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:


1 Answer 1


Nope. What happened to your product term, $\sum_{i,j}x^{(i)}_j\cdot y_i$? You'll find this clearer if you write it in matrix notation rather than scalar notation. Your joint density is then

$f(\mathbf{y}\,|\,\mathbf{\beta}, \sigma^2) = \frac{1}{(2\pi\sigma^2)^{n/2}} \exp{-\{\frac{1}{2\sigma^2}(\mathbf{y}-\mathbf{X\beta})'(\mathbf{y}-\mathbf{X\beta})\}}$

  • $\begingroup$ $\xi(\theta) $ and $T(y)$ are vectors. So, I think the sum in the initial takes for k=2 the the elements of the vectors and multiplies them and sums it up. $\endgroup$
    – random_guy
    Jul 11, 2014 at 18:40
  • 1
    $\begingroup$ The product term is still missing. A set of sufficient statistics for the problem are $\mathbf{y'y}$, $\mathbf{X'y}$. I don't see the second (vector) statistic anywhere. $\endgroup$
    – Dennis
    Jul 11, 2014 at 18:44
  • $\begingroup$ So, thanks for your comments, but I guess I need to think a bit more about it. I don't understand, what I am doing wrong. At first, I try to do it with matrix notation. Maybe I see your point then. $\endgroup$
    – random_guy
    Jul 11, 2014 at 18:49
  • $\begingroup$ Write out $(\mathbf{y}'-\mathbf{\beta'X'})(\mathbf{y}-\mathbf{X\beta})$ $\endgroup$
    – Dennis
    Jul 11, 2014 at 18:57
  • $\begingroup$ Okay, so I get now $T(y) = (y'y, y'X)$ and $\xi(\theta) = (-\frac{1}{2\sigma^2},\frac{1}{\sigma^2}\beta) $. Is this right? $\endgroup$
    – random_guy
    Jul 11, 2014 at 19:03

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