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Intuitively, it should be one by the definition of conditional probability. Also $$P(A|A) = \frac{P(A\cap A)}{P(A)} = 1$$ But this formula breaks down when $P(A) = 0$. If $P(A) = 0$ is it still the case that $P(A|A) = 1$? This question arose because I'm a grad student and someone mentioned they like to give their students a check of understanding such as what $P(A|A)$ is. Another person brought up the question of what if $P(A) = 0$.

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    $\begingroup$ The answer is that it depends. If you are wanting to work in a scenario where you are allowed to consider such a conditional probability, then it will equal $1$. If you are wanting to work in a scenario where you are not allowed to consider such conditional probabilities, then it is fine to leave it as undefined. There are many different ways to define things such as probabilities and conditional probabilities in order to allow for or ignore such things. See here for example. $\endgroup$
    – JMoravitz
    Apr 25, 2023 at 21:38
  • $\begingroup$ @JMoravitz when would you want to work in a scenario where you are allowed to consider such a conditional probability? $\endgroup$
    – Andrew
    Apr 26, 2023 at 1:07
  • $\begingroup$ @AndrewZhang for instance when conditioning on certain values of continuous random variables, or when sampling from multidimensional space and you want to look at a particular slice of that space rather than the whole space. As an example, when picking a point uniformly at random from positions on the globe, given that we selected a point on the equator what is the probability we picked a point in the western hemisphere? The question makes sense that we should be able to ask it, and it has an intuitive answer of $\frac{1}{2}$. We get to choose the rules of math so that we can actually answer $\endgroup$
    – JMoravitz
    Apr 26, 2023 at 3:28

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If you open the textbook "Grimmet, Probability and Random processes", you see that conditional probability $\mathbb{P}(A|B)$ is being defined only under assumption that $\mathbb{P}(B)>0$. In the other case you will get division by zero in this formula, so value of it will be undefined.

From the other side, from graduate level, you can think about conditional probability as conditional expectation of indicator. In that case you can take a look at this explanation and see how this thing can be interpreted in case of event with zero probability, in particular case when we talk about the event $B=\{\xi=x\}$ where $\xi$ is a continuous random variable.

To clarify why did I provide this link I want to make an addition. Suppose we have two continuous random variables, $\xi$ and $\eta$ and they have joint probability density function, $f_{(\xi, \eta)}(x,y)$. Discussing conditional expectations, notation $\mathbb{P}(\xi \leq x | \eta = y) := \int\limits^{x}_{-\infty}\frac{f_{(\xi, \eta)}(v,y)}{f_\eta(y)}dv = \int\limits^{x}_{-\infty}f_{\xi|\eta}(v|y)dv$ is being widely used. But as your said in comments even in that case we just use other notation and even in that case $\mathbb{P}(\xi \leq x | \eta = y) \ne \frac{\mathbb{P}(\xi \leq x, \eta =y)}{\mathbb{P}(\eta=y)}$, because we cannot divide by zero. I added this example not to mislead you more, but to show, that sometimes people use this notation but refer to other formula, which has different meaning.

But from my point of view, I agree with @JMoravitz : in particular case, when $B=A$ your professor can define $\mathbb{P}(A|A):=1$ even if $\mathbb{P}(A)=0$. And he can say that it's just an agreement, like we did with equality with conditional density on the right hand side. Maybe it will be useful for him in some situations or make some proofs of theorems shorter.

I hope now my answer is clearer but feel free to ask any questions in comments.

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  • $\begingroup$ It’s not defined no matter how you approach it, measure theory or not. $\endgroup$
    – Andrew
    Apr 26, 2023 at 1:03
  • $\begingroup$ @AndrewZhang It is defined so long as we say it is defined. That's how defining things works. That is the only requirement to defining things. You might not have defined it, and your teachers or books you have read may not have defined it... but whether or not something can be defined is another matter entirely. Whether or not such a definition continues to be used by other people depends on if such a definition is useful, and it can be useful here. $\endgroup$
    – JMoravitz
    Apr 26, 2023 at 3:31
  • $\begingroup$ The article you linked to says: "It is tempting to define the undefined probability $P ( A ∣ X = x )$ using this limit, but this cannot be done in a consistent manner. In particular, it is possible to find random variables $X$ and $W$ and values x, w such that the events ${ X = x }$ and ${ W = w }$ are identical but the resulting limits are not." Which seems to me to imply that even with this limit definition we can not define such a conditional probability $\endgroup$
    – Aphyd
    Apr 26, 2023 at 3:43
  • $\begingroup$ @Aphyd yes, I didn't say that it can be defined this way, just said that this is one of interpretations. When you will learn conditional expectations you will see that people tend to use notation $\mathbb{P}(A|\xi=x)$ just to denote that expression with conditional density on the right hand side. Note, that it's just a notation. If you have Grimmet's book you can take a look at page 117, 5th edition. But I agree with JMoravitz, in particular case, when $B=A$ your professor can define $P(A|A):=1$ even if $P(A)=0$. And he can say that it's just a notation, like that with conditional density. $\endgroup$
    – perepelart
    Apr 26, 2023 at 6:12
  • $\begingroup$ @Aphyd, I made the addition in the answer and added more concrete formulas so you can understand about which things I told. $\endgroup$
    – perepelart
    Apr 26, 2023 at 7:31

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