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From this website, i see that i can multiply two pdf to get the pdf of joint distribution. http://www.math.uah.edu/stat/dist/Joint.html

However, I cannot get the answer in wiki if I multiply two Cauchy distributions. https://en.wikipedia.org/wiki/Cauchy_distribution $$f(x, y; x_0,y_0,\gamma)= { 1 \over 2 \pi } \left[ { \gamma \over ((x - x_0)^2 + (y - y_0)^2 +\gamma^2)^{1.5} } \right]$$

for example,

$$(\frac{1}{\pi}\frac{1}{1+x^2})(\frac{1}{\pi}\frac{1}{1+y^2})\neq{1 \over 2\pi} \left[ { 1 \over (x^2 + y^2 +1)^{1.5} } \right]$$

Why?

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    $\begingroup$ Since the r.h.s. is a joint pdf of two dependent r.v.'s, and the l.h.s. is a joint pdf of two independent r.v.'s. $\endgroup$ – NCh Jul 2 '17 at 1:53
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NCh is right. The formula given in the Wikipedia article is for a 2 dimensional analogue of the Cauchy distribution, and the article states explicitly that the two coordinates are not statistically independent. The formula at the Utah site is for two independent 1 dimensional Cauchy random variables. My advice is, forget about the 2 dimensional analogue of the Cauchy distribution given in Wikipedia, at least for the time being.

You are possibly mislead by a false analogy with normal or gaussian random variables, where 2 independent gaussians are jointly gaussian. This is a very special property of the gaussian distribution; you can read about it in https://en.wikipedia.org/wiki/Maxwell%27s_theorem .

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  • $\begingroup$ That means If I am handing two independent cauchy variables, l.h.s is the correct pdf? $\endgroup$ – fairytale Jul 2 '17 at 5:19
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    $\begingroup$ You mean the left hand side of the 2nd displayed equation in your original post? Yes. $\endgroup$ – kimchi lover Jul 2 '17 at 12:06
  • $\begingroup$ Thanks @kimchi lover $\endgroup$ – fairytale Jul 2 '17 at 12:09

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