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I'm working out a derivation of a posterior distribution that's bivariate normal, and I've gotten to the point where I know it's proportional to:

$$\exp\left[- \frac { x_1^2 (\delta^2 + \sigma_2^2) + x_2^2 (\delta^2 + \sigma_1^2) - 2 \delta^2 x_1 x_2 - 2\sigma_2^2 \nu x_1 - 2 \sigma_1^2 \nu x_2 }{2 \left ( \sigma_1^2 \sigma_2^2 + \sigma_1^2 \delta^2 + \sigma_2^2 \delta^2 \right)} \right]$$

In the above, the variable vector is $(x_1, x_2)$. It seems to me like it should be straightforward, then, to find the mean vector $\vec \mu$ and the variance-covariance matrix $\mathbf \Sigma$, but I don't know of any way to do it other than literally doing algebraic manipulation and creating a system of equations for the five variables I need to find ($\mu_{x_1}, \mu_{x_2}, \sigma_{x_1}^2, \sigma_{x_2}^2, \rho$) and then solving it, but a preliminary look into that seems to be a huge mess.

Is there any other way to find the values of my coefficients?

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1 Answer 1

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Compare the expression with the simplified version of the bivariate normal (see this for example, page $3$).

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  • $\begingroup$ By "simplified" do you mean the one at the end of slide 9 of that link (same as the one on the Wikipedia page)? That seems like it's still the same problem of a five-variables system of equations. And also I do not know the multiplicative constant, I only know the formula I mentioned there. $\endgroup$
    – Red
    Commented Aug 6, 2019 at 21:41
  • $\begingroup$ Do you have the numerical values? If yes, please post those. I can try to show you how to do it. $\endgroup$
    – PTDS
    Commented Aug 6, 2019 at 21:46
  • $\begingroup$ There aren't numerical values, I want to solve this analytically for this 2-variable case and then generalise it for the N-variable case. $\endgroup$
    – Red
    Commented Aug 6, 2019 at 21:51
  • $\begingroup$ In your formulation, there is no constant term in the exponent. Did you omit it by choice or is there really no constant term in the exponent you calculated? $\endgroup$
    – PTDS
    Commented Aug 6, 2019 at 22:15
  • $\begingroup$ By choice, yes, it should be derivable from everything else and it made getting to this point at all in my derivation a much easier and smoother affair. $\endgroup$
    – Red
    Commented Aug 6, 2019 at 22:18

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