From the compound sampling model where:

$Y_i | \theta_i \sim Poi(\theta_i)$

The marginal distribution of $\theta_i$ is $G$, non-parametric.

We get that the Bayes estimate of $\theta_i$ under squared error loss is the posterior mean:

$\theta_i = \mathbf{E}(\theta_i|\mathbf{y}) = \mathbf{E}(\theta_i|y_i)=\frac{\int(u^{y_i+1}e^{-u}/y_i!)dG(u)}{\int(u^{y_i}e^{-u}/y_i!)dG(u)}=\frac{(y_i+1)p(y_i+1)}{p(y_i)} \quad (\star)$

Where $p(y)= \int f_i(y_i|\theta_i)dG(\theta_i)$ is the marginal distribution of $y$.

It is claimed that $\star$ is monotonic in $y$ in the text I am reading about this.

If we try to compare the function $\theta_i(y)$ and $\theta_i(y+1)$ we need to show $$\frac{\int(u^{y_i+1}e^{-u}/y_i!)dG(u)}{\int(u^{y_i}e^{-u}/y_i!)dG(u)} \leq \frac{\int(u^{y_i+2}e^{-u}/(y_i+1)!)dG(u)}{\int(u^{y_i+1}e^{-u}/(y_i+1)!)dG(u)} $$

The $!$ terms cancel but I am unable to finish the proof.


1 Answer 1


this is a mere consequence of the Cauchy-Schwarz inequality: $$\left(\int f(u)g(u)\text{d}u\right)^2 \le \int f(u)^2\text{d}u \int g(u)^2\text{d}u$$ when $$f(u)=u^{y_i/2}\qquad g(u)=u^{y_i/2+1}$$


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