Comparing Fisher Information of sample to that of statistic Let $X_1,...,X_n$ be Bernoulli($p$) where $p$ is unknown, and $n>2$, and let $T=X_1+X_2$. My task is to calculate the information about $p$ in the entire sample and compare it to the information about $p$ given by the statistic. 
After a few lines of work, I obtain the following expression for information contained in the sample: $$I_X(p)= n*E_p[(\frac{X_1p^{-1}(1-p)+X_1+1}{1-p})^2] $$
To calculate this, I first calculated the information given by one observation, and then multiplied that information by $n$.
Now, after some work, I obtained the following expression for  information contained in $T$: $$I_T(p)=E_p[(\frac{Tp^{-1}(1-p)-1}{1-p})^2]$$
Now, I know from previous questions that $T$ is sufficient for $p$. Hence, $I_X(p)$ should equal $I_T(p)$. However, I have no idea how I am to compare these two quantities, because one has an $n$, and the other has an $X_2$ (in the $T$). Any advice?

Fisher Information Matrix (FIM), is the negative of the Expectation of the Hessian of the log likelihood function, namely
\begin{equation}
 I(p) = - E H(p)
\end{equation}
where $I(p)$ is the FIM and $H(p)$ is the Hessian. Your likelihood function for independent samples is
\begin{equation}
 L(p) = f(x_1\vert p) \ldots f(x_n \vert p)
\end{equation}
where 
\begin{equation}
 f(x_i \vert p) 
 =
 p^{x_i} (1 - p)^{1 - x_i}
\end{equation}
Both equations above give us the likelihood function
\begin{equation}
 L(p)
 =
 p^{ \sum x_i} (1 - p)^{n - \sum x_i}
\end{equation}
Log likelihood becomes 
\begin{equation}
 l(p) = \log L(p) 
 =
 \log \big( p^{ \sum x_i} (1 - p)^{n - \sum x_i} \big)
\end{equation}
Denoting $\bar{x} = \frac{1}{n} \sum x_i$ for sake of simple presentation, one gets: 
\begin{equation}
 l(p)
 =
 n\bar{x} \log(p)
 +
 n(1 - \bar{x})\log(1-p)
\end{equation}
The score function (gradient of the log likelihood) is
\begin{equation}
 s(p) 
 =
 l'(p)
 =
 n
 \frac{\bar{x}}{p}
 -
 n
 \frac{(1- \bar{x})}{1 - p}
\end{equation}
The Hessian becomes
\begin{equation}
 H(p) = s'(p)
 =
 - \frac{n(1-2p)\bar{x} + np^2}{p^2(1-p)^2}
\end{equation}
The FIM becomes
\begin{equation}
 I(p)= -E H(p)
 =
 E\big[ \frac{n(1-2p)\bar{x} + np^2}{p^2(1-p)^2} \big]
 =
 \big[ \frac{n(1-2p)E\bar{x} + np^2}{p^2(1-p)^2} \big]
\end{equation}
But $E\bar{x} = p$ so
\begin{equation}
 I(p) 
 =
 E\big[ \frac{n(1-2p)\bar{x} + np^2}{p^2(1-p)^2} \big]
 =
 \big[ \frac{n(1-2p)p + np^2}{p^2(1-p)^2} \big]
 =
 \frac{n}{p(1-p)}
\end{equation}
Now, 
\begin{equation}
 T = X_1 + X_2
\end{equation}
and the distribution of $T$ is 
\begin{equation}
 g(T=t\vert p) = Pr(T = t) = C_2^t p^t(1-p)^{n-t}
\end{equation}
Using Bayes theorem, we know that
\begin{equation}
 f(x_1 \ldots x_n \vert T = t, p) = \frac{f(x_1 \ldots x_n,t=T \vert p)}{g(T= t \vert p)}
\end{equation}
Let's get $f(x_1 \ldots x_n,t=T \vert p)$, the likelihood function becomes 
\begin{equation}
 f(x_1 \ldots x_n , T = t\vert  p) = 
 p^{t + \sum\limits_{i=3}^n x_i}(1-p)^{n - t - \sum\limits_{i=3}^n
 x_i}
\end{equation}
\begin{equation}
 f(x_1 \ldots x_n \vert T = t, p) = \frac{f(x_1 \ldots x_n,t=T \vert p)}{g(T= t \vert p)}
 =
 \frac{p^{t + \sum\limits_{i=3}^n x_i}(1-p)^{n - t - \sum\limits_{i=3}^n
 x_i}}{C_2^t p^t(1-p)^{n-t}}
\end{equation}
which is
\begin{equation}
 f(x_1 \ldots x_n \vert T = t, p) 
 =
 \frac{p^{ \sum\limits_{i=3}^n x_i}(1-p)^{ - \sum\limits_{i=3}^n
 x_i}}{C_2^t}
 =
 \frac{1}{C_2^t}
 \Big(\frac{p}{1-p}\Big)^{\sum\limits_{i=3}^n
 x_i}
 =
 h(p)
\end{equation}
which is a function of $p$. 

Hence, $T$ is NOT a sufficient statistics of $p$. However, if $T$ is the sum of ALL your samples, then you'd get a sufficient statistic.

