Inconsistency of Bayesian Premium Given:

*

*The amount of a claim, $X$ is uniformly distributed on the interval $[0,\theta]$

*The prior density of $\theta$ is $\pi(\theta) = \frac{500}{\theta^2}, \theta > 500$
Two claims, $x_1=400$ and $x_2=600$ are observed. The posterior distribution is
$$f(\theta|x_1,x_2)=3\bigg(\frac{600^3}{\theta^4}\bigg), \theta > 600$$
Calculate the Bayesian premium, $E(X_3|x_1,x_2)$
Why the resulted Bayesian premium is different under these two methods:

*

*Calculate the posterior distribution of $X_3$, I will get $X_3|x_1,x_2$ is uniformly distributed in $[0,800]$ hence,  $E(X_3|x_1,x_2)=400$

*Using the formula $\int_{600}^\infty E[X_3|\theta] f(\theta|x_1,x_2) d\theta$ i will get $E(X_3|x_1,x_2)=450$

For the first method, it makes no conceptual sense that the posterior distribution should have bounded support. The observations that $x_1=400$ and $x_2=600$ allow you to rule out $\theta < 600$, but any value of $\theta > 600$ still is possible, so intuitively, the distribution of $X_3$ should have positive support for all values of $X_3$.
Now, turning to where your computation goes wrong, note that $f(X_3|\theta) = \frac1\theta\times 1\{X_3 \leq \theta\}$. In your calculation, you have forgotten the crucial $1\{X_3 \leq \theta\}$ piece. Including it, you should find that
$$f(X_3 | x_1,x_2) = 
\int_{\max(X_3,600)}^\infty\frac1\theta 3\left(\frac{600^3}{\theta^4}\right)\mathrm d\theta$$
