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How do we interpret P(A|B|C) ?

by sahil_time
Tags: interpret, pa|b|c, probability
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sahil_time
#1
Feb15-14, 02:02 PM
P: 108
How do we interpret P(A|B|C) ?

is P(A|B|C) = P(A|B)?

First of all, do such probabilities P(A|B|C) even make sense?

If they do,

Then, P(A|B|C) means, that given C has occurred and given that B occurs which depends on C, what is the probability that A occurs which depends on B?
Will this not be the same as saying, that since C has occurred, so P(A|B|C) = P(A|B) simply?

Example: A = I am alive.
B = Building fell.
C = Earthquake struck.

Q1) so is not P(A|B|C) = P(A|B) simply?

Q2) and so P(A,B,C) = P(A|B|C)P(B|C)P(C) = P(A|B)P(B|C)P(C) ?
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Simon Bridge
#2
Feb16-14, 01:06 AM
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Where did you see that notation - what is the context?

P(A|B) would be the probability of event A given that evet B has happened.

For three interdependent events, draw the ven diagram and label the different sections for what events happened.

Example: A = I am alive.
B = Building fell.
C = Earthquake struck.

Q1) so is not P(A|B|C) = P(A|B) simply?
Well there may or may not have been an Earthquake.
First you have to say what you want to describe, then use the math to describe it.

... that you survived given the building fell down would be P(A|B) ... regardless of whether or not there was an Earthquake.

P(A,B,C) would be the probability that you survived when the earthquake struck and the building came down.

Given that an earthquake struck, the building comes down with probability P(B|C)

If the building came down in an earthquake, then the probability you survived was P(A|(B,C))

Q2) and so P(A,B,C) = P(A|B|C)P(B|C)P(C) = P(A|B)P(B|C)P(C) ?
I don't think so.

P(A,B,C)=P((A,B)|C)P(C)
sahil_time
#3
Feb17-14, 12:54 AM
P: 108
Quote Quote by Simon Bridge View Post
Where did you see that notation - what is the context?

Context : I was thinking how would we expand P((A|B),C), so using

P(X,Y) = P(X|Y)P(Y), yields us P((A|B),C) = P((A|B)|C)P(C) . So i was thinking

how to interpret such a thing ?

sahil_time
#4
Feb17-14, 01:09 AM
P: 108
How do we interpret P(A|B|C) ?

Quote Quote by Simon Bridge View Post
I don't think so.
I was thinking, that suppose P(I am alive|Building Fell) = 0.1 . Now it does not matter what

caused the building to fall, could be an earthquake or weak structure or bomb fell and so on.

What matters is that the building fell, so that is why intuitively P(A|B|C) = P(A|B).

Meaning P(I am alive|Building Fell | Earthquake Struck) = P(I am alive| Building Fell | Meteor Struck)

= P(I am alive| Building Fell | Bomb fell) = P(I am alive| Building Fell) ?

So it does not matter what caused the building to fall, as long as i am concerned with My

probability of BEING ALIVE given that the BUILDING FELL. Is this right?

So we can say P(A|B|C|D|E....) = P(A|B) .
jbriggs444
#5
Feb17-14, 05:54 AM
P: 928
Quote Quote by sahil_time View Post
I was thinking, that suppose P(I am alive|Building Fell) = 0.1 . Now it does not matter what

caused the building to fall, could be an earthquake or weak structure or bomb fell and so on.

What matters is that the building fell, so that is why intuitively P(A|B|C) = P(A|B).
What if C is "you lived", "you died", or "you were on the roof wearing a parachute"? The condition given by C can be reflected in the value of P(A|B).

Intuitively P(A|B) is the average for A over all possible situations that conform to B. P(A|B|C) would then be the average over those situations that also conform to condition C.
sahil_time
#6
Feb17-14, 01:55 PM
P: 108
Quote Quote by jbriggs444 View Post
What if C is "you lived", "you died", or "you were on the roof wearing a parachute"? The condition given by C can be reflected in the value of P(A|B).

Intuitively P(A|B) is the average for A over all possible situations that conform to B. P(A|B|C) would then be the average over those situations that also conform to condition C.

P(A|B) implies that we are now in the domain of B where we have to calculate the amount of A. It

says that B has taken place, and hence we are in its Domain (Like in Venn Diagrams). Now as long

as we are concerned with A, it is directly concerned with B. Now since B has happened, it does

not matter what led to B's "Happening", and hence we can say that A depends solely on B and

not on anything else? Not true?


Also, to point out, just how would we interpret an event such as P(A|B|C) ? Is it even valid in the

strict mathematical sense?


Thank You.
Stephen Tashi
#7
Feb17-14, 02:22 PM
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Quote Quote by sahil_time View Post
how would we interpret an event such as P(A|B|C) ? Is it even valid in the strict mathematical sense?
In the strict mathematical sense (measure theory) it is complicated to define P(A|B), much more complicated that thinking about Venn diagrams. However, once you have defined P(A|B), you have defined a new probability space where P(A|B) denotes the probability of A in that space. Applying the definition of conditional probability again to that new space, you can define the probability of the event A in a space where C is a condition. So interpreting P(A|B|C) to mean P( (A|B) | C) does make sense.

P( (A|B)|C) amounts to the same thing as P(A | (B and C) ), so you usually see it written that way rather than P(A|B|C).
Simon Bridge
#8
Feb17-14, 07:51 PM
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Quote Quote by sahil_time View Post
Context : I was thinking how would we expand P((A|B),C), so using
Can you turn that notation into a question that the expansion is supposed to answer?

Quote Quote by sahil_time View Post
I was thinking, that suppose P(I am alive|Building Fell) = 0.1 . Now it does not matter what caused the building to fall, could be an earthquake or weak structure or bomb fell and so on.
Except, in the way you initially set up the situation, it does matter: you said so when you defined C.

The way you set it up P(A|B|C) is asking a different question to P(A|B).
Which is the nub of your original question right?

I think you will understand it better if you work from the question you want the math to ask/answer instead of asking what the notation is asking about.

What matters is that the building fell, so that is why intuitively P(A|B|C) = P(A|B).
If X is the event A|B, not sure what that would mean but anyway: what you have written is that P(X|C)=P(X)

Maybe.
But is that generally the case?

Stephen Tashi has given you the strict mathematical interpretation of the notation.
Quote Quote by Stephen Tashi
P( (A|B)|C) amounts to the same thing as P(A | (B and C) ), so you usually see it written that way rather than P(A|B|C).
You do have to be careful about notation to avoid ambiguity and confusion.

What usually happens is the events that you want to know about go on the LHS of the conditional bar, and the events you already know about on the RHS. There is only one conditional bar - otherwise the meaning can get murky.

Focussing on the question:

So your relatives know that there was an earthquake and the building you were in collapsed, how hard should they look?

The probability you are still alive would be P(A|(B,C)) ... this would be the same as P(A|B) if P(A|C)=1. (?)

i.e. if the building didn't get you, maybe the earthquake did?
sahil_time
#9
Feb18-14, 12:31 PM
P: 108
I think i probably got it figured, Thanks to your replies.

Here is my non-formal solution.

Imagine a Bag that contains 16 Blue balls and 4 Red balls.

B-16
R-4

Now what will be P(R|B|R) ? (This is analogous to my original question)

Now Suppose we go the formal way,

P(R|(B|R)) = P(R,(B|R)) / P(B|R)...................(1)

Now P(R|(B|R)) = Number of Red balls left / Total number of balls left

= 3/18 (Since one Blue and one Red is used up)

= 1/6

P(B|R) = 16/19

Therefore, substituting in (1)

P(R,(B|R)) = 8/57

Now P(R,(B|R)) absolutely makes no sense whatsoever, because we do not know whether to

change sample space for the Red ball that is picked with the Blue one.

So we make sense of it by interchanging bracket and ask what will be P((R,B)|R), which makes

sense. But if P((R,B)|R) is indeed equal to P(R,(B|R)) then we need to find its value.

So, P((R,B)|R) = P(B|R) * P(R|(B,R))

If also as previously mentioned P(R|(B,R)) = P(R|(B|R)) = 1/6

Then

P((R,B)|R) = 16/19 * 1/6

= 8/57

Hence P(R,(B|R)) = P((R,B)|R) IF P(R|(B,R)) = P(R|B|R)

And all this makes perfect sense.

Am i right?

Thank You everyone again.
Stephen Tashi
#10
Feb18-14, 01:45 PM
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Quote Quote by sahil_time View Post
Now P(R|(B|R)) = Number of Red balls left / Total number of balls left
Am i right?
You aren't being clear, even in an informal sense. To speak of the probability of an event unambiguously requires that you specify the "probability space" where that event is defined. You aren't being specific about which events are defined in which spaces. You seem to be defining the event R| (B|R) as some space of outcomes involving taking several random samples from the urn "without replacement".

Suppose we define the sample space to consist of outcomes ("points") of that give the color of a ball picked in a single draw. If the event "R" denotes the ball being red and the event "B" denotes the ball being blue then P(R|B) is zero since the ball isn't both red and blue.

Suppose we define the sample space to consist of outcomes that give the colors of two balls selected in a first and second draw from the urn. Then the outcomes are the ordered sets of letters like (r,r), (r,b),(b,r),(b,b). If "R" denotes the event that the first balls is red then it stands for the set consisting of the two "points" { (r,r), (r,b) }. To determine P(R|B), you have to say what you mean by the event B. Do you mean B to be the event consisting of the points where the second ball drawn is blue? That would be the set of outcomes{(r,b),(b,b)}.

To define the conditional probability P(A|B) we must employ events A,B that are in the same probability space to begin with. It isn't sufficient to define the event B as "the ball is blue" because unless you are talking about a space where there is only one ball drawn, the phrase "the ball is blue" is ambiguous.
Simon Bridge
#11
Feb18-14, 02:36 PM
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I think he's trying to ask: The probability that the second ball drawn is red, knowing that the first one drawn was blue... how would he write that down?
sahil_time
#12
Feb18-14, 05:41 PM
P: 108
Quote Quote by Stephen Tashi View Post
Do you mean B to be the event consisting of the points where the second ball drawn is blue?
I apologize for the ambiguity, what i meant for P(B|R) is that the first ball drawn is RED, and without

replacement, the second ball drawn is BLUE. So what is the probability that the second ball is BLUE

given the first ball is RED.
Stephen Tashi
#13
Feb18-14, 06:18 PM
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If the eventual goal is to interpret P(A|B|C) , we need a probability space where A,B,C are each sets of outcomes in that space. If we want an event to be a set of outcomes described by what happens on the 1,2,...n-th draw then an outcome ( a "point") in the space must be an ordered set that represents n draws from the urn.

So for 3 draws we need a space where the outcomes are (r,r,r),(r,r,b),(r,b,r),(r,b,b)... etc.

if (R|B)|R is supposed to mean"
"The probability ( that the 3rd ball chosen is red given the probability the second ball chosen is blue) given the third ball ball chosen is red" then the notation (R|B)|R is terrible notation because the first R in the expression refers to a different set than the second R in the notation.

A better notation for that event would be to let an outcome be represented by variables (x_1,x_2,x_3) and use the notation ( x3=R | x2= B) | x1 = R.

It's well known that [itex] P(A \cap B) [/itex] is defined differently than [itex] P(A|B) [/itex] and these two probabilities need not be equal. However, the set of outcomes involved in the sets within the notation [itex] P(A \cap B) [/itex] and [itex] P(A | B) [/itex] are the same sets. Both notations refer to computing the probability of the set of outcomes [itex] A \cap B [/itex]. The reason the two probabilities need not be equal is because different probability measures are involved, not because the two sets of outcomes are different.

The fact that one can write down a string of symbols such as (A|B)|C or A|(B|C) does not imply that such a string of symbols has a pre-defined meaning in mathematics. (Defining conditional probability and defining the notation for conditional probability are not the same task.) The definition of the conditional probability P(A|B) doesn't tell you what symbolism like P( (A|B)|C) means any more that it tells you what symbolism like P(A**B++/C) would mean.

If you want to define the notation P((A|B) |C) in a way that is consistent with the usual notation then both (A|B) and C would have to be interpreted as sets. I don't think this is a good method. I prefer the approach of defining P( (A|B) |C) to fit with the verbal interpretation of P(X|Y). By that interpretation P( (A|B)|C) = "(the probability of the event A given the event B) given the event C", which, verbally, is equivalent to "the probability of the event A given both the events B and C". So P( (A|B)|C) = P(A| (B intersection C) ).

The verbal interpretation of P( A| (B|C)) might be more controversial.


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