Probability function for discrete functions

[member 587159]

My textbook says that if $X: \Omega \to \mathbb{R}$ is discrete stochast (I.e., there are only countably many values that get reached), then it suffices to know the probability function $p(x) = \mathbb{P}\{X =x\}$ in order to know the distribution function $\mathbb{P}_X: \mathcal{R} \to \mathbb{R}: A \mapsto \mathbb{P}\{X \in A\} = \mathbb{P}(X^{-1}A)$Indeed, if $S:= \{x : p(x) > 0\}$, then for $A \in \mathcal{R}$, it follows that $$\mathbb{P}\{X \in A\} = \sum_{x \in S \cap A}p(x)$$

But how do they get this formula?

I tried the following:

$$\mathbb{P}\{X \in A\} = \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S\cap A}\{a\} \cup\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$

$$= \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in S \cap A}\{a\}\right)\right) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) $$
$$=\sum_{a \in S \cap A}p(a) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right)$$

But how do I show that the probability on the right is zero? I can't use $\sigma$-additivity on uncountable disjoint unions.

EDIT: $\mathcal{R}$ is the smallest sigma algebra that contains the usual topology on the real numbers, i.e. it are the Borel parts of the reals.

 

[StoneTemplePython]

$$=\sum_{a \in S \cap A}p(a) + \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right)$$

But how do I show that the probability on the right is zero? I can't use $\sigma$-additivity on uncountable disjoint unions.

I'll take a shot in the dark here-- I don't totally follow your notation or even what a "distribution function" is (generally means CDF but...). The problem should be easy if you can find a way to focus your attention on countable sets.

In any case, your problem reminded me of a quote from Kolmogorov that I like: "Behind every theorem lies an inequality."

- - - -

You have some special structure in that probabilities are always real non-negative and sum to one. The axioms immediately give that

$\Pr\{B\} + \Pr\{B^c\} = 1$

supposing you are allowed to use that, perhaps you can try re-running your argument over something complementary?

The goal would be to combine the two via linearity, show that your probabilities sum to one, and you end up with

$0\leq \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) \leq 0$

(or equivalently: take advantage of positive definiteness after subtracting 1 from each side of your equation -- but the idea is to show that the thing in question is bounded above and below by zero, and hence is zero.)

- - - -
Hopefully this helps or at least provides some inspiration toward a way to get the result you want via an inequality.

 

[member 587159]

I'll take a shot in the dark here-- I don't totally follow your notation or even what a "distribution function" is (generally means CDF but...). The problem should be easy if you can find a way to focus your attention on countable sets.

In any case, your problem reminded me of a quote from Kolmogorov that I like: "Behind every theorem lies an inequality."

- - - -

You have some special structure in that probabilities are always real non-negative and sum to one. The axioms immediately give that

$\Pr\{B\} + \Pr\{B^c\} = 1$

supposing you are allowed to use that, perhaps you can try re-running your argument over something complementary?

The goal would be to combine the two via linearity, show that your probabilities sum to one, and you end up with

$0\leq \mathbb{P}\left(X^{-1}\left(\bigcup_{a \in A\setminus S}\{a\}\right)\right) \leq 0$

(or equivalently: take advantage of positive definiteness after subtracting 1 from each side of your equation -- but the idea is to show that the thing in question is bounded above and below by zero, and hence is zero.)

- - - -
Hopefully this helps or at least provides some inspiration toward a way to get the result you want via an inequality.

Thanks. I defined probability distribution at the beginning of the post. I will take some time to digest your answer.

 

[mathman]

The statement looks obvious. Since the positive probability elements are countable (elements of S), to get P(A), add up all the probabilities of those elements in A which have positive probabilities.

 

[member 587159]

The statement looks obvious. Since the positive probability elements are countable (elements of S), to get P(A), add up all the probabilities of those elements in A which have positive probabilities.

I don't quite understand what you mean. Can you elaborate?

 

[mathman]

I don't quite understand what you mean. Can you elaborate?

I am not sure what I need to say. S is a subset (countable) of R containing all the points with positive probability. A is a subset of R. The probability of A is the sum of the probabilities of all points of A which have positive probability. Since S consists of all points of probability, the intersection of S and A contains all the points of A needed to define the probability of A.

 

[member 587159]

I am not sure what I need to say. S is a subset (countable) of R containing all the points with positive probability. A is a subset of R. The probability of A is the sum of the probabilities of all points of A which have positive probability. Since S consists of all points of probability, the intersection of S and A contains all the points of A needed to define the probability of A.

I think I understand what you mean. Let me write it out:

$$P(X \in A) = P(X\in A\cap X(\Omega))$$
$$= \sum_{x \in A \cap X(\Omega)} P(X=x)$$
$$=\sum_{x\in A \cap S} P(X=x) + \sum_{x \in A \cap X(\Omega) \setminus S} P(X=x)$$

And the last sum is 0. Does this make sense? In the second equality I used that the image of X is at most countable and for the last equality that $S \subseteq X(\Omega)$

 

[mathman]

Your notation throws me. That's why I prefer words. What\ is\ X(\Omega)? X does not have to be countable - only the subset of X consisting of points of positive probability.

 

[member 587159]

Your notation throws me. That's why I prefer words. What\ is\ X(\Omega)? X does not have to be countable - only the subset of X consisting of points of positive probability.

$X(\Omega) = Im(X)$, the image of the function $X$. I.e. all the values that X attain. And clearly this is countable, because by assumption this is countable (the variable is discrete)

 

[mathman]

I think I understand what you mean. Let me write it out:

$$P(X \in A) = P(X\in A\cap X(\Omega))$$
$$= \sum_{x \in A \cap X(\Omega)} P(X=x)$$
$$=\sum_{x\in A \cap S} P(X=x) + \sum_{x \in A \cap X(\Omega) \setminus S} P(X=x)$$

And the last sum is 0. Does this make sense? In the second equality I used that the image of X is at most countable and for the last equality that $S \subseteq X(\Omega)$

In your original definition of X, it appears that X is countable, so there are only a countable number (if any) points in the second sum where each point has P(X=x)=0.

 

[member 587159]

In your original definition of X, it appears that X is countable, so there are only a countable number (if any) points in the second sum where each point has P(X=x)=0.

Every term in the last sum is zero, because the sum runs over elements not in S (so by definitions points with probability 0)

 

[mathman]

Every term in the last sum is zero, because the sum runs over elements not in S (so by definitions points with probability 0)

The point I was making is that if the number of points in X were not countable, the last term would be problematical. Summations have meaning only if the number of terms is countable.