5
$\begingroup$

In Miranda and de Cooman's chapter 3, "Structural judgements", in Augustin et al.'s Introduction to Imprecise Probability, example 3.4 on p. 65 shows that independence in the selection (type-2 independence) does not imply strong independence (type-3 independence) for lower previsions. One of the numerical values in the example seems incorrect to me. I'm not sure whether there is a typo (there is an obvious trivial typo at the top of the same page, so this seems possible), or whether I don't fully understand the example (quite plausible).

There are two possible configurations, or compositions, for a pair of urns containing red ($R$) and green ($G$) balls, but we don't know which configuration is actual. Given a configuration, draws from each urn are random, i.e. independent.

Configuration 1:    Urn 1: $\{R, R, G\}$     Urn 2: $\{R,R,G\}$

Configuration 2:    Urn 1: $\{R, G, G\}$     Urn 2: $\{R,G,G\}$

Let $X_k$ be the r.v. representing the outcome of the draw from the $k$th urn. The text claims that given this setup, the lower prevision $\underline P$ of $X_1=R$ & $X_2=G$ is \begin{equation} \underline P(X_1 = R, X_2=G) = \frac{4}{9} . \end{equation} It seems to me that this value should be 2/9. Since the draws are independent within each configuration, the probability (linear prevision) corresponding to configuration 1 is \begin{equation} P(X_1 = R, X_2=G) = P(X_1=R) \times P(X_2=G) = \frac{2}{3} \times \frac{1}{3} = \frac{2}{9} . \end{equation} Similarly, the probability corresponding to configuration 2 is \begin{equation} \frac{1}{3} \times \frac{2}{3} = \frac{2}{9} . \end{equation} Since the lower prevision is the lower envelope of the probabilities in this example, it seems to follow that $\underline P(X_1 = R, X_2=G) = 2/9$. However, I suspect that I've interpreted the example incorrectly in some respect or have some more basic misunderstanding.

(I believe that if 2/9 were the correct value here, the example would still illustrate the point that independence in the selection doesn't imply strong independence. For that point all that's needed is that $\underline P(X_1 = R, X_2=G) > 1/9$.)

$\endgroup$

2 Answers 2

7
$\begingroup$

I am one of the authors of the book. Indeed it is a typo, sorry it caused some misunderstanding... feel free to contact me if you have further questions.

$\endgroup$
1
  • 1
    $\begingroup$ Welcome to Mathematics Stack Exchange! Your contact details belong in your profile. $\endgroup$
    – Glorfindel
    May 30, 2017 at 7:38
3
$\begingroup$

Yes, indeed, $\tfrac 29$ appears to be the correct value for that setup.

It looks to be a poty.

$\endgroup$
1
  • 2
    $\begingroup$ Agreed. I double checked the treatment in the book, and indeed 2/9 should be the correct value here. $\endgroup$ May 30, 2017 at 7:23

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.