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0 pointskgwgk6y ago0 comments

What did you mean with the following paragraph then?

> Let's consider two independent rolls, X and Y. You can't compute the joint distribution P(Y, (X|E)), it just doesn't make sense as the two "variables" are defined over different spaces.

Do you agree that you cannot compute the joint distribution P(Y,X) either because the two variables are defined over different spaces?

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bonoboTP6y ago

I meant them to be defined on the same space. It's a single experiment, the outcome of which are two rolls that happen to be independent.

kgwgkOP6y ago

If you mean that the space for this single experiment composed of two rolls (random variables X and Y) is the cartesian product of {x=1,x=2,x=3,x=4,x=5,x=6} and {y=1,y=2,y=3,y=4,y=5,y=6}, then I agree.

But the fact that each variable alone is defined on the "same" sample space {1,2,3,4,5,6} is irrelevant.

The situation is no different from the joint probability for random variables X and Z corresponding to a single experiment consisting of a dice roll and a coin toss, where the relevant space is the cartesian product of {x=1,x=2,x=3,x=4,x=5,x=6} and {z=1,z=2}.

And it is also similar for the situation you asked about, with a random variable Y and a "conditional" random variable X|Even. The relevant space is the cartesian product of {y=1,y=2,y=3,y=4,y=5,y=6} and {x=2,x=4,x=6}.

bonoboTP6y ago

Let's consider something with less independence, because it makes things harder to notice. Temperature indoors T1, temperature outdoors T2, IsOvercast O.

Let's say T2|O=1 is a "conditional random variable". Let's consider the average temperature indoors and outdoors. What would ((T1|O=1) + T2)/2 even mean? How could you use the two "variables" in the same expression? What is even their joint distribution? They are defined over different spaces!

This means, we must always carefully condition all variables used together on the exact same things. So ((T1|O=1) + (T2|O=1))/2 is valid. But then why do this on all variable instances that we use? It would be very tedious. At some point we want to get to a distribution (or some function of a distribution, like the expectation or variance), so it's much simpler to say for example P((T1 + T2)/2 | O=1), which is just a good old conditional distribution. Conditioning is an operation on a distribution and in my mind the bar (|) is really a slot in the P() notation and is short for P(A,B)/P(B). A bar popping up elsewhere (like in expectations) must be directly determined by the distribution (a random variable is not).

Overall, since you cannot mix differently conditioned "conditional random variables" in a single expression, you may just as well put your conditioning on the side of the whole expression in the P().

Not sure if my point is coming across...

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bonoboTP6y ago

I meant them to be defined on the same space. It's a single experiment, the outcome of which are two rolls that happen to be independent.

kgwgkOP6y ago

But the fact that each variable alone is defined on the "same" sample space {1,2,3,4,5,6} is irrelevant.

bonoboTP6y ago

Let's consider something with less independence, because it makes things harder to notice. Temperature indoors T1, temperature outdoors T2, IsOvercast O.

Overall, since you cannot mix differently conditioned "conditional random variables" in a single expression, you may just as well put your conditioning on the side of the whole expression in the P().

Not sure if my point is coming across...

1 more reply

j / k navigate · click thread line to collapse