5
$\begingroup$

I've seen the way to "prove"/"derive" for a continuous random variable $Y$ the probability of event $A$ given the event $\{Y=y\}$ is the intuition that the event $\{Y=y\} = \lim_{\delta \to 0} \{y - \frac{\delta}{2} \le Y\le y + \frac{\delta}{2}\}$. What I've seen then is:

To find $P(A|Y=y)$ is to first evaluate $$P\left(A|Y\in\left(y-\frac{\delta}{2}, y+\frac{\delta}{2}\right)\right)$$ and then take the limit as $\delta \to 0$.

But doesn't this method assume that

$$\lim_{\delta\to 0}P\left(A|Y\in\left(y-\frac{\delta}{2}, y+\frac{\delta}{2}\right)\right)\overset{?}{=}P\left(A|\lim_{\delta\to 0}\bigg\{Y\in\left(y-\frac{\delta}{2}, y+\frac{\delta}{2}\right)\bigg\}\right)=P(A|Y=y) \ ?$$

I might be skirting on the edges of measure theory (which I am not familiar with) but is there an intuitive way to make this claim seem intuitive/convincing? Or is there a result which shows why this is valid to do, if it is? Addition: Thanks to @Masoud - how would you know the limit always exists, if it does?


Example:

I'd like to add an example which lead me to this. Suppose $N_{t_1, t_2}$ represents the number of occurrences of a phenomenon in the time interval $(t_1, t_2]$ for $0<t_1<t_2$. Suppose you are given the last occurrence was at time $s$. Let $X$ be the time until the next occurrence of the phenomenon, starting at time $s$. We need to find the distribution of $X$. (For those of you familiar this is related to the well know poison process).

Then, the probability $$\mathbb P (X>t|\text{Last occurrence was time }s)=\mathbb{P}(N_{s,s+t} = 0|\text{Last occurrence was time }s)$$

and this point, since there's no way to definition the conditional null set in terms of $N_{t_1, t_2}$, I saw the source do $$\mathbb{P}(N_{s,s+t} = 0|\text{Last occurrence was time }s) \overset{?}{=} \lim_{h\to 0} \mathbb P(N_{s+h,s+t+h} = 0| N_{s+h, s}=1)$$

On what basis can we apply such a limit? I couldn't find a definition to state when you can do something like this. In this case, the limit is being applied in both the condition to help with the null set, and the set for which we're finding the probability too!

Thanks in advance

$\endgroup$
8
  • 1
    $\begingroup$ I think another good question is: Is it guarantee that the limit always exist?. I use $P(A|Y=y)=E(1_A |Y=y )$ since $E(1_A |Y=y )$ does not care about $P(Y=y)=0$. $\endgroup$
    – Masoud
    Apr 13, 2020 at 5:23
  • $\begingroup$ @Masoud Yes, I will add that, thanks for adding value to the question! Although I wonder, how would you find $E(1_A |Y=y)$? Wouldn't the expectation just give you the probability back - sorry if this is a silly question. $\endgroup$
    – user523384
    Apr 13, 2020 at 6:18
  • 1
    $\begingroup$ The deeper question is "how are you defining $P(A \mid Y=y)$ (or more broadly, the conditional expectation $E[ \cdot \mid Y=y]$) in the first place?" It may be that in your context, this limit is your definition of that quantity. In measure theory, one shows existence of conditional expectations using the Radon-Nikodym derivative; under some conditions, there is some correspondence between that abstract definition and the limit-based definition that you are describing, for instance see here. $\endgroup$
    – angryavian
    Apr 13, 2020 at 6:18
  • $\begingroup$ @angryavian Thanks! So you mean the definition could be the limit from the "outside"? Does this limit always exist? $\endgroup$
    – user523384
    Apr 13, 2020 at 6:21
  • $\begingroup$ Also, is saying something like $\{Y=y \} = \lim_{\delta \to 0} \{ y - \frac{\delta}{2} \le Y \le y + \frac{\delta}{2} \}$ valid in the first place? Because I think the event $\{ Y = y \}$ is actually "code" for $\{\omega \in \Omega: Y(\omega) = y \}$ $\endgroup$
    – user523384
    Apr 13, 2020 at 6:26

1 Answer 1

2
$\begingroup$

You are right to be suspicious of the limit interchange.

Here's a simple counterexample:

Let $Y$ be drawn uniformly from the interval $[0,1]$, and be $y=\frac12$. Be $A$ the event $Y\in\mathbb Q$.

Obviously $P(Y\in\mathbb Q\mid Y=\frac12)=1$.

On the other hand, the rational number in an interval are a null set, therefore for any $\delta>0$, we have $P(Y\in\mathbb Q\mid Y\in(\frac12-\frac\delta2,\frac12+\frac\delta2))=0$. And therefore also the limit for $\delta\to 0$ vanishes.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer! I understand. What would be a way to make this limit procedure well defined? Are there some conditions or some adjustments we can do? $\endgroup$
    – user523384
    Apr 13, 2020 at 7:33
  • $\begingroup$ @user523384: I'm not sure a general derivation of probabilities for discrete sets from probabilities for continuous sets is possible; for example, in an uniform distribution, you'd expect $P(Y=a\mid Y\in\{a,b\})=1/2$, however you can make that conditional probability arbitrary without changing any of the interval probabilities (because those discrete probabilities enter the interval probabilities with measure 0 anyway). $\endgroup$
    – celtschk
    Apr 13, 2020 at 8:15

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .