Suppose $Y$ is a random variable in the exponential family, with pmf/pdf $$f(y) = \exp\left[\sum_{j=1}^{s}\theta_jT_j(y)-B(\theta)+c(y) \right]$$ for $y \in \Omega$ (the support of $Y$), and where $$\theta = \begin{bmatrix} \theta_1 \\ \theta_2 \\ \vdots \\ \theta_s\end{bmatrix}$$ is the canonical parameter.

I wish to show that

$$\Omega = \{y:\exp[c(y)] > 0\}\text{.}$$

where $\Omega$, the support of $Y$, is defined by $\{y: f(y) > 0\}$.

The Google searching I've found either just mentions this as a fact, or states that it's obvious. And I don't see why it is.

If we set $f(y) > 0$, we obtain $$\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]\exp[-B(\theta)]\exp[c(y)] > 0$$ or $$\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]\exp[c(y)] > 0\text{.}$$ The problem is, though, how does this imply that $\exp[c(y)] > 0$? Why is it that we ignore $\exp\left[\sum_{j=1}^{s}\theta_jT_j(y)\right]$?

  • $\begingroup$ Can you find a number $x$ such that $\exp\{x\} \leq 0$? $\endgroup$ – jbowman Oct 3 '18 at 1:24
  • $\begingroup$ @jbowman No, as long as $x \in \mathbb{R}$ (not allowing for $-\infty$). $\endgroup$ – Clarinetist Oct 3 '18 at 1:30
  • 1
    $\begingroup$ So $\exp[c(y)] > 0$ because it can't be otherwise. $\endgroup$ – jbowman Oct 3 '18 at 1:34
  • $\begingroup$ @jbowman I was doubting myself when I thought of something similar, because it seemed too simple... thanks! $\endgroup$ – Clarinetist Oct 3 '18 at 1:54
  • $\begingroup$ Been there, done that! $\endgroup$ – jbowman Oct 3 '18 at 1:59

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Browse other questions tagged or ask your own question.