Suppose $X_i$'s are independent, where $i=1,2,...,n$. The density function of $X_i$ is $f_{\theta_i}(x)$. Let Fisher information of $X=(X_1, ...,X_n)$ be $I_X$. Then $$I_X = \sum_{i=1}^{n}I_{X_i},$$ where $I_{X_i}$ is the Fisher information of distribution $f_{\theta_i}(x)$.
My question does $\lim_{n\rightarrow\infty}\frac{I_{X}}{n}$ exists? If so, deonte the limit as $I_{\infty}$. Does $\frac{I_{X}}{n}$ converge to $I_{\infty}$ in probability?