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Propensity Interpretation

Karl Popper's Propensity Interpretation aimed to solve the problem how to assign probability to the outcome of an individual experiment. While the Classical and the Frequency Interpretations try to reduce the notion of probability to other, already known concepts, the Propensity Interpretation identifies probability with a new quantity, called ``propensity'', expressing the measure of the ``probabilistic causal tendency'' of the system to behave in a certain way.

The standard objection against propensity says that it does not provide an admissible interpretation of probability. Consider two events $A$ and $B$ which are causally related, therefore $p(B\vert A)>p(B)$. $p(B\vert A)$ can be interpreted as propensity, for we can speak meaningfully of the tendency of a cause to produce the effect. Consider now $p(A\vert B)$ which is also a meaningful concept, in the sense that it can be easily expressed by the Bayes rule: $p(A\vert B)=$ $\frac{p(A\wedge B)}{p(B)}$. However, $p(A\vert B)$ cannot be interpreted as propensity, the argument goes, because it does not make sense to talk about the causal tendency of the effect to have been produced by one cause or another.

This usual objection is, however, not acceptable, in my view, because it is based on a misinterpretation of conditional probability, in general. Conditional $p(B\vert A)$ is nothing but the ratio $\frac{p(A\wedge B)}{p(A)}$ and, in general, it has nothing to do with ``the tendency of a cause $A$ to produce the effect $B$.'' As in any other interpretation of probability, the correlation $p(A\vert B)>p(A)$ is only a necessary but not a sufficient condition for a causal relationship. The origin of the misunderstandings is that conditional probability is often saddled with completely unjustified meaning: it does not mean, for example, the value for which the probability of event $B$ changes when event $A$ happens, or it is not equal to the probability of event $B$ if the system is prepared such that event $A$ occurs with probability 1, etc. In case of dicing, for example, the conditional probability $p\left(<4>\vert<\textrm{even}>\right)$ is nothing but the rate $\frac{p\left(<4>\right)}{p\left(<\textrm{even}>\right)}=\frac{1}{3}$. But, it does not mean that $p\left(<4>\right)=\frac{1}{3}$ if the die is prepared such that $p\left(<\textrm{even}>\right)=1$. In this case, the conditional probability $p\left(<4>\vert<\textrm{even}>\right)$ would not be a well-defined notion, because the `` $p\left(<\textrm{even}>\right)=1$'' preparation does not correspond to a unique condition: if the die is biased in such a way that $p\left(<2>\right)=1$, then $p\left(<\textrm{even}>\right)=1$, and $p\left(<4>\right)=0$. While in another case, if it is biased such that $p\left(<4>\right)=1$, then, again, $p\left(<\textrm{even}>\right)=1$, but $p\left(<4>\right)=1$.

There is, on the contrary, a more difficult problem with propensity. In Propensity Interpretation, probability - propensity - is a separate quantity, which is not expressed in terms of other, empirically defined quantities. How, then, is the numerical value of propensity determined? We have no starting point for the empirical test of the value of propensity. Consequently, there is no empirical basis for such a proposition as ``the probability of getting <Heads> is $\frac{1}{2}$'', and the whole talk about probabilities loses empirical control. We do not even know whether propensities satisfy Kolmogorov axioms, or not.


next up previous
Next: Subjective Interpretation Up: Difficulties of the standard Previous: Relative Frequency Interpretation
lalo
2003-10-23