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I am trying to answer the following question

Likelihood is given by

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I understand that

Gaussian

So I thought I could simply substitute $y_i$ for $x_i$ in the gaussian formula and proceed from there. However if I do this then the negative sign is the wrong way around. What am I missing?

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You must be handling the sign wrong somehow

$$\log \left(\frac{1}{2\pi\sigma^2}\right)^{1/2}= \frac{1}{2}\log \left(\frac{1}{2\pi\sigma^2}\right) = -\frac{1}{2}\log (2\pi\sigma^2)$$ and $$\log \exp \left(-\frac{1}{2 \sigma^{2}}\left(\mu-y_{i}\right)^{2}\right) = -\frac{1}{2 \sigma^{2}}\left(\mu-y_{i}\right)^{2} $$ so the loglikelihood has minus signs and the minus loglikelihood doesn't

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  • $\begingroup$ I missed that ($\mu -y_i)^2 = (y_i-\mu)^2$ $\endgroup$
    – Kirsten
    Commented Jan 2, 2021 at 4:46

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