# Least Square Error and how to find it

I am working on this problem and having trouble starting.

Consider the regression model $$y_t = \beta_1 y_{t-1}+e_t$$ where $$e_t$$ is the white noise with zero-mean and variance $$\sigma^2_e$$. Assume that we observe $$y_1, y_2, ... , y_n$$ and consider the model above for $$t = 2, 3, ..., n$$. Show that the "least squares estimator" of $$\beta_1$$ is $$\hat{\beta} = \frac{\sum_{t=2}^n y_ty_{t-1}}{\sum y^2_{t-1}}$$

I am not familiar with regression analysis and I have came across this problem somewhat unexpectedly, so I tried to learn what least squares estimators are and I ended up seeing things such as MMSEs and regression lines.

Honestly I did not really get anywhere besides MAYBE I finde the MSE then use calculus to minimize it?

where the MSE I think should be

\begin{align} MSE[y_t,y_{t-1}] & = E[(y_t-\beta_1y_{t-1})^2] \\ & = E[y_t^2]-2\beta_1E[y_ty_{y-1}]+\beta^2_1E[y_{t-1}^2] \end{align}

I also considered the MLE method

$$L[\beta_1] = \Pi_{i=2}^{n}\left(\beta_1y_{i-1}+e_i\right)$$

but the $$+e_t$$ term threw me off and I am not sure if this seems like the right path.

I would really appreciate your help.

Consider the regression model (AR(1) model without constant term): $$y_t=\beta_1y_{t-1}+e_t.$$ In order to estimate $$\beta_1$$, we can apply Least Squares Method, i.e. determine $$\beta_1$$ for which $$\sum_{t=2}^{n}e_{t}^{2}=\sum_{t=2}^{n}(y_t-\beta_1y_{t-1})^2\rightarrow min.$$ By taking the derivative of $$\sum_{t=2}^{n}(y_t-\beta_1y_{t-1})^2$$ with respect to $$\beta_1$$, we have that $$\frac{\partial \sum_{t=2}^{n}(y_t-\beta_1y_{t-1})^2}{\partial\beta_1} =-2\sum_{t=2}^{n}(y_t-\beta_1y_{t-1})y_{t-1}.$$ Then solving the following equation for $$\beta_1$$ $$-2\sum_{t=2}^{n}(y_t-\beta_1y_{t-1})y_{t-1}=0,$$ we find that $$\beta_1=\frac{\sum_{t=2}^{n}y_ty_{t-1}}{\sum_{t=2}^{n}y_{t-1}^2}$$