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
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    $\begingroup$ 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

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