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I have an exercise where I have to use a probabilistic model to fit the data. Here is the original problem (Problem 1): enter image description here

(Problem 2 with solution) At the lecture we learnt to derive Bayesian linear regression solution. We have dataset $\textbf{x} = (x^{(1)},...,x^{(N)}) $ and $\textbf{y} = (y^{(1)},...,y^{(N)}) $, parameters $\textbf{w} = (w_0,w_1)^T $.

We also have $\phi(x) = (1,x)^T$ and $ y = h^w(x) + \epsilon$ with $\epsilon \sim N(0,\sigma^2)$

Then we have the Bayesian solution : enter image description here

The question is how to use insight from Problem 2 to solve Problem 1. As I realize problem 1 is just problem 2 version without any predictors $x$ (in which $\mu = w_0$ and $e^{(i)} = \epsilon$ ). My best guess would be $\mu = y^{(N+1)} = \text{Mean}(y^{(i)})$ but I'm not sure how to prove it

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  • $\begingroup$ Is it necessary to use Problem 2 as a 'substep' in the solution? For example, since the shift by a constant of a normal distribution is again a normal distribution we have $p(y_i) = N(\mu, 1)$ and so we can just write down the likelihood and minimize $- \log L$ directly which will lead you to $\mu = \text{mean of the $y_i$}$ i believe... $\endgroup$ Nov 2 '17 at 10:59
  • $\begingroup$ Not necessarily. Actually I think what u just said is exactly what I'm looking for. Could you elaborate/derive what is the log-likelihood form and how to minimize it to get the answer ? That would be great ! $\endgroup$ Nov 2 '17 at 11:02
  • $\begingroup$ I have to go to lunch now :-) If you can't solve it on your own within the next 45 minutes then I will write a complete answer... deal? $\endgroup$ Nov 2 '17 at 11:10
  • $\begingroup$ Yes sir. I'll try to :) $\endgroup$ Nov 2 '17 at 11:20
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Well I think I got the answer for this. Thanks to Fabian Werner and based on this wikipedia article https://en.wikipedia.org/wiki/Maximum_likelihood_estimation:

So we have $p(y_i) = N(\mu,1)$ and log-likelihood $$\log(L(\mu,\sigma)) = (-n/2)log(2\pi)-(1/2)\sum(y^{i}-\mu)^2$$ Taking derivatives: $$ 0 = \frac{\partial}{\partial\mu}\log(L(\mu,\sigma)) = 0 - \frac{-2n(\bar{y} - \mu)}{2}$$ And we get the answer $\mu = \bar{y} $ (Mean of y)

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  • $\begingroup$ So, what about the other two questions? ;-) $\endgroup$ Nov 2 '17 at 12:20

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