0
$\begingroup$
  1. If we take the example of rolling dice once where $\Omega=\{1,2,3,4,5,6\}$, can we "create" a random variable, $X$, that only takes values $X=2$ and $X=4$, each with probabilities $\frac{1}{6}$ that sum to $\frac{2}{6}$? In other words, must we assign a probability distribution to a random variable that sums to $1$ by definition? I've been introduced to measure theory, so hopefully you can also point me to which part of the measure-theoretic definition of a random variable necessitates this.
  2. Does a random variable have to assign a value to each outcome, $\omega$?
$\endgroup$
  • $\begingroup$ The answer is absolutely no! It is possible to have X only 2 and 4 but then P(2)+P(4) =1. Random variables are defined on a probability space. The definition of a probability distribution is that of you integrate over all possible disjoint outcomes must integrate to 1 in the continuous or sum to 1 in the discrete case. $\endgroup$ – Michael R. Chernick Jan 12 '17 at 18:56
1
$\begingroup$

There are several approaches to probability, but all them have in common that the probability of the whole sample space equals 1. For a random variable, that means that the sum of the probability function is 1.

One approach is axiomatic: a probability is a measurable function of the sample space on the interval [0,1] with some properties, and one of them is that the measure on the whole sample space is 1.

From the frequentist approach and using your die as example: The probability of each result is the ratio between outcomes yielding such a result and the total number of outcomes when number of trials became large (or tends to infinite). Sum of all probabilities equals the probability of getting a number, that is it's the number of all outcomes divided by the number of trials, but since every trial gives an outcome (every time you roll your die you get a number), global probability will be 1.

If you modify you random variable in a way that you only register some outcomes of the die (e.g. rejecting all values different from 2 or 4) you just need to normalize your probabilities.

Edit about the 2-4 experiment after comment:

If a variable just takes values X=2 and X=4, every time we get a value of that variable it will be 2 or 4. If we are doing it with a real die, we can just roll the dice until we get 2 or 4, or we can relabel all faces so that each show a 2 or a 4. Then, if we roll de dice a lot of times, we wil get a lot of 2s and 4s, we will be able to compute the proportion of 2 and 4, if the number of trials tend to infinite, those proportions will be probabilities, and sum of both proportions will be 1. If 2 and 4 are equally probable, each probability will equal 1/2.

Anyway, we can use our die in a different way. We can roll the dice and write down three different possible results:

  • 2
  • 4
  • Not a 2 nor a 4

That would be a (quite weird) categorical random variable with 3 possible outcomes (not just 2). In a normal die, probabilities will be 1/6, 1/6 and 4/6. Please notice that probability of 2 and 4 doesn't sum 1, but that's just because 2 and 4 is not the whole probability space.

$\endgroup$
  • $\begingroup$ Can you elaborate on that last point? Does $P(X=2)=\frac{1}{2}$ then? I'm having difficulty understanding how we can then just ignore trials of the experiment if it's not $2$ or $4$. $\endgroup$ – Jack Sep 17 '16 at 23:31
  • $\begingroup$ @Pere I wonder if this statement is true: "There are several approaches to probability, but all them have in common that the probability of the whole sample space equals 1." It's an excellent point - that there are several approaches to probability - but is Kolmogorov's second axiom ("sum to one") necessary or just desirable mathematically and for interpretation? $\endgroup$ – Graeme Walsh Sep 17 '16 at 23:45
  • $\begingroup$ @GraemeWalsh I'm afraid you need somebody else with deeper knowledge than me to discuss the motivation of Komogorov's axioms. For this answer, intended for basic level, I just took the approaches "probability is what Kolmogorov's axioms define" and "probability is proportion when number of trials goes to infinity as Laplace said", assuming that both approaches are equivalent for practical purposes - at least at basic level. $\endgroup$ – Pere Sep 17 '16 at 23:53
  • $\begingroup$ @Pere That's perfectly reasonable. I understand. $\endgroup$ – Graeme Walsh Sep 18 '16 at 0:02
-1
$\begingroup$

if you measure in percent the sum is 100.

the question can be rephrased independent of probability theory. do the fractions of some entity sum to one? The answer is that if the entity is a collection then no. for example the sum of the fractions of two pies is two no matter how they are sliced. but you can always think of the whole set as a single entity and compute the fraction of the total. that's called normalization. normalized probability distributions sum to one by definition.

you can arbitrarily exclude un-normalized probability distributions from the definition.

$\endgroup$
  • $\begingroup$ Your final sentence about "un-normalized probability distribution" seems to undo your entire post by suggesting the answer could be "no, probabilities don't have to sum to unity--that's called an 'un-normalized' distribution." We seem to be back where we began. $\endgroup$ – whuber Jan 12 '17 at 17:39

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