What is ρ in the Bivariate Normal distribution?

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In going from the the Normal distributions for the horizontal axis, \(\mathcal{N}(\mu_h,\,\sigma_h^2)\), and vertical axis, \(\mathcal{N}(\mu_v,\,\sigma_v^2)\) a new equation was postulated with a parameter \(\rho<math>.     <math> f(h,v; \mu_h, \mu_v, \sigma_h, \sigma_v, \rho) = \frac{1}{2 \pi \sigma_h \sigma_v \sqrt{1-\rho^2}} \exp\left( -\frac{1}{2(1-\rho^2)}\left[ \frac{(h-\mu_h)^2}{\sigma_h^2} + \frac{(v-\mu_v)^2}{\sigma_v^2} - \frac{2\rho(h-\mu_h)(v-\mu_v)}{\sigma_h \sigma_v} \right] \right) \)

First a bit of explanation about what \(\rho\) is. Assuming that two variables are correlated, a simple correlation to propose is that the two variables are linearly correlated. Thus for some point \(i\) the equation of interest is:

   \(v_i = v_0 + \beta h_i\)

Where \(v_0\) is the intercept along the vertical axis, and \(\beta\) is the slope of the line. Given the locations \((h_i, v_i)\) of the shots in the group on the target, the coefficients \(v_0\) and \(\beta\) are calculated to give a "best" fit to the data.



For the population of shots, if there is a linear relationship between the horizontal and vertical positions of a shot, then point \((\mu_h, \mu_v)\) would be on the line. Thus around \((\mu_h, \mu_v)\) \(\beta\) would not only be the slope of the line, but it would also be a proportionality constant.

\(\beta = \frac{(v-\mu_v)}{(h-\mu_h)}\)