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
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, $\eta(\theta)T(y) = -\frac{(y-\theta)^2}{2}$.
To rewrite the normal distribution function in the desired form, we can start by expressing the exponent in the exponent of the exponential term as a square.
Let's expand the square in the exponent of the given pdf:
\begin{align*}
(y - \theta)^2 &= y^2 - 2y\theta + \theta^2 \\
&= \theta^2 - 2y\theta + y^2 \\
&= \left(\theta^2 - 2\theta y + y^2\right)
\end{align*}
Now, we can rewrite the given pdf as follows:
\begin{align*}
f_{\theta}(y) &= \frac{1}{\sqrt{2\pi}}e^{-\frac{(y-\theta)^2}{2}} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(\theta^2 - 2\theta y + y^2)} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}(y^2 - 2\theta y + \theta^2)} \\
&= \frac{1}{\sqrt{2\pi}}e^{-\frac{1}{2}y^2 + \theta y - \frac{1}{2}\theta^2} \\
&= \frac{1}{\sqrt{2\pi}}e^{\theta y - \frac{1}{2}y^2 - \frac{1}{2}\theta^2}
\end{align*}
By comparing the final expression with the desired form of $\displaystyle f_{\theta }(y)$, we can see that:
\begin{align*}
h(y) &= \frac{1}{\sqrt{2\pi}} \\
\eta(\theta) &= \theta \\
T(y) &= y \\
B(\theta) &= \frac{1}{2}(y^2 + \theta^2)
\end{align*}
Therefore, we have $\displaystyle \eta(\theta) T(y) = \theta y$.