Basic Hellinger Distance One definition of Hellinger Distance is $$L_{H}: (1/2) E_{\theta}([\sqrt{\frac{f(x|\delta)}{f(x|\theta)}}-1])^{2}$$
My book has that  for $x \sim N(\theta,1)$
$$L_{H}(\theta,\delta)=1-\exp(-(1/8)(\delta-\theta)^{2})$$
I am not sure the easiest way to understand this. Firstly, the derivation, and why it makes sense and what it represents. I don't really understand what the second distribution is. And if the form given is best or if one should use a better form. I assume I just use normal for the second since I am not sure. The final form looks similar to the forms given on the wiki
 A: Hellinger distance will give you the distance between densities in the same family, but which have different parameters. Regarding the derivation, perhaps this will help:
\begin{align*}
E_{\theta}\left[\left(\sqrt{\frac{f(x|\delta)}{f(x|\theta)}}-1\right)^{2}\right] &= E_{\theta}\left[\left(\frac{f(x|\delta)}{f(x|\theta)} +1 - 2\sqrt{\frac{f(x|\delta)}{f(x|\theta)}} \right)^{2}\right] \\
&= \int \left(\frac{f(x|\delta)}{f(x|\theta)} +1 - 2\sqrt{\frac{f(x|\delta)}{f(x|\theta)}} \right)^{2} f(x|\theta) \text{d}x \\
&= \int f(x|\delta)\text{d}x +\int  1 f(x|\theta)\text{d}x  - 2 \int \sqrt{\frac{f(x|\delta)}{f(x|\theta)}}  f(x|\theta) \text{d}x \\
&= 1 + 1 - 2\int \sqrt{\frac{f(x|\delta)}{f(x|\theta)}}  f(x|\theta) \text{d}x \\
&= 2\left(1 - \int \sqrt{f(x|\delta) f(x|\theta)}   \text{d}x \right).
\end{align*}
Edit: I'll take it a bit further:
\begin{align*}
\int \sqrt{f(x|\delta) f(x|\theta)}   \text{d}x &= \int (2\pi)^{-1/2} \exp\left[-\frac{1}{4}\left((x- \delta)^2 + (x - \theta)^2  \right)\right]   \text{d}x \\
&= \int (2\pi)^{-1/2} \exp\left[-\frac{1}{4}\left( 2x^2 + \delta^2 + \theta^2 - 2x\delta - 2 x \theta   \right)\right]   \text{d}x \\
&= \int (2\pi)^{-1/2} \exp\left[-\frac{1}{4}\left\{ 2( x - (\theta - \delta)/2)^2 + \frac{1}{2}(\delta - \theta)^2 \right\} \right]   \text{d}x \\ 
&= \exp\left[-\frac{1}{8}(\delta - \theta)^2 \right] \int (2\pi)^{-1/2} \exp\left[-\frac{( x - (\theta - \delta)/2)^2}{2}  \right]   \text{d}x \\ 
&= \exp\left[-\frac{1}{8}(\delta - \theta)^2 \right].  
\end{align*}
The trick is to "complete the square" so that we can get the integral of a density (which is $1$).
