Let random sample $(X_{1},...X_{n})$ is taken from a population with mean $\mu $ and variance $\sigma ^{2}$. Compare suggesting estimators for $\mu $ according to mean squared error.
Suggesting estimators are , $T_{1}=\frac{3X_{1}-X_{2}-X_{3}+4X_{n}}{5}$ and $T_{2}=\frac{X_{1}}{2}-\frac{X_{2}}{3}+\frac{X_{n}}{6}$
Here is my solution : $E(X)=\mu $ and $var(X)=\sigma ^{2}$
$E(T_{1})=E(\frac{3X_{1}-X_{2}-X_{3}+4X_{n}}{5})=\frac{5\mu }{5}=\mu $ and $var(T_{1})=var(\frac{3X_{1}-X_{2}-X_{3}+4X_{n}}{5})=\frac{27}{25}\sigma ^{2}$
$MSE(T_{1};\mu )=Var(T_{1})+(E(T_{1})-\mu ))^{2}=\frac{27}{25}\sigma ^{2}$
$E(T_{2})=E(\frac{X_{1}}{2}-\frac{X_{2}}{3}+\frac{X_{n}}{6})=\frac{1}{3}\mu $ and $var(T_{2})=var(\frac{X_{1}}{2}-\frac{X_{2}}{3}+\frac{X_{n}}{6})=\frac{7}{18}\sigma ^{2}$
$MSE(T_{2};\mu )=Var(T_{2})+(E(T_{2})-\mu ))^{2}=\frac{7}{18}\sigma ^{2}+\frac{4}{9}\mu ^{2}$
But how can I compare these two mean squared error. I do not know the value of $\mu ^{2}$
$MSE(T_{1};\mu )<MSE(T_{2};\mu )$ or $MSE(T_{2};\mu )<MSE(T_{1};\mu )$?
The method of evaluating two estimators conparing their MSE is applied if both are biased, or in case of both unbiased comparing their variances.
In your case you have $T_1$ unbiased and $T_2$ biased thus $T_1$ is preferred
EDIT
As I told before, the usual procedure is to prefer the unbiased estimator Vs the biased one, but obviously different decisions can be taken w.r.t. the various values of $\mu$. For example if $\mu=0$ $T_2$ is preferred.
Setting $\sigma^2=1$ without loss of generality, observe the following drawing
As you can see $T_2$ is preferred to $T_1$ when the "squared true mean" $\mu^2$ is less than 1.56.