Finding variance from a beta distribution Suppose $[X\mid \theta]\sim\operatorname{Uniform}([0, \theta])$ and $B$ is is a beta distributed random variable.  Additionally, $\max(X_1,…,X_n)/θ\sim \operatorname{Beta}(n+1,1)$
Show that $\hat{θ}= \frac{n+2}{n+1} \max(X_1,…,X_n)$ and find the variance of $\hat\theta$ given $\theta$.

I found: $E[B]= \frac{n+1}{n+2}$ and $V[B]= \frac{n+1}{(n+2)^2(n+3)}$
How do I find the variance of $\hat{\theta}\mid \theta?$
 A: I will assume $\hat{\theta} = \frac{n+2}{n+1} \max(X_1, \ldots, X_n)$ is a definition. (I am confused why you ask "show $\hat{\theta} = \frac{n+2}{n+1} \max(X_1, \ldots, X_n)$" when $\hat{\theta}$ is not defined elsewhere.)
If $(X_i \mid \theta)$ are Uniform$[0, \theta]$ and conditionally independent given $\theta$, then the conditional distribution of $$\max(X_1, \ldots, X_n)/\theta$$ given $\theta$ is a $\text{Beta}(n,1)$ distribution, which has mean $\frac{n}{n+1}$ and variance $\frac{n}{(n+1)^2 (n+2)}$.
If you multiply $\max(X_1, \ldots, X_n)/\theta$ by the scalar $\frac{n+2}{n+1} \theta$, you get $\hat{\theta}$ which has mean $\frac{n(n+2)}{(n+1)^2} \theta$ and variance $\frac{n(n+2)}{(n+1)^4} \theta^2$.

If you had an extra sample $X_0$ then
$$\max(X_0, \ldots, X_{n})/\theta$$
given $\theta$ has a $\text{Beta}(n+1, 1)$ distribution, which has mean $\frac{n+1}{n+2}$ and variance $\frac{(n+1)}{(n+2)^2 (n+3)}$ which you computed.
Multiplying by this random variable by $\frac{n+2}{n+1} \theta$ yields $\hat{\theta}$ which has mean $\theta$ and variance $\frac{1}{(n+1)(n+3)} \theta^2$.
A: I will assume you meant $B\sim\operatorname{Beta}(n+1,1).$
Recall that $\operatorname{var}(B) = \operatorname E\big(B^2\big) - \big( \operatorname E(B) \big)^2$.
\begin{align}
& \operatorname E(B^2) = \int_0^1 x^2\cdot f_B(x)\,dx \\[8pt]
= {} & \int_0^1 x^2\cdot \frac{\Gamma(n+2)}{\Gamma(n+1)\Gamma(1)} \cdot x^n\,dx \\[8pt]
= {} & (n+1) \int_0^1 x^{n+2}\,dx = \frac{n+1}{n+3}.
\end{align}
So the variance is
$$
\frac{n+1}{n+3} - \left( \frac{n+1}{n+2} \right)^2 = \frac{(n+1)(n+2)^2 - (n+1)^2}{(n+3)(n+2)^2} = \frac{n+1}{(n+3)(n+2)^2}.
$$
