I am confronted with the exercise below. I have no given solution, so I hope someone can tell me whether my solution is right or wrong.
I want to show the deterministic linear model $\quad \boldsymbol y = \boldsymbol X \boldsymbol \beta + \boldsymbol u,\quad \boldsymbol u \sim N(0,\sigma^2 \boldsymbol I) \quad$ corresponds to an exponential model of the following form:
$f(y; \boldsymbol \theta) = C(\boldsymbol \theta)h(y) \mathrm{exp} \left\{\sum\limits_{j=1}^k \xi_j(\boldsymbol \theta)T_j(y) \right\} \\$ with $\boldsymbol \theta = (\beta', \sigma^2)'$.
Clearly, $\boldsymbol y \sim N(\boldsymbol X \boldsymbol \beta,\sigma^2 \boldsymbol I) $. So I started as follows:
$f(y; \boldsymbol \theta)= \prod\limits_{i=1}^n \frac{1}{{\sigma \sqrt {2\pi } }}\mathrm{exp} \left\{ -\frac{1}{2\sigma^2} (y_i- \sum\limits_{j=1}^kx_{j}^{(i)} \beta_j)^2 \right\} $
After some rearrangement, I got:
$f(y; \boldsymbol \theta)= \frac{1}{{{(2\pi)^{n/2} } }}\mathrm{exp} \left\{ -\frac{1}{2\sigma^2} \sum\limits_{i=1}^n (\sum\limits_{j=1}^kx_{j}^{(i)}\beta_j)^2 + \mathrm{ln}(\sigma^{n} )\right\} \\ \qquad \quad \quad \mathrm{exp} \left\{ -\frac{1}{2\sigma^2}\sum\limits_{i=1}^n y_{i}^2 + \frac{1}{\sigma^2}\sum\limits_{i=1}^ny_i\sum\limits_{j=1}^kx_{j}^{(i)}\beta_j \right\} $
So, I concluded it corresponds to the exponential model with the functions
$h(y)=\frac{1}{{(2\pi)^{n/2} }} \\ C(\boldsymbol \theta)=\mathrm{exp} \left\{ -\frac{1}{2\sigma^2} \sum\limits_{i=1}^n (\sum\limits_{j=1}^kx_{j}^{(i)}\beta_j)^2 + \mathrm{ln}(\sigma^{n} )\right\}\\ \xi(\boldsymbol \theta) = (-\frac{1}{{2\sigma^2 } },\quad \frac{1}{{\sigma^2 }} \boldsymbol \beta) \\ \\ T(y) =( \boldsymbol y'\boldsymbol y,\quad \boldsymbol y' \boldsymbol X ) $.
EDIT: First, I had the wrong solution but found the right one with the help of Dennis. I did not bear in mind that T has to be the sufficient statistic. So, I had $\boldsymbol X$ in $\xi$ but could easily change it with matrix notation:
