Fisher's information for two independent random variables If $X$ and $Y$ are two independent random variables, with regular distributions, how can I prove
$I_{x,y}(\theta) = I_x(\theta) + I_y(\theta)$       ?
Thanks!
I tried:
$$ {\rm E}_\theta \left[\left( \frac {\partial}{\partial\theta} \log f_\theta^{xy} (x,y)\right)^2 \right] ={\rm E}_\theta \left[\left( \frac {\partial}{\partial\theta} \log f_\theta^{x} (x)\right)^2 \right]  + {\rm E}_\theta \left[\left( \frac {\partial}{\partial\theta} \log f_\theta^{y} (y)\right)^2 \right]  + {\rm E}_\theta \left[\left( \frac {\partial}{\partial\theta} \log f_\theta^{y} (y)\right) \left( \frac {\partial}{\partial\theta} \log f_\theta^{x} (x)\right) \right]  $$
and the last element should be equal to zero.
 A: First of all if $X$ and $Y$ are two independent random variables, then
$$f_{(X,Y)}(x,y;\theta)=f_X(x;\theta)\cdot f_Y(y;\theta)$$
$$\log (f_{(X,Y)}(x,y;\theta))=\log(f_X(x;\theta))+ \log(f_Y(y;\theta))$$
As It was mentioned under certain regularity conditions:
$$\mathcal{I}(\theta) =\operatorname{E} \left[\left. \left(\frac{\partial}{\partial\theta} \log f(X;\theta)\right)^2\right|\theta \right] = - \operatorname{E} \left[\left. \frac{\partial^2}{\partial\theta^2} \log f(X;\theta)\right|\theta \right]\,$$
So
$$
\begin{eqnarray}
\mathcal{I}_{XY}(\theta) &=&- \operatorname{E} \left[\left. \frac{\partial^2}{\partial\theta^2} (\log(f_X(x;\theta))+ \log(f_Y(y;\theta))\right|\theta \right]\,=\\&=&- \operatorname{E} \left[\left. \frac{\partial^2}{\partial\theta^2} \log(f_X(x;\theta))\right|\theta \right]\,- \operatorname{E} \left[\left. \frac{\partial^2}{\partial\theta^2}  \log(f_Y(y;\theta))\right|\theta \right]\,=\\
&=&\mathcal{I}_X(\theta)+\mathcal{I}_Y(\theta)
\end{eqnarray}
$$
A: Suppose $X$ and $Y$ are independent random variables whose distribution depends on some parameter $\theta$. Then the density of $(X,Y)$ is the product of the marginal densities, i.e. $$
f_{(X,Y)}(x,y;\theta)=f_X(x;\theta)\cdot f_Y(y;\theta)\tag{1}
$$ for all $(x,y)$ and $\theta$. The Fisher information of $(X,Y)$ is
$$
I_{(X,Y)}(\theta)={\rm E}\left[\left(\frac{\partial}{\partial\theta}\log f_{(X,Y)}(X,Y;\theta)\right)^2\right].
$$
Under certain regularity conditions (which ensures that we may interchange ${\rm E}$ and $\frac{\partial}{\partial\theta}$) we have that 
$$
{\rm E}\left[\frac{\partial}{\partial\theta}\log f_{(X,Y)}(X,Y;\theta)\right]=0
$$
and hence
$$
I_{(X,Y)}(\theta)=\mathrm{Var}\left(\frac{\partial}{\partial\theta}\log f_{(X,Y)}(X,Y;\theta)\right).
$$
Use this to conclude.
A: First we have that X is independent of Y, which implies that so is their score function, given by:
$ \frac {\partial}{\partial\theta} \log f_\theta^{x} (x)$ and $ \frac {\partial}{\partial\theta} \log f_\theta^{y} (y)$
We have that the likelihood function for X and Y is given by:
$L(\theta)= f_\theta^{xy} (x,y) =f_\theta^{x} (x)f_\theta^{y} (y)$
And Fisher Information is given by:
$ {\rm E}_\theta \left[\left( \frac {\partial}{\partial\theta} \log f_\theta^{xy} (x,y)\right)^2 \right] = {\rm Var} \left[\frac {\partial}{\partial\theta} \log f_\theta^{xy} (x,y) \right] = {\rm Var} \left[\frac {\partial}{\partial\theta} \log f_\theta^{x} (x)f_\theta^{y} (y) \right] = {\rm Var} \left[\frac {\partial}{\partial\theta} \log f_\theta^{x} (x) + \frac{\partial}{\partial\theta} \log f_\theta^{y} (y) \right]$
Now, since the score functions are independent:
${\rm Var} \left[\frac {\partial}{\partial\theta} \log f_\theta^{x} (x) + \frac{\partial}{\partial\theta} \log f_\theta^{y} (y) \right] = {\rm Var} \left[\frac {\partial}{\partial\theta} \log f_\theta^{x} (x)\right] + {\rm Var} \left[\frac{\partial}{\partial\theta} \log f_\theta^{y} (y) \right] = I_X(\theta) + I_Y(\theta)$
Concluding the proof. 
Remember: This property do not require that the log of distribution function is twice differentiable in $\theta$. 
