3
$\begingroup$

I've been trying to get into the chapter 4 in Lehmann's Theory of point estimation, but I can't seem to understand his presentation of the Bayesian setup. He starts of by the introduction below and after a few examples of uses of Bayesian estimators he outlines the idea (after dots in my photo). I don't know what he means by: $EL(\Theta,d)$.

In my opinion there should be two expectations there since we want to find d to minimize (1.1), I can't see how minimizing the one above being sufficient. I've tried with "Law of total expectation" and Fubini but nothing has been satisfactory. I have a similar problem with a theorem which comes right after the second paragraph.

enter image description here

$\endgroup$
7
  • 1
    $\begingroup$ What is "Lehmanns book"? Can you make the figure larger, so it's easier to read? $\endgroup$ May 16 '16 at 16:37
  • 1
    $\begingroup$ The reference to posterior risk at the end makes it clear that the expression $EL(\Theta,d)$ is the expected loss of state $\Theta$ with decision $d$. $\endgroup$
    – whuber
    May 16 '16 at 17:29
  • $\begingroup$ @whuber well the bayes risk (1.1) is $E_{\Lambda}E_{\theta} L[\theta,\delta]$, then why is this minimize given $EL(\Theta,d)$ is? I suppose they mean this is $E_{\Lambda}L(\Theta,d)$ btw $\endgroup$
    – user123124
    May 16 '16 at 17:55
  • 2
    $\begingroup$ Yes; Lehmann took pains to emphasize that $\Theta$ is governed by the distribution $\Lambda$, so this is the distribution involved in the expectation. The question he is answering is, how small can you make the risk by means of your choice of $d$? You might be over-interpreting this situation, which involves nothing more complicated than that idea. $\endgroup$
    – whuber
    May 16 '16 at 18:07
  • 1
    $\begingroup$ "Overlooked" in what sense? Lehmann explicitly takes expectations, both in the prior and posterior cases he describes. $\endgroup$
    – whuber
    May 16 '16 at 19:30
8
$\begingroup$

This is Fubini's theorem in action: when minimising in $\delta$ $$\mathbb{E_{\Lambda}} \{\mathbb{E}_{\theta}[L(\theta,\delta)]\}=\int_\Theta\int_\mathcal{X} L(\theta,\delta(x))\text{d}P_\theta(x)\text{d}\Lambda(\theta)=\int_\mathcal{X} \int_\Theta L(\theta,\delta(x))\text{d}\Lambda_x(\theta)\text{d}P(x)$$where $\Lambda_x$ denotes the posterior distribution of $\theta$ conditional on $x$, one minimises in $d$ for each value of $x$ the posterior expected loss $$\int_\Theta L(\theta,d)\text{d}\Lambda_x(\theta)$$and set $$\delta(x)=\arg\min_d \int_\Theta L(\theta,d)\text{d}\Lambda_x(\theta)$$assuming all quantities are finite.

$\endgroup$
2
  • $\begingroup$ So $\Lambda$ is prior after the first equality? The subscripts confuses me, I tought there expressionen for prior and post where almost the same(of the form in your answer), and that one just changed the distribution of $\Lambda$ in post. $\endgroup$
    – user123124
    May 17 '16 at 4:34
  • 1
    $\begingroup$ $\Lambda$ denotes the prior and $\lambda_x$ the posterior, while $P_\theta$ denotes the sampling distribution and $P$ the marginal. $\endgroup$
    – Xi'an
    May 17 '16 at 5:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.