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How to show risk function = (3-theta)^2? I tried using the mean squared error formula to no avail.

critterTI
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  • I believe it is the true mean in the normal distribution X~N – critterTI Jun 27 '21 at 10:47
  • Where is this example taken from? I find it curious that $\mathbb{E}_\theta(X-\theta)^2$ does not depend on the distribution of $\theta$ and that $\theta$ remains in the equation of interest (in yellow) even though we are integrating over it... – Richard Hardy Jun 27 '21 at 14:04
  • Hi: In the second one, there is no $X$ so the RV is constant and the integral degenerates to 1. $E_{\theta}(3-\theta)^2 = \int (3- \theta)^2 f(x) dx = (3 - \theta)^2$. – mlofton Jun 27 '21 at 15:31
  • @mlofton, since we have $\mathbb{E}_\theta$ rather than $\mathbb{E}_X$, why are you integrating w.r.t. $X$? (Trying to understand what is going on.) – Richard Hardy Jun 27 '21 at 15:54
  • Hi Richard: Maybe it's a notational issue ? It's been too long but Ferguson's text would explain the notation. I forget the title of it. All I know is that since the estimator is not a function of the data, the risk is not random so there's nothing to integrate with respect to. – mlofton Jun 27 '21 at 16:17
  • @mlofton, I have checked a couple of other threads by now and got the impression that $\mathbb{E}\theta$ refers to integration w.r.t. $X$ just as you did, not $\theta$. I find this notation quit unfortunate, but it seems to be not uncommon. In another thread there was $\mathbb{E}{X|\theta}$ which is much clearer. – Richard Hardy Jun 27 '21 at 16:24
  • Ok. Glad it worked out. This is the text I was referring to. It requires serious effort but it explains the concept of risk and loss functions quite nicely. The notation can definitely be a killer. https://www.amazon.com/Mathematical-Statistics-Decision-Theoretic-Approach/dp/0122537505/ref=sr_1_5?dchild=1&keywords=thomas+ferguson+decision+theory&qid=1624810892&sr=8-5 – mlofton Jun 27 '21 at 16:24

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