Inequality involving a 0-probability event

Let $S=\{x_1,x_2,...,x_N\}$ be a set of points sampled i.i.d. from some distribution $P$. Let $A(x) \in \{true, false\}$.

If $\exists x \in S$ s.t. A(x) is true, then does it follow that $P_{x\sim P}[A(x)]>0$?

• Is $X$ a discrete random variable or a continuous random variable? – Dilip Sarwate Mar 11 '15 at 3:48
• It can be either. I'm less clear when P is a continuous distribution. – axk Mar 11 '15 at 4:20

No.

For example, suppose the distribution is standard normal, and define $A(x)$ as true for $x = 1$ and false for all other $x$. Then the probability that $A(x)$ is true for a new draw from the distribution is $0$, because the probability that a random standard normal deviate equals $1$ is also $0$.

The trick here is that although $A(x)$ is not false for all $x ∈ ℝ$, it is false almost everywhere on $ℝ$. Thus, it is almost surely false.

Yes.

OK, I am a little provocative here. Kodiologist’s answer above is ok: a normal sample $x_1, \dots, x_n$ with $A(x_1)$ true and $A(x)$ false for all $x \ne x_1$ gives some kind of counterexample.

However, the following is true as well: given $A(x)$ (measurable) taking values true/false, if $\mathbb P(A(X)) = 0$ when $X$ follows the specified distribution, then $P(A(X_1) \text{ or } \dots \text{ or } A(X_n)) = 0$ when $X_1, \dots, X_n$ are independent with this distribution.

That is, while technically Kodiologist is right, the probability of being in this situation if the predicate $A(\cdot)$ is fixed beforehand is 0...

I’ll defend this paradox:

• The point is that you cannot use a sample $x_1, \dots, x_n$ to make inference on a predicate $A(\cdot)$ which is defined after the sampling. If $A(\cdot)$ is defined before the sampling, and if for some sampled $x_i$ $A(x_i)$ holds, you’ll never get wrong in concluding that $\mathbb P(A(X)) > 0$.

• Imagine that there is a bet to be made on $\mathbb P(A(X))$. The predicate $A$ is fixed (but we don’t see its definition) and we sample a finite number of values $x_i$ in a normal distribution (say), and we observe that for at least one of the sampled values $A(x_i)$ holds. I bet that $\mathbb P(A(X)) > 0$. Is anyone willing to bet anything on the contrary?

• I am not telling that the statement "$\exists x$ in the support of the distribution of $X$ s.t. $A(x)$ holds $\Rightarrow \mathbb P(A(X)) > 0$" is true. It could not be more wrong. This is all about the fact that $x$ is assumed to be sampled in the distribution of $X$.

• Wouldn't this argument result in concluding that conditional probabilities based on events of probability zero can be ignored because the events they are conditioned on almost surely do not happen? – whuber Mar 11 '15 at 21:21
• You're basically right: I chose a pathological $A$ and the conjecture would hold for suitably well-behaved classes of $P$ and $A$. But since @axk put no restrictions on these, I gave an answer for the general case. – Kodiologist Mar 11 '15 at 22:11
• @whuber I certainly wouldn’t discard a question on $\mathbb P(X>0 | Y =y)$ where $(X,Y)$ is, say, a normal vector. Honestly I don’t see anything disputable in my answer. It’s not about pathological predicate The point is that you cannot use a sample $x_1, \dots, x_n$ to make inference on a predicate $A(\cdot)$ which is defined after the sampling. If $A(\cdot)$ is defined before the sampling, and if for some sampled $x_i$ $A(x_i)$ holds, you’ll never get wrong in concluding that $\mathbb P(A(X)) > 0$. – Elvis Mar 12 '15 at 8:50
• I inserted this clarification in my answer. – Elvis Mar 12 '15 at 8:54
• Thank you @Elvis and whuber. It does answer my intended question, and seeing both Elvis' answer Kodiologist's answer together with this discussion helped me see what I was missing. So basically the predicate is fixed beforehand and measurable, I just didn't realise that this was important when I formulated the question. – axk Mar 12 '15 at 15:29