i$ – juaninf Aug 28 '21 at 10:29
  • but I do not know how they got that the probability density function of $z$ is 1/4. – juaninf Aug 28 '21 at 10:31
  • Uniform on $[a,b]$ means the pdf is $$f(x) = \begin{cases}\frac{1}{b-a} & \text{ if } x\in[a,b]\ 0 &\text{otherwise}\end{cases}$$ so for $a=-1,b=1$ you have $\frac{1}{b-a} = \frac{1}{2}$. (Twice, because both $x_i$ and $y_i$ are uniform there, and independent) – Clement C. Aug 28 '21 at 21:39

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