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Continuous distributions assign probability 0 to individual values. But, according to DeGroot, it doesn't mean that it is impossible for the random variable to take individual values. So, why not make continuous probability density functions(pdf) not defined at individual values? Also, if I have continuous pdf I cannot calculate probability of individual events but only those probabilities that says that the random variable can take a range of values, but why?

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  • $\begingroup$ Strictly speaking the pdf of the continuous distribution is not a function, but a class of functions defined and equal up to sets of measure zero. Therefore, given a point $x=a$ and a representative of that class $f$, there is the possibility of $f$ not being defined at $x=a$. In that sense, what you want is already part of the theory. On the other hand, the probability of the event $\{\omega:\ X(\omega)=a\}$ is defined and is the integral of the pdf on $\{a\}$, which is zero. Yet another aspect is the interpretation of that probability being zero, which doesn't mean the value $a$ is nottaken. $\endgroup$
    – Alamos
    Apr 30 '15 at 11:02
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There are just way too many numbers on the real line for all to a have a finite probability.

If you have a situation in which you have a finite number N of outcomes, you can assign to each a finite probability. As you increase the number N of possible outcomes, you can still do so. Since the probabilities still have to add up to 1, the "typical" probability for a certain outcome will be of the order of 1/N. So, as you increase the number of possible outcomes, the probability of most outcomes will need to be lower and lower.

In fact, it is possible to assign finite probabilities to each element of a countably infinite set (like the natural numbers). Think of the probability distribution $p_n = \frac{1}{2^n}$ for example. This would assign 50% probability to the 1st outcome, 25% to the second and so on. The probabilities still add up to 1, and for each outcome has a finite, non-zero, probability. But you see that most outcomes actually have an arbitrarily small probability.

The thing with the real line is that there are uncountably more outcomes. It isn't possible to assign a finite number to each element of an uncountable set without the sum going to infinity, unless most probabilities are 0 (Maybe somebody can provide a proof of this, I don't have it now).

Now, I haven't read DeGroot, and perhaps I am missing context, but it is interesting to think what does it mean for a random variable "to take individual values". For a continuous random variable, a value would be a irrational number, as these far outnumber the rationals. What kind of system actually selects a particular irrational number? How would an instrument tell you the particular irrational chosen? It cannot print out the full decimal expansion, because the digits are infinite. Only for a few cases we have an expression for it, like $e$ or $\sqrt 2$ ( This video from Vi Hart could be helpful).

In general, even if perhaps you can think of the system to actually be in one particular position, the whole point of working with probabilities is that you can only access to partial knowledge. You can measure the value of the variable up to a certain experimental error (the size of your ruler, the accuracy of your thermometer etc), which is to say, that the value is contained in a certain interval.

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  • $\begingroup$ finite $\leftrightarrow$ non-zero. $\endgroup$
    – Alamos
    Apr 30 '15 at 11:41

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