comparing MSE of estimations of binomial random variables $X$ is a binomial random variable defined over 12 Bernoulli trials with a success probability of  $p$ in each (i.e. $X\sim\operatorname{Bin}(12,p)$. Consider $\hat p=\frac X{10}$
Determine the range for which the mean squared error of $\hat p =\frac X{10}$  is worse than the mean squared error of $\hat p=\frac X {12}$.
I have calculated $\operatorname{MSE}\left(\frac X{12}\right)$ and $\operatorname{MSE}\left(\frac X{10}\right)$ that are equal to $\frac 1{12}\left(p-p^2\right)$ and $ \frac{12p-8p^2}{100}$ respectively.
But comparison of them gives such a stupid result $4p^2> -44p$
Where is my mistake? Do you have any idea?
Thanks in advance.
 A: I will write $\tilde p = \frac1{10}X$ and $\hat p=\frac1{12}X$ to avoid confusion. Recall that the mean-squared error of an estimator $\hat\theta$ of a parameter $\theta$ is
$$\operatorname{MSE}\left(\hat\theta,\theta\right) = \mathbb E\left[(\hat\theta, \theta)^2\right] = \operatorname{Var}\left(\hat\theta\right) + \operatorname{Bias}\left(\hat\theta,\theta\right)^2, $$
where
$$ \operatorname{Bias}(\hat\theta) = \mathbb E\left[\hat\theta\right] - \theta.$$
We can further simplify this as 
$$\begin{align*} \operatorname{MSE}(\hat\theta) &= \mathbb E\left[\hat\theta^2\right] - \mathbb E\left[\hat\theta\right]^2 + \left(\mathbb E\left[\hat\theta\right]-\theta\right)^2\\
&= E\left[\hat\theta^2\right] - \mathbb E\left[\hat\theta\right]^2 + \mathbb E\left[\hat\theta\right]^2-2\theta\mathbb E\left[\hat\theta\right]+\theta^2\\
&= E\left[\hat\theta^2\right] -2\theta\mathbb E\left[\hat\theta\right]+\theta^2.
\end{align*}$$
So we compute
$$ \begin{align*}
\operatorname{MSE}(\tilde p) &= E\left[\tilde p^2\right] -2 p\mathbb E\left[\tilde p\right]+p^2\\
&= \left(\frac1{10}\right)^2\mathbb E[X^2] - 2p\cdot\frac1{10}\mathbb E[X] + p^2\\
&= \frac1{25}p(3-2p),
\end{align*}
$$
and
$$\begin{align*}
\operatorname{MSE}(\hat p) &= E\left[\hat p^2\right] -2 p\mathbb E\left[\hat p\right]+p^2\\\\
&= \left(\frac1{12}\right)^2\mathbb E[X^2] - 2p\cdot\frac1{12}\mathbb E[X] + p^2\\
&=\frac1{12}p(1-p).
\end{align*}
$$
The values of $p$ for which $\operatorname{MSE}\left(\tilde p\right)>\operatorname{MSE}\left(\hat p\right)$ satisfy
$$ \frac1{25}p(3-2p) > \frac1{12}p(1-p),$$
which are
$$(-\infty, -11)\cup(0,\infty).$$
Hence $\operatorname{MSE}\left(\tilde p\right)>\operatorname{MSE}\left(\hat p\right)$ for all $p\in(0,1)$.
