Reading de Finetti's "Foresight", there is an issue I would like to clarify:

We have $n$ mutually exclusive events, $E_1, E_2, \dots, E_n$, that we believe to occur with probability $p_1, p_2, \dots, p_n$ respectively. Now we want to bet on those events, so if for example $E_1$ materialized and the others not, we would make a gain of $(1-p_1)S_1 - p_2S_2 - \dots -p_nS : = G_1$, that is we wager $p_1S_1 + p_2S_2+ \dots + p_nS_n$ to win either $S_1, S_2$, or $S_n$.

This gives a system of linear equations, with the determinant of the coefficient matrix necessarily equal to zero. Otherwise, we could construct a Dutch Book.

$$ \begin{vmatrix} 1-p_1 & -p_2 & -p_3 &\dots & -p_n \\ -p_1 & 1-p_2 & -p_3 & \dots & p_n \\ \vdots & \vdots & \vdots & & \vdots \\ -p_1 & -p_2 & -p_3 & \dots & 1-p_n \end{vmatrix} = 1 - \sum_{i = 1}^n p_i = 0 $$

De Finetti now claims that, if the determinant is equal to $0$, the expected value of the gains is equal to $0$, or $\sum_{i=1}^n p_iG_i = 0$. I verified the result for the first three dimensions, but have failed to generalize it so far.

Any input would be highly appreciated!


1 Answer 1


Although this can be done by simplifying the double summation, it might be more revealing to do the algebra with matrix notation.

Let $p=(p_1,p_2,\ldots, p_n)^\prime$, $S=(S_1,S_,\ldots, S_n)^\prime$, and $G=(G_1,G_2,\ldots,G_n)^\prime$ represent the information as column vectors. Write $\mathbf{1}=(1,1,\ldots, 1)^\prime$ and let $\mathbb{I}_n$ be the $n \times n$ identity matrix.

We know

$$p^\prime \mathbf{1}=1,$$

we are given

$$G = (\mathbb{I}_n - \mathbf{1}p^\prime)S,$$

and wish to compute the number

$$p^\prime G.$$

Substituting the foregoing into this expression gives

$$p^\prime G = p^\prime(\mathbb{I}_n - \mathbf{1}p^\prime)S = p^\prime S - (p^\prime \mathbf{1}) (p^\prime S) = p^\prime S - (1)(p^\prime S) = 0,$$


  • $\begingroup$ Once again, I am indebted to you, @whuber. Thank you very much indeed! Your skill and expertise is really inspiring, I hope that with the years I shall come somewhat closer to your level of knowledge! $\endgroup$ Commented Jul 25, 2017 at 17:16
  • 1
    $\begingroup$ I have long been grateful for having had the opportunity, very early in my career, of taking a challenging and rigorous course in linear algebra. Such courses are difficult to find, but nevertheless I would warmly recommend the experience for anyone wishing to work in a quantitative field. $\endgroup$
    – whuber
    Commented Jul 25, 2017 at 17:21
  • $\begingroup$ Yes, I see the necessity as well. Despite receiving the equivalent of As in my LinAlg and Linear Models classes, I picked up "Matrix Algebra from a Statistician's Perspective" and read in it, whenever I have some spare time. Would you have further recommendations? $\endgroup$ Commented Jul 25, 2017 at 22:06
  • 1
    $\begingroup$ There must be thousands of current linear algebra texts, and the one I used is now very old (but not outmoded!), so I'm in no position to make recommendations. There is an ongoing stream of reviews in The American Statistician that over the years has covered such textbooks from a statistical perspective, so that might be a good place to do some research. $\endgroup$
    – whuber
    Commented Jul 25, 2017 at 22:13

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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