To rewrite the given pdf in the desired form, we need to express it as the product of two functions: $h(y)$ and $e^{\eta(\theta)T(y)-B(\theta)}$.
In this case, we have:
$h(y) = \frac{1}{\sqrt{2\pi}}$
$\eta(\theta) = -\frac{(y-\theta)^2}{2}$
$T(y) = 1$
$B(\theta) = 0$
Therefore, $\eta(\theta)T(y) = -\frac{(y-\theta)^2}{2}$.
The normal distribution \mathcal{N}(\theta ,1) with with mean \theta and known variance \sigma ^2=1 has pdf
\displaystyle \displaystyle f_{\theta }(y) \displaystyle = \displaystyle \frac{1}{\sqrt{2\pi }}e^{-\frac{(y-\theta )^2}{2}}.
Rewrite f_{\theta } in the form
\displaystyle \displaystyle f_\theta (y) \displaystyle = \displaystyle h(y) e^{\eta (\theta )T(y)-B(\theta )}\qquad \text {where } \eta (\theta ),\, T(y):\mathbb {R}\to \mathbb {R},
and enter the product \eta (\theta )T(y) below.
\eta (\theta ) T(y)=\quad
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