--edit-- Bayes risk is defined as: $$E_\theta[R_T(\theta)]=\int_\Omega R_t(\theta)p(\theta)d \theta$$ where $p$ is the pdf of $\theta$, $R_t$ is the risk function of the estimator
I'm stuck with a question, 9.27 in Bain & Engelhardt about Bayes risk. The loss function is given by $L(t;\theta)=(t-\theta)^2$. I have the following estimators of $\theta$: $\hat{\theta}_3, \hat{\theta}_4$ with respective risk functions of: $R(\hat{\theta}_3;\theta)=4/9+\theta^2/9$, $R(\hat{\theta}_4;\theta)=2/9+\theta^2/9$. So I have to compute Bayes risk for these. It is given that $\theta \sim Beta(2,1)$ I start with $\hat{\theta}_3$:
$$E[R(\hat{\theta}_3;\theta)]=E[4/9+\theta^2/9]=4/9+(1/9)E(\theta^2)=4/9+(1/9)[E(\theta^2)-\mu^2]+(1/9)\mu^2=4/9+(1/9)\sigma^2+(1/9)\mu^2$$
Where $\mu=E(\theta), \sigma^2=Var(\theta)$ We know from $Beta(2,1)$ that: $\mu^2=(2/3)^2=4/9$ and $\sigma^2=1/18$ Thus we have:
$$E[R(\hat{\theta}_3;\theta)]=4/9+(1/9)\sigma^2+(1/9)\mu^2=4/9+1/162+4/81=(72+1+8)/162=81/162=1/2$$
While the solutions say $13/27$. The difference is very small. But the thing is, I followed the same procedure with the other estimator and again my answer was almost right, but not precisely right according to the solutions. So I suspect something is wrong with my approach. Very thankful to the person who can point me to the error.
Your prior function is
\begin{equation} Beta(2,1) = 2\theta \end{equation}
Inserting this into the formula for Bayes risk gives:
\begin{equation} \int_{0}^{1} R_t (\theta) 2\theta d\theta \end{equation}
For example for the third estimator you have:
\begin{equation} R_t (\theta_3) = \frac{4}{9} + \frac{1}{9} \theta^2 \end{equation}
Inserting this in the integral gives:
\begin{equation} \int_{0}^{1} \frac{8}{9}\theta + \frac{2}{9}\theta^3 d\theta = \frac{1}{2}\end{equation}
edit: $2\theta \times (\frac{4}{9}+\frac{1}{9} \theta ^2)=\frac{8}{9}\theta + \frac{2}{9}\theta^3 $ previously claimed: $2\theta \times (\frac{4}{9}+\frac{1}{9} \theta ^2)=\frac{8}{9}\theta + \frac{1}{9}\theta^3 $