Comparing 2 Point of Estimate Question


  $X_1$ and $X_2$ are random variables that satisfy $\operatorname{E}[X_1] = \operatorname{E}[X_2] = \mu$ and $\operatorname{Var}[X_1] = \operatorname{Var}[X_2] = 1$.  Specify the range for $\mu$ so that the point estimate $\hat \mu_1 = \frac{X_1 + X_2}{3} + 3$ has a lower mean squared error than the point estimate $\hat\mu_2 = \frac{X_1 + X_2}{2}$.


Hi, I was trying to calculate the $u$ that satisfies the stated condition. However, 

I found $\frac{1}{2} + \frac{(u-9)^2}{3} &lt; \frac{1}{2} + 0^2$  (By using mse(T) = var(T) + bias(t)^2)

But there is no value for u for this inequality, so I wonder if I have a mistake in some point or not.
Note under the assumption of independence, $$\operatorname{MSE}[\hat \mu_1] = \operatorname{Var}\left[\frac{X_1 + X_2}{3} + 3\right] + \operatorname{E}^2\left[\frac{X_1 + X_2}{3} + 3 - \mu\right] = \frac{2}{9} + \left(3-\frac{\mu}{3}\right)^2$$ whereas $$\operatorname{MSE}[\hat \mu_2] = \operatorname{Var}\left[\frac{X_1 + X_2}{2}\right] + \operatorname{E}^2\left[\frac{X_1 + X_2}{2} - \mu\right] = \frac{1}{4}.$$
should find u u which satisfies:
MSE[(X1+X2)/3+3]E{[(X1-u)+(X2-u)]^2+2(X1-u)+(X2-u)+(9-u)^2&lt;(1/4)E[(X1-u)^2+(X2-u)^2+2(X1-u)(X2-u)] or:
(1/9){E[(X1-u)+(X2-u)]^2+2E0+0+E(9-u)^2}&lt;(1/4)[1+1+2COV(X1,X2)] or:
(1/9){E[(X1-u)^2+(X2-u)^2+2(X1-u)(X2-u)]+0+(9-u)^2}&lt;(1/4)[2+EX1X2-(EX)^2] or:
(1/9)[1+1+2COV(X1,X2)]+(9-u)^2}&lt;(1/4)[2+(EX)^2-(EX)^2] or:
(1/9){[2+0]+(9-u)^2}&lt;(1/4)[2+0] or:
(1/9)*[2+(9-u)^2]&lt;(1/2) or:
2[2+(9-u)^2]&lt;9 or:
(9-u)&lt;+-sqrt(4.5-2)
thus solutions are all u, that satisfies:
 |u-9|