Interpretation problems of linear model with no predictors Let $Y=X\beta$, where $X=(\begin{matrix}1 & 1 & 1\end{matrix})^T$, $Y=(\begin{matrix}6 & 5 & 4\end{matrix})^T$ and $\beta=(\begin{matrix}\beta_0 \end{matrix})$. Now $X\beta=(\begin{matrix}\beta_0 & \beta_0 & \beta_0\end{matrix})^T=(\begin{matrix}6 & 5 & 4\end{matrix})^T$ which does not make sense to me. How can $\beta_0$ equal different constants? The model should only have $E(Y)$ as $X\beta$. How can a $3\times 1$ matrix $Y$ be expressed as a single constant $E(Y)$?
If this mystery is cleared, it should be done in a way that explains also the $OLS$ for $\beta$. $\beta_{OLS}$ is $(X^TX)^{-1}X^Ty=15$. So $E(Y)=\bar y=15$. However, to me it looks like that $\bar y=5$. Could someone clear this confusion about the predictorless model?

The linear regression model is
$$
Y_i = \beta_0 + \beta_1 X_1 + \dots + \beta_k X_k + \varepsilon_i
$$
with $\varepsilon_i$ being i.i.d. Gaussian noise with mean equal to zero. The model estimates the conditional mean of $Y_i$. Intercept-only model is
$$
Y_i = \beta_0 + \varepsilon_i
$$
the model would estimate the mean on $Y_i$, the result would be equal to the arithmetic average of $Y_1, Y_2,\dots, Y_n$. You can write the same model as you did with $X = (1 ~1~ 1)^T$. The model has only one parameter $\beta_0$ that is equal to the average of $Y_i$'s. If you make predictions from such a model, it will predict the average for each $Y_i$ value. Recall that the same arithmetic mean minimizes the squared error, the same loss function is used by linear regression, so the single parameter model needs to predict the average.
I don't know how did you get the 15 result, but the answer is 5.
julia> X = [1, 1, 1]
julia> Y = [6, 5, 4]
julia> inv(X'X)X'Y
5.0
julia> sum(Y)/length(Y)
5.0

