Comparing the likelihood of two realizations of a continuous variable

Question

Suppose we both know X distributes N(0,1). I draw a number from it, I observe it and tell you: it is either 2 or 100000 (this statement is true)

Are both realizations equally likely? Or is P(2)>P(100000)?

Edit: There was some confusion about the question, so I will try to explain it better. I am trying to figure out a more complex situation, and I may have confused you with the example. I will stick to this example because it explains better what I found challenging. You draw an $x$ from an $X$ that distributes $N(0,1)$. Then, someone tells you (truthfully) that either $x=2$ or $x=100000$. I want to compute the $E(x)$. Normally, $E(x)=P(2|x \in {2,100000}) \times 2+P(100000|x \in {2,100000}) \times 100000$. I am finding challenging to compute $P(2|x \in {2,100000})$. How do I compute Bayes Rule conditioning on an event that has probability 0? If I take $\varepsilon$ intervals around $2$ and $100000$ it seems I should use the density function for X. However, they are both equally likely (both unconditional probabilities are 0). Which option is right?

Answer 1

I would say that this question doesn't have a well defined answer, for the following reasons:

1. Imagine that I draw $X$ by drawing $Y_i \sim N(0,1)$, for $i=1, 2, \ldots$, until I first get $Y_j \ne 2$, then I let $X=Y_j$

What is the distribution of $X$? It has a standard normal distribution!

So, if I'm feeding these to your friend, who tells you truthfully that the realized $x$ either equals 2 or 100000, it won't be 2.

2. What if I'm your friend, drawing $X \sim N(0,1)$, without reject $2$'s as we did above, but I always look at the realized $x$, and tell you that it either equals $x$ (which it actually does) or $x/50000$ (which it actually does not). Under this process, if told that $x$ either equals $2$ or $100000$, again it won't be 2.

For a famous example of the problem with conditioning on events of probability zero, see the Borel-Kolmogorov paradox.

Now, I'm not saying that you cannot condition on an event of probability zero. I think people do it, by "conditioning on the levels of densities" as described in this post:

Probability, conditional on a zero probability event

But you would have to decide if that makes sense for your use case.

I can imagine someone asking, "But if some friend of yours told you that truthfully the number they drew from a standard normal distribution was either 2 or 100000, which would you think it was?"

Well, first, the hypothetical seems extremely implausible given that they are both exactly integers. It would seem more plausible if they were irrationals with those integer parts (although I know any fixed irrationals would also have probability zero of being drawn).

But second, no human has ever actually drawn from a standard normal distribution, nor is it even humanly possible to do so.

Instead, anyone who has drawn a number using any computer or other device that only had finitely many possible outputs (such as from memory limitations, or time & output speed limitations), drew from a PMF. For example, if you use a computer to draw a float32 from some pseudo-random number generator approximating a standard normal distribution, then both 2 and 100000 have some positive probabilities (assuming both can be exactly represented in float32). In this case, 2 certainly would be much more likely (assuming the friend drew until it was either 2 or 100000, otherwise we need to know the data generation procedure to rule out situations like #2 above), and I expect that the conditional probability would be basically the same as conditioning on the levels of the PDF.