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I am currently trying to learn about the exponential family of distributions. I have the example $Y \sim \exp(\lambda)$, where the density is $f_\theta (y) = \theta e^{- \theta y}$ for $y \ge 0$ and $\theta > 0$. Here, we have that $\text{supp}(f_\theta) = [0, \infty)$ doesn't depend on $\theta$. The density $f_\theta$ has the required form with $c(\theta) = \theta$, $T(y) = -y$, $d(\theta) = \ln(\theta)$, $S(y) = 0$. But why do we associate the negative symbol in the exponent of $e^{- \theta y}$ to $T(y) = -y$ rather than $c(\theta) = \theta$? So, in other words, why would we not instead have $c(\theta) = -\theta$ and $T(y) = y$?

The Pointer
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    Look up "natural parameter." – whuber Apr 16 '21 at 15:46
  • Is it because the theta is in front at the beginning of the density (instead of - theta)? – Three Diag Apr 16 '21 at 15:47
  • @whuber I don't see anything that says that the natural parameter must be negative. What exactly are you referring to here? – The Pointer Apr 16 '21 at 16:16
  • @Xi'an What do you mean by this? – The Pointer Apr 16 '21 at 17:12
  • @Xi'an So you're saying that both $T(y) = -y$ and $c(\theta) = \theta$ and $c(\theta) = -\theta$ and $T(y) = y$ are valid? I'm not exactly sure what whuber meant by his comment about the natural parameter, but I thought he was making the point that only $T(y) = -y$ and $c(\theta) = \theta$ is valid. Am I misunderstanding him? – The Pointer Apr 16 '21 at 17:27
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    Both representations define one and the same exponential family,$$A(\theta)\exp{c(\theta)T(y)}$$If $\theta$ is the natural parameter then $c(\theta)=\theta$ and hence $T(y)=-y$. – Xi'an Apr 16 '21 at 17:35
  • @Xi'an Ahh, ok, I see. Thanks for the clarification. – The Pointer Apr 16 '21 at 17:38

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