$f_Y (y; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp(-\frac{(y-\mu)^2}{2\sigma^2 })$ is the normal distribution pdf
I am trying to get this in the form $$Y∼f_Y (y;θ,ϕ)= \exp\left[\frac{yθ-b(θ)}{a(ϕ)}+ c(y,ϕ)\right]$$
my notes did this in one step
$$f_Y (y; \mu, \sigma^2) = \exp\left[\frac{y \mu - \mu^2 /2 }{\sigma^2}-\frac{1}{2}\left(\frac{y^2}{\sigma^2} + \log(2 \pi \sigma^2 ) \right)\right]$$
Then its pretty obvious that $b(\theta ) = \mu^2 /2, \theta = \mu , \phi = \sigma^2 $.
I'm just strugging with the last step
my attempt:
$$f_Y (y; \mu, \sigma^2) = \frac{1}{\sqrt{2\pi \sigma^2}} \exp(-\frac{(y-\mu)^2}{2\sigma^2 })$$
$$= \exp\left( - \frac{1}{2} \log(2 \pi \sigma)- \frac{(y-\mu)^2}{2 \sigma^2}\right)$$
$$= \exp\left( - \frac{1}{2} \log(2 \pi \sigma)- \frac{y^2 - 2y\mu + \mu^2}{2 \sigma^2}\right)$$
$$= \exp\left(\frac{2 y \mu - \mu^2}{2\sigma^2} - \frac{1}{2}\left( \frac{y^2}{\sigma^2}+\log(2 \pi \sigma^2)\right)\right)$$
I got $b(\theta ) = \mu^2, \theta = 2\mu , \phi = 2\sigma^2 $.
They apparently divided by 2. Is that neccessary? Can I leave my answer as is?
If your starting point is that $f_Y (y; \mu, \sigma^2) = f_Y (y; \theta, \phi)$ then you are requiring that $\mu = \theta$ and $\sigma^2 = \phi$ from the outset, which means you do have to write that first term as $$\frac{y \mu - \mu^2 /2 }{\sigma^2}$$ so that $\theta$ matches up with $\mu$.
In other words, you're not free to set $\theta = 2\mu$ the way you have done.