Choose parameters ,such that MSE of an estimator is constant

Question

I have an estimator

:

$X = (X_1,X_2,...,X_n)$ are iid and have distribution $B(1,\theta)$

$T(X) = X_1 + X_2 + ... + X_n$

I need to find such value of constants $\alpha$ and $\beta$ s.t MSE of estimator is constant.

$MSE(\hat \theta) = Var(\hat \theta) + (bias(\hat \theta))^2$;

I calculated $Var(\hat \theta) = \frac{(n*\theta)}{(n + \alpha + \beta)^2}$

And $Bias(\hat \theta) = E(\hat \theta) - \theta = \frac{(n + \alpha)}{n + \alpha + \beta} - \theta$

But if I plug in the results into the formula of MSE I get horrible algebraic calculations.

I have two questions:

1)Are my calculations of Variance and Bias correct?

2)Is there clever trick of solving the exercise?Under clever trick I mean the solution without horrible algebraic calculations.If yes give me a hint.

EDIT: $MSE(\hat \theta) = \frac{(n*\theta)}{(n + \alpha + \beta)^2} + (\frac{(n + \alpha)}{n + \alpha + \beta} - \theta)^2$

Let $A = (n + \alpha + \beta)$

Then $MSE(\hat \theta) = \frac{(n*\theta)}{A^2} + (\frac{(n + \alpha)}{A} - \theta)^2 = \frac{n*\theta + (n + \alpha)^2 + 2(n + \alpha) \theta A + \theta^2 A^2}{A^2}$

And from this place, I suppose I need to apply some clever trick, because otherwise I will be involved in horrible algebraic calculations.

Answer 1

Here is the relevant excerpt from my book Bayesian Choice (2001, Chapter 2):

Consider $x\sim{\mathcal B}(n,\theta)$ when $\theta$ is to be estimated under the quadratic loss, $$ \mathrm{L}(\theta,\delta) = (\delta-\theta)^2. $$ Bayes estimators are then given by posterior expectations (see Section 2.5) and, when $\theta \sim{\mathcal B}e \left({\sqrt{ n} \over 2}, {\sqrt{ n} \over 2} \right)$, the posterior mean is $$ \delta^ \ast (x) = {x+ \sqrt{n}/2 \over n+ \sqrt{ n}}. $$ Moreover, this estimator has constant risk, $\mathrm{R}(\theta,\delta^*) = 1/4(1+\sqrt{n})^2$. Therefore, integrating out $\theta$, $\mathrm{r}(\pi) = \mathrm{R}(\theta,\delta^*)$ and $\delta^*$ is minimax according to Lemma 2.9. Notice the difference with the MLE, $\delta_0(x) = x/n$, for the small values of $n$, and the unrealistic concentration of the prioraround $0.5$ for larger values of $n$.