showing that $\bar{X}$ is inadmissible by comparing with $\max(\bar{X},2)$ under squared error loss function suppose $X_1,X_2,\ldots,X_n$ be a random sample of $N(\theta,1), \theta>2$.
how can I show $\bar{X}$ is inadmissible estimator 
Compared to $\max(\bar{X},2)$  under Squared error loss function
 A: $%preamble
 \newcommand\Var{\mathrm{Var}}
 \newcommand\E{\mathrm{E}}
 \newcommand\Bias{\mathrm{Bias}}
 \newcommand\htheta{\hat\theta}
$
Let X be a (integrable) random variable and $c$ a constant, then flesh out the derivation  
$$\E[(X-c)^2] = \E[((X-\E[x])-(c-\E[x]))^2] = \Var(X)+(\E[x]-c)^2$$
Then, in particular, reason and memorize that if $\hat\theta$ is an estimator for $\theta$ then
$$\E[(\htheta-\theta)^2] = \Var[\htheta] + \Bias(\htheta,\theta)^2$$
In the case of $\htheta_1 = \bar{X}$, determine $\Var[{\htheta_1}]$ as a function of $n$ and verify $\Bias(\htheta_1,\theta)^2 = 0$. 
It should be somewhat intuitive that estimator $\htheta_2 = \max(2,\htheta_1)$ is biased but has lower variance, but not immediately clear that the total MSE is lower.

Recall the important rules of total expectation and total variance, that if X and Y are random variables (with finite expectation and variance) then
$$\E[X] = \E[\E[X|Y]] $$
$$\Var[X] = \Var[\E[X|Y]] + \E[\Var[X|Y]]$$

Let $I$ a subset of the probability space (an event). Then, using the above rules and an indicator random variable, see that
$$\E[X] = \E[X|I] P[I] + \E[X|I^c] P[I^c]$$
$$\Var[X] = (\Var[X|I] P[I] + \Var[X|I^c] P[I^c]) + (\E[X|I]-\E[X|I^c])^2 P[I]P[I^c]$$

In particular, let $I = (2,\infty)$, and use the above for $\htheta_1$ and $\htheta_2$ to determine expressions for their variance and squared bias in terms conditioned on $I$ and $I^c$. Since both estimators are the same conditional on $I$, you should see opportunity to make the necessary inequality.
