# What is the cardinal of all discrete probability distributions (discrete and continous)

So I should build a map between something countable, I suppose, and all discrete probability distributions. The same for continous. My attempt:
As for every discrete probability distribution there is a table of probabilities $p_i$ for some $x_i$ we can list all these tables. Set of pairs $<x_i,p_i>,\; i=1,2,\ldots$ is obvioulsy countable and as uniou of countable sets is sountable we conclude that set of all discrete probability distributions is countable. But I don't have any clue about continous distributions. What can I map them with? Hints?

• What is the real problem here? Your stated problem, starting with "So", reads like a step in a larger program of yours. What is it? By the way, the set of discrete measures on the reals is not countable. The collection of all point masses at singletons, $\{\delta_x: x\in\mathbb R\}$, where $\delta_x(A)=1$ if $x\in A$ and $0$ otherwise, has the cardinality of $\mathbb R$. – kimchi lover Nov 25 '17 at 4:12
• "So" reffers to the header. I don't understand your example. Could you explaint it in more detail? – ioleg19029700 Nov 25 '17 at 4:52

## 1 Answer

The answer to the question apparently being asked is No: the cardinality of the set of discrete distributions is uncountable.

Let $P_a$ denote the Bernoulli distribution with expectation $a$, that is, $P_a$ is the probability law for which $P_a(X=1)=a$ and $P_a(X=0)=1-a.$ The set of all Bernoulli distributions is $\mathcal B = \{P_a: a\in[0,1]\}$. All members of $\mathcal B$ are discrete distributions. Note that $P_a\ne P_b$ if $a\ne b$, that is, the distributions in $\mathcal B$ are all unequal to each other. But the cardinality of $\mathcal B$ is equal to that of the set $[0,1]$. Since $\mathcal B$ is a subset of the set of all discrete distributions, this letter set is not countable, either.

One might ask, whether there exist only countably many triangles in the plane, because a triangle is uniquely specified by its vertices, and each triangle has only 3 vertices. Or if there are only countably many line segments in the line, because each segment has only two endpoints. Or only countably many points in the Cartesian plane, because each point is a singleton set $\{(x,y)\}$ of only two coordinates, and hence countable.

I hope I have misunderstood this question.