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Let's say I have 3 Random Variables $X_1, X_2, X_b$ where "$b$" stands for "background". Each one of them is Gaussian with $N(\mu_i, \sigma^2_i)$ for $ i\in\{1,2,b\}$. I will assume $\mu_b=0$. Now I make $N$ experiments which measure the variables $X_1+X_b, X_2+X_b$ (where $X_b$ is measured at the same time for both of them) and I want to estimate all the $\mu_i$'s and $\sigma_i$'s.

I know how the estimate the means by taking the average of the results (because $\mu_b = 0$). Also I can easily estimate $\sigma_j^2 + \sigma_b^2$ because if the fact that $X_j+X_b = N(\mu_j,\sigma_j^2 + \sigma_b^2)$... but I need another estimator so I can get all the values for $\sigma$'s!

I thought about using the off-diagonal elements of the co-variance matrix (which are supposed to be $\sigma_b^2$) but I get huge problem when they are negative. Can someone help me find the missing estimator?

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  • $\begingroup$ But I know that $V(Y_1) - V(Y_2) = \sigma_1^2 - \sigma_2^2 $ and then using your equation I find that $\sigma_2^2 = \frac{V(Y_1−Y_2) - V(Y_1)+V(Y_2)}{2} $ and I think the right side can be negative for some values of $Y_1,Y_2$ :( $\endgroup$ – RyArazi Feb 20 '17 at 10:18
  • $\begingroup$ I know they are not equal. But using the fact that $V(Y_j) = \sigma_j^2 + \sigma_b^2$ I can get after some algebra what I wrote above... and it's not always negative (I think) $\endgroup$ – RyArazi Feb 20 '17 at 12:00
  • $\begingroup$ If my Comment is not helpful, please disregard it and try something else. $\endgroup$ – BruceET Feb 20 '17 at 17:27

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