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In elementary probability (i.e. with no measure theory involved) we condition our probabilities on events. This does seem intuitive because in the basic cases, we are just asking 'given an event has occurred, what is the probability of another event?'.

However, now I am developing a measure-theoretic view of probability, where I see that we are conditioning on sigma algebras rather than events.

I have read somewhere that the latter approach is the correct way to proceed, and the former approach is simply economical notation.

So the question is, how does one intuitively understand conditioning on sigma algebras?

Dhruv Gupta
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1 Answers1

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A probability conditioned on an event $\Pr (A ~|~ B)$ and on a sigma algebra $\Pr(A ~|~ \mathcal{H})$ are two different objects. The first is a value and the second is a function. You can think of conditioning on a sigma algebra as the "complete" picture (it also helps to avoid some issues when in $\Pr(A ~|~ B)$ when the probability $\Pr(B)=0$).

Consider the case $\mathcal{H}=\{\emptyset,B,B^c,X\}$. Since $\Pr(A ~|~ \mathcal{H})$ is a function (more accurately, I should write $\mathbb{E}[1_A ~|~ \mathcal{H}]$) you may ask what is the value of this function at the point $B$ and obtain $\Pr(A ~|~ B)$.

Nick A. R.
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  • @ Nick A.R. : What you say here about Pr(B) = 0 is interesting because I remember my professor in first year grad school saying: "Don't worry. You'll learn how to deal with zero in a measure theory class". But, what I don't understand is how can one condition on a sigma algebra when it contains an event and its complement. It is purely notational and what it really means is that one of the specific events in the sigma algebra will be conditioned on. – mark leeds Oct 01 '23 at 21:41