# Pattern Recognition and Machine Learning: Maximizing log likelihood with respect to Beta

On page 29 in Christopher Bishop's Pattern Recognition and Machine Learning book he gives the following two equations

1.62

$$\ln p(t|\pmb{x}, \pmb{w}, \beta) = - \frac{\beta}{2} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 + \frac{N}{2} \ln \beta - \frac{N}{2} \ln(2\pi)$$

He then goes on to explain that maximizing this with respect to w would result in the following by basically removing everything that is not dependent on $$w$$ which makes sense...

$$\pmb{w}_{ML} = \frac{1}{2} \sum_{n=1}^N \{ y(x_n, w) - t_n \}^2$$

Immediately following this the claim is made that maximizing with respect to $$\beta$$ gives the following...

$$\frac{1}{\beta_{ML}} = \frac{1}{N} \sum_{m=1}^N \{ y(x_n, w) - t_n \}^2$$

and I can't quite get there by the same logic, how would you arrive at this?

• Did you try differentiating the log-likelihood with respect to $\beta$ and setting the derivative to $0$? May 15 '19 at 3:35
• I think I was wrongly putting a term in the summand which was screwing me up. I believe my answer arrives there correctly. Does it look right @MinusOne-Twelfth ?
– Joff
May 15 '19 at 4:35

I was wrongly assuming that the $$\frac{N}{2} \ln \beta$$ was part of the summand. Taking the derivative and setting it to 0 gave me the proper outcome.
\begin{aligned} &\frac{\partial}{\partial \beta} \Big\lbrack -\frac{\beta}{2} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 + \frac{N}{2} \ln \beta - \frac{N}{2} \ln(2\pi) \Big\rbrack \\ &\frac{\partial}{\partial \beta} \Big\lbrack -\frac{\beta}{2} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 \Big\rbrack + \frac{\partial}{\partial \beta} \Big\lbrack \frac{N}{2} \ln \beta \Big\rbrack \\ 0 = &-\frac{1}{2} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 + \frac{N}{2\beta}\\ -\frac{N}{2\beta} = &-\frac{1}{2} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 \\ \frac{1}{2\beta} = &\frac{1}{2N} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 \\ \frac{1}{\beta} = &\frac{1}{N} \sum_{n=1}^N \{ y(x_n, \pmb{w}) - t_n \}^2 \\ \end{aligned}