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 $\underline P(X_1 = R, X_2=G) > 1/9$.)


2 Answers 2


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.

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

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

It looks to be a poty.

  • 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

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