https://i.stack.imgur.com/BZauH.png
I'm interested in part (g), specifically. I know that the risk function is the expected value of the loss function, but I'm confused as to how to evaluate it. When I replace $\theta$ with $\tilde \theta$, I get an expression which is pretty difficult to evaluate the expected value of... so perhaps I'm not going about this the right way. How would you calculate this risk function?
Find risk function for $\bar X$. $$\mathbb E[X_1]=\frac{1-p}{p}=\theta, \ \text{Var}[X_1]= \frac{1-p}{p^2}=\theta^2+\theta. $$ Then $$\mathbb E[\bar X]=\theta, \ \text{Var}[\bar X]= \frac{\theta^2+\theta}n.$$ Risk runction for $\bar X$ is $$ \mathbb E\left[\frac{(\bar X-\theta)^2}{\theta^2+\theta}\right] = \frac{\text{Var}[\bar X]}{\theta^2+\theta} = \frac1n. $$ Find risk function for $\tilde \theta=\frac{n+1}{n}\bar X$. $$ \mathbb E\left[\frac{\left(\frac{n+1}{n}\bar X-\theta\right)^2}{\theta^2+\theta}\right] =\frac{\mathbb E\left[\left(\frac{n+1}{n}\bar X-\theta\right)^2\right]}{\theta^2+\theta}=\frac{\text{Var}\left[\frac{n+1}{n}\bar X-\theta\right]+\left(\mathbb E\left[\frac{n+1}{n}\bar X-\theta\right]\right)^2}{\theta^2+\theta} = \frac{\frac{(n+1)^2}{n^2}\text{Var}\left[\bar X\right]+\frac{\theta^2}{n^2}}{\theta^2+\theta} =\frac{(n+1)^2}{n^3}+\frac{\theta}{n^2(\theta+1)} $$