Given:

1. The amount of a claim, $$X$$ is uniformly distributed on the interval $$[0,\theta]$$
2. 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:

1. Calculate the posterior distribution of $$X_3$$, I will get $$X_3|x_1,x_2$$ is uniformly distributed in $$[0,800]$$ hence, $$E(X_3|x_1,x_2)=400$$
2. 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: $f(X_3|x_1,x_2)=\int_{600}^\infty \frac{1}{\theta} 3\bigg(\frac{600^3}{\theta^4}\bigg) d\theta$ Will resulted in $f(X_3|x_1,x_2)=\frac{1}{800}$ Hence $E[X_3|x_1,x_2]=400$ Note: I have checked that using the CDF, $X_3|x_1,x_2$ is indeed uniform $[0,800]$ Dec 6, 2022 at 16:22
• For the second method: $\int_{600}^\infty E[X_3|\theta]f(\theta|x_1,x_2)d\theta=\int_{600}^\infty \frac{\theta}{2} 3\bigg(\frac{600^3}{\theta^4}\bigg) d\theta=450$ Dec 6, 2022 at 16:24
• For those who are curious, the support in posterior of $\theta$ become $\theta>600$ since we observed $x_2=600$ and $x_2$ should be contained in interval $[0,\theta]$ Dec 6, 2022 at 16:30

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 $$f(X_3 | x_1,x_2) = \int_{\max(X_3,600)}^\infty\frac1\theta 3\left(\frac{600^3}{\theta^4}\right)\mathrm d\theta$$
• Appreciate your explanation of $X_3$ support Dec 6, 2022 at 17:34