Saddlepoint approximation for Exponential family I read the following in a book:
The saddlepoint approximation of an exponential family density function is $$\tilde P(y;\mu,\phi) = \frac{1}{\sqrt{2\pi \phi V(y)}}exp(-\frac{d(y, \mu)}{2\phi})$$
Where $d(y, \mu)$ is the unit deviance; $V(y) = \frac{d\mu}{d\theta}$ evaluated at $\mu=y$.
I managed to see this for individual distributions (e.g. Poisson) but not for the general form. How do I differentiate the unit deviance twice and get that it's equal to $V(y)$?
 A: We want to approximate the $b(y, \phi)$ in the following representation of an expo-family pdf:
$$\mathcal{P}(y;\mu,\phi) = b(y,\phi)e^{-\frac{d(y,\mu)}{2\phi}}
$$
Where the b-function is the regular $c(y,\phi)$ of the expo-family + absorbing the $t(y,y)$ part of the unit deviance $d(y,\mu)$, and we do this using Laplace's method / Saddle point approximation.
So - $f(y) = e^{-\frac{d(y,\mu)}{2\phi}}$, i.e. the function without the "bounding constant" (b-function).
$h(y) = \ln f(y)=-\frac{d(y,\mu)}{2\phi} = -2\frac{t(y,y)-t(y,\mu)}{2\phi}=-\frac{y\theta(\mu)|_{\mu=y}-b(\theta(\mu))|_{\mu=y}-y\theta(\mu)+b(\theta(\mu))}{\phi}$.
We need to calculate $f(\hat y)\sqrt{2\pi\frac{1}{|h''(\hat y)|}}$, where $\hat y$ is the mode (stationary point) of $h(y)$. (This is the Saddle-Point/Laplace approx. for the bounding constant).
$h'(y) = -\frac{1}{\phi}(\theta(\mu)|_{\mu=y} + y \frac{d\theta}{d\mu}|_{\mu=y}- \frac{db}{d\theta} \frac{d\theta}{d\mu}|_{\mu=y}-\theta(\mu))
$
Now $\frac{d\theta}{d\mu} = \frac{1}{V(\mu)}$, and $\frac{db}{d\theta}=\mu$, evaluated at $\mu=y$ we get that the two middle terms cancel out. And we get that for the mode, $\theta(\mu)|_{\mu=\hat y} = \theta(\mu)$. This means that $h(\hat y) = 0$, and $f(\hat y)$ = 1.
Taking the 2nd derivative we get:
$h''(y) = -\frac{1}{\phi}\frac{d\theta}{d\mu}|_{\mu=y} = -\frac{1}{\phi V(y)}
$
Hence the approximation to the f function integral comes out to be: $f(\hat y)\sqrt{2\pi\frac{1}{|h''(\hat y)|}} = \sqrt{2\pi \phi V(y)}$, and so the b-function is approximated by $\frac{1}{\sqrt{2\pi \phi V(y)}}$
