I have the classical regression model

$$y = \beta X + \epsilon$$ $$\epsilon \sim N(0, \sigma^2)$$

where $X$ is taken to be fixed (not random).

If $\hat\beta$ is the OLS estimate for $\beta$, then it is known that $(y^T y, X^T y)$ pair is a complete sufficient statistic for $x_0^T \beta$, for some input $x_0$.

Can we conclude that $(y^T y, X^T y)$ is also a sufficient statistic for $\beta$, and why? I think for this to work $X^T X$ should be full rank. I mean a 1 to 1 transformation of a sufficient statistic is still a sufficient statistic, but it is still a sufficient statistic for $x_0^T \beta$. Based on what are we going to conclude the sufficiency of $\hat\beta$ for $\beta$ itself?

I have the classical regression model

$$y = \beta X + \epsilon$$ $$\epsilon \sim N(0, \sigma^2)$$

where $X$ is taken to be fixed (not random), and $\hat\beta$ is the OLS estimate for $\beta$.

It is known that $(y^T y, X^T y)$ pair is a complete sufficient statistic for $x_0^T \beta$, for some input $x_0$.

Can we conclude that $(y^T y, X^T y)$ is also a sufficient statistic for $\beta$, and why? I think for this to work $X^T X$ should be full rank. I mean a 1 to 1 transformation of a sufficient statistic is still a sufficient statistic, but it is still a sufficient statistic for $x_0^T \beta$. Based on what are we going to conclude the sufficiency of $\hat\beta$ for $\beta$ itself?