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I am reading this paper, on page 4 section 2.3 it says that the joint distribution of s1, s2 is uniform on a square when s1 and s2 are uniform distributions symmetric across 0 and they have same distributions. Isn't this problem same as the square of a uniform distribution which is symmetric across 0, Like another question here? I'm confused why the joint distribution is uniform on square here.

$$p(s_i) = \begin{cases} \frac{1}{2\sqrt{3}}, & \text{if} \ |s_i| \leq \sqrt{3} \\0,& \text{otherwise} \end{cases} $$

Ruthvik Vaila
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1 Answers1

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This has nothing to do with symmetry. If $S_1$ is uniform on the interval $[a_1, b_1]$, and $S_2$ is uniform on the interval $[a_2, b_2]$, and the two random variables are independent, then the joint distribution of $(s_1, s_2)$ is uniform on the rectangle with vertices $(a_1, a_2)$, $(a_1, b_2)$, $(b_1, a_2)$, and $(b_1, b_2)$.

This can be seen by noting that the joint density is the product of the two marginal densities. $$p_{S_1,S_2}(s_1, s_2) = p_{S_1}(s_1) p_{S_2}(s_2) = \begin{cases}\frac{1}{b_1-a_1} \cdot \frac{1}{b_2 - a_2} & a_1 \le s_1 \le b_1, a_2 \le s_2 \le b_2 \\ 0 & \text{otherwise} \end{cases}$$

angryavian
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  • In my case joint density is square of one of the density(independence) rite? If yes, then it's basically a r.v transformation as in another link that i gave rite? – Ruthvik Vaila Oct 15 '17 at 18:02
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    @RuthvikVaila Not at all. The joint distribution of of $S_1,S_2$ has density $1/12 = (1/2\sqrt{3})^2$ and is supported on the 2-dimensional region ${(s_1,s_2) : |s_1| \le \sqrt{3}, |s_2| \le \sqrt3}$. On the other hand, $S_1^2$ has density $1/2\sqrt{3s}$ and is defined on the interval $[0,3]$. – Trevor Gunn Oct 15 '17 at 18:12
  • I see, i got the dimensions confused. – Ruthvik Vaila Oct 15 '17 at 18:18