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I have measured two variables which depend on the same set of four parameters. I want to know the parameters which best explain my measurements. Of course, I cannot solve for four unknowns from just two equations. But assuming my parameters to be constrained, maybe assuming a uniform prior over a certain range or a Gaussian prior with mean and standard deviation for each of the parameters, I feel it should be possible to get a most likely set of parameters determining my measurements.

How would I proceed? What are the technical terms I have to google to do what I want?

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It is possible indeed. As you are looking for a bayesian approach, you may want to read about "maximum a posteriori" (MAP) which is essentially the whay you suggest in your question.

Basic idea: find the values of the parameters $w_1, w_2, w_3, w_4$ which maximize $P(w_1, w_2, w_3, w_4 | x, y)$ where $x$ and $y$ are your measured variables.

From Bayes rule: $P(w_1, w_2, w_3, w_4 | x, y) = P(x, y | w_1, w_2, w_3, w_4) P(w_1, w_2, w_3, w_4) / P(x,y)$. You may ignore here the normalization factor $P(x,y)$. Now you may choose - as you said - some prior $P(w_1, w_2, w_3, w_4) $ and also find the likelihood function $P(x, y | w_1, w_2, w_3, w_4)$ either because you know how your measured variables depend on the parameters, or because you assume some appropiate model for that relationship. Afterwards, you select the values for $w_i$ for maximizing, and this set of parameters is the MAP solution for your problem.

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  • $\begingroup$ That seems to point me in the right direction, thanks for explicitely writing down the terms. I have problems to construct the likelihood function, though. I know how the measurement depends on the parameters, meaning $x=f_1(w_1,w_2,w_3,w_4)$ and $y=f_2(w_1,w_2,w_3,w_4)$. It seems I have to assume a distribution for $x$ and $y$. Can I just choose to maximise some inverse quadratic loss $1/((x-f_1(w1,w2,w3,w4))^2+(y-f_2(w1,w2,w3,w4))^2)$ weighted with $\prod P(w_i)$ over $w_i$? Or is that something different? $\endgroup$
    – dodi
    Commented Mar 5, 2019 at 16:05
  • $\begingroup$ It seems to me that if you do that, you would be replacing the likelihood for those inverse quadratic losses and I am not sure about what would that mean. What you can do is, for example, if you have noise measurements (which is often the case), consider that $x = f_1(w_1, ..., w_4) + Z$ using some for the noise $Z$. Then you can get the likelihood. For simplicity in this example I will assume $Z$ is Gaussian with $0$ mean and that the conditional measured variables $x|w_1, ..., w_4$ and $y|w_1, ..., w_4$ are independent. Then $ P(x,y|w_1, ..., w_4) = \mathcal{N}(\mu,\,\Sigma)$ where... $\endgroup$
    – Javi
    Commented Mar 5, 2019 at 20:13
  • $\begingroup$ ... $\mu = [f_1(w_1, ... , w_4) \quad f_2(w_1, ... , w_4)]$ and $\Sigma$ is the 2x2 diagonal matrix where $\sigma_x^2$ and $\sigma_y^2$ are the elements in the diagonal. For further simplicity one may assume that the variance in the measurements are equal for each variable ( $\sigma_x^2 = \sigma_y^2 = \sigma^2$ ) $\endgroup$
    – Javi
    Commented Mar 5, 2019 at 20:17
  • $\begingroup$ Just in case... I wrote the likelihood using a mean vector and a covariance matrix, but since I assumed independence that is the same as the product of the distributions $\endgroup$
    – Javi
    Commented Mar 5, 2019 at 20:21
  • $\begingroup$ Got it I think, cheers! $\endgroup$
    – dodi
    Commented Mar 5, 2019 at 22:45

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