Derive the ML-estimation for the parameter $\beta$ I think I know in general how to derive the maximum likelihood estimation for a parameter given a distribution. But I can't wrap my head around this one!
We have the observations: $(y_1, x_1),...,(y_n, x_n)$ from a random sample where: $Y_i \sim N(\beta x_i, 1)$ for $i = 1,2,...,n$. We are interested in the expected value of $y_i$ given $x_i$.
How do i derive the ML-estimate for the parameter $\beta?$
Don't know how to begin when the parameter $\beta$ shows up in that way.
EDIT 1
I set:
$$\tilde{\mu} = x_i\beta$$
The Likelihood is:
$$\prod_{i=1}^{n} \frac{1}{\sqrt{2\pi\sigma^2}}e^{-\frac{1}{2}(\frac{y_i - \tilde{\mu}}{\sigma})^2} \iff (\frac{1}{\sqrt{2\pi\sigma^2}})^n e^{-\sum_{i=1}^{n}\frac{1}{2\sigma^2}(y_i - \tilde{\mu})^2}$$
Then the log-likelihood is:
$$-\frac{n}{2}\ln{2\pi} -\frac{n}{2}\ln{\sigma^2} -\frac{1}{2\sigma^2}\sum_{i=1}^{n}(y_i - \tilde{\mu})^2$$
If we take the derivative of the log-likelihood we get:
$$-\frac{1}{2\sigma^2}\frac{d}{d\tilde{\mu}}(\sum_{i=1}^{n}(y_i - \tilde{\mu})^2)$$
Which is:
$$-\frac{1}{2\sigma^2}\frac{d}{d\tilde{\mu}}(\sum_{i=1}^{n}(y_i^2 - 2y_i\tilde{\mu} + \tilde{\mu}^2))$$
Should i now insert $\tilde{\mu} = x_i\beta$?
If i do i get:
$$-\frac{1}{2\sigma^2}\frac{d}{d\beta}(\sum_{i=1}^{n}(y_i^2 - 2y_i x_i\beta + x_i^2\beta^2))$$
What should i do now, this approach below seems bad?
$$-\frac{1}{2\sigma^2}\sum_{i=1}^{n}\frac{d}{d\beta}(y_i^2 - 2y_i x_i\beta + x_i^2\beta^2)$$
$$-\frac{1}{2\sigma^2}\sum_{i=1}^{n} - 2y_i x_i + 2 x_i^2\beta$$
$$\frac{1}{2\sigma^2} 2n\bar{x} (n\bar{y}  - n\bar{x}\beta)$$
And set to zero for finding the estimate:
$$\frac{1}{2\sigma^2} 2n\bar{x} (n\bar{y}  - n\bar{x}\beta) = 0$$
$$\bar{y} = \bar{x}\beta$$
$$\frac{\bar{y}}{\bar{x}} = \beta$$
Which is wrong.
EDIT 2
I do try my best to understand so please explain if you downvote.

The derivation is basically fine until here. After this step, you introduce $\bar x, \bar y$ and the result is not correct.
$$0=-\frac{1}{2\sigma^2}\sum_{i=1}^{n} - 2y_i x_i + 2 x_i^2\beta.$$
It's given that $\sigma=1$. We can multiply through by the leading fraction. The rest is rearranging.
$$\begin{align}
0 &= \sum_{i=1}^{n}  y_i x_i -  x_i^2\beta\\
 &= \sum_{i=1}^{n}  y_i x_i - \sum_{j=1}^n  x_j^2\beta \\
\sum_{i=1}^{n}  y_i x_i &= \beta \sum_{j=1}^n  x_j^2 \\
\frac{\sum_{i=1}^{n}  y_i x_i}{ \sum_{j=1}^n  x_j^2} &= \beta
\end{align}$$
