Using the MSE of $c\theta$ to infer $MSE_{c \theta_{hat}}=MSE_{\theta_{hat}}/1+\lambda$

Question

Suppose $\newcommand{\Bias}{\rm Bias}\hat\theta$ is an unbiased estimator of $\theta$ and $\operatorname{Var} \hat\theta =\lambda \theta^2$ for constant $\lambda$. Find the MSE of c$\theta$, where $c$ is a constant. From there, find the value of c that minimizes the mean squared error and deduce that $MSE_{c \hat\theta}=MSE_{\hat\theta}/1+\lambda$

So here's I've done so far. I think my main issue is confusing the c with the $\lambda$. I'm not so sure.

The formulas I'll be using are:

$MSE_{c \hat\theta}=\operatorname{Var} (c \hat\theta)+\Bias^2$ where $\Bias=c\theta-\theta$ hence:

$MSE_{c \hat\theta}=c\theta^{2}+((c-1)\theta)^2=(c^{2}-c+1)\theta$

To find a $c$ value that minimizes MSE, we just take the derivative at $c=0$ where we get $2c-1=0$ and $c=1/2$.

Now moving on to the fraction. I think $MSE_{ \hat\theta}=\operatorname{Var}_{\hat\theta} +\Bias^2=\lambda\theta^2$

From there we plug into the formula and get $(c^2-c+1)\theta^{2}=\lambda\theta^2/1+\lambda$, which I know isn't true.

What did I misinterpret and how can I change it?

Answer 1

If $\operatorname{Var} \hat\theta =\lambda \theta^2$ for constant $\lambda$, then wouldn't $\operatorname{Var} (c\hat\theta) =c^2\lambda \theta^2$ for constant $c$? Leading to $MSE_{c \hat\theta}=c^2\lambda\theta^{2}+((c-1)\theta)^2=(c^{2}(\lambda+1)-2c+1)\theta^2$. The derivative w.r.t. $c$ is thus $2\theta^2(c(\lambda+1)-1)$, which has a single root at $c=1/(\lambda+1)$. With that value of $c$, we have $MSE_{c \hat\theta}=(1/(\lambda+1))^2\lambda\theta^{2}+(((1/(\lambda+1))-1)\theta)^2=\lambda\theta^2/(\lambda+1)=MSE_{\hat\theta}/(\lambda+1)$.