# Finding a probability density from an exponential family using a moment generating function

I would like to find the density of a sum of i.i.d. random variables $$\bar{Y}=\frac{1}{\nu}(Y_1+...+Y_\nu)$$, where the density of these random variables is $$f_Y(y;\theta)=e^{\theta y - b(\theta)}f_0(y)$$. In turn, the density $$f_0(y)$$ is specified using the m.g.f. $$M_0(\xi)=e^{b(\xi)}$$ and a differentiable function $$b(\xi)$$.

Edit: Found an additional bit of information that the resulting distribution has to be from the exponential family $$f_\bar{Y}(y;\theta, \phi)= \exp{(\frac{y\theta-b(\theta)}{a(\phi)}+c(y, \phi))}$$, where $$a(\phi)=\nu^{-1}$$.

Attempt 1:

Omitting the details, I found the m.g.f. of $$f_Y(y,\theta)$$ to be equal to $$M_Y(\xi)=e^{b(\xi + \theta)-b(\theta)}$$.

Then, I computed the m.g.f. of the sum of $$Y_i$$'s using this property, which resulted in $$M_{\bar{Y}}(\xi)=\left(M_Y(\frac{1}{\nu}\xi)\right)^\nu=e^{\nu\left(b(\frac{1}{\nu}\xi+\theta)-b(\theta)\right)}$$ I cannot find a way to transform this into $$f_{\hat{Y}}(y; \theta)$$ and would greatly appreciate a pointer or a suggestion on how to proceed.

Attempt 2:

What seems like a possible solution is to recast $$M_{\bar{Y}}$$ into the form of $$M_Y$$ and reconstruct the density from that. For example, if one were to introduce substitutions like $$\tilde{b}(\theta)=\nu b(\theta)$$ and $$\tilde{\xi}=\frac{1}{\nu}\xi$$, then the expressions of the two m.g.f.'s would be identical. This means that $$M_{\bar{Y}}(\tilde{\xi})= \int_{-\infty}^{\infty}e^{\tilde{\xi}y}e^{\theta y-\tilde{b}(\theta)}f_0(y)dy= \int_{-\infty}^{\infty}e^{\tilde{\xi}y}e^{\theta y-\nu b(\theta)}f_0(y)dy.$$ I am not entirely sure how to deal with $$\tilde{\xi}$$ at this point... Is it a permissible operation to transform $$y$$ in the integral into $$\tilde{y}=\frac{1}{\nu} y$$? It would then follow that $$\int_{-\infty}^{\infty}e^{\tilde{\xi}y}e^{\theta y-\nu b(\theta)}f_0(y)dy= \int_{-\infty}^{\infty}e^{\xi\tilde{y}}e^{\nu\theta \tilde{y}-\nu b(\theta)}\nu{}f_0(\nu\tilde{y})d\tilde{y}= \int_{-\infty}^{\infty}e^{\xi\tilde{y}}e^{\nu\theta \tilde{y}-\nu b(\theta)+\ln(\nu{}f_0(\nu\tilde{y}))}d\tilde{y}.$$ The density would then be expressed as $$f_\hat{Y}(y;\theta, \nu)=e^{\nu(\theta \tilde{y}-b(\theta))+\ln(\nu{}f_0(\nu\tilde{y}))}$$ and it would have the form in the desired form mentioned in the Edit above.

Would this make sense? My math background is not really so strong, so I have a feeling that I could have made some illegitimate operations here...