Is this conditional probability has derived correctly?

In summary, the conversation discusses the problem of tracing relations for events of the same type, with the goal of expressing P(A_1 A_2...A_n|X) using the inclusion-exclusion property of probability. The conversation also touches on the notation and assumptions being used, including the assumption of a large set of equalities and the possibility of dealing with independent events.
  • #1
sabbagh80
38
0
Hi friends,

The problem:
Assume the events [itex] A_i [/itex] for [itex] i=1,2,...,n [/itex] are of the same type. please trace the following relations.

[itex] P(X│A_1 A_2…A_n )= \frac{P(X)}{P(A_1 A_2…A_n )} P(A_1 A_2…A_n│X)= [/itex]

[itex] \frac{P(X)}{P(A_1 A_2…A_n )}[ (-1)^{n-1} P(⋃_{i=1}^{n} A_i │X)+(-1)^{n-2} {n \choose 1} P(A_1│X)+ [/itex]
[itex] (-1)^{n-3} {n \choose 2}P(A_1 A_2│X)+…+{n \choose n-1}P(A_1 A_2…A_{n-1}│X) ] [/itex]

[itex] = \frac{1}{P(A_1 A_2…A_n )} [ (-1)^{n-1} P(X|⋃_{i=1}^{n}A_i ) P(⋃_{i=1}^{n}A_i)+ (-1)^{n-2} {n \choose 1}P(X│A_1 )P(A_1 )+(-1)^{n-3} {n \choose 2} P(X│A_1 A_2 )P(A_1 A_2 )+ [/itex]
[itex] …+{n \choose n-1}P(X│A_1 A_2…A_{n-1} )P(A_1 A_2…A_{n-1}) ] [/itex]

[itex] =(-1)^{n-1} \frac{P(⋃_{i=1}^{n}A_i )}{P(A_1 A_2…A_n )} P(X|⋃_{i=1}^{n}A_i )
+(-1)^{n-2} \frac{P(A_1 )}{P(A_1 A_2…A_n )} {n \choose 1} P(X│A_1 )+(-1)^{n-3} \frac{P(A_1 A_2 )}{P(A_1 A_2…A_n )} {n \choose 2} P(X│A_1 A_2 )+[/itex] [itex]…+{n \choose n-1} \frac{P(A_1 A_2…A_{n-1})}{(P(A_1 A_2…A_n)} P(X│A_1 A_2…A_{n-1} ) [/itex]

Is everything OK? thanks a lot in advance for your involvement!
 
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  • #2
sabbagh80,

That doesn't make sense. It's probably because you intend notation like [itex]{n \choose 2}P(A_1 A_2│X) [/itex] to mean [tex] \sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_i A_j|X) [/tex].

And I don't understand why the term [itex] P(\cup_{i=1}^n A_i|X) [/itex] would need a factor of [itex] (-1)^{n-1} [/itex] with it.
 
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  • #3
Stephen Tashi,
By [itex] {n \choose 2}P(A_1 A_2│X) [/itex], I don't mean:
[itex] \sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_i|X)p(A_j|X) [/itex]
I mean [itex] \sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_iA_j|X)[/itex]

I just used the inclusion exclusion property of probability.
I want to write [itex] P(A_1 A_2…A_n│X) [/itex] according to the above property.
 
  • #4
I was editing my message as you replied. See my revised version.

The point is that the use of the specific indices 1 and 2 in your expression does not convey the fact that you want the indices to vary over all possible pairs of distinct indices. (You do want that, right? Otherwise the [itex] n \choose 2 [/itex] doesn't make sense.)
 
  • #5
sabbagh80 said:
I just used the inclusion exclusion property of probability.
I want to write [itex] P(A_1 A_2…A_n│X) [/itex] according to the above property.

That is the correct idea. Just get the notation fixed.
 
  • #6
As I mentioned in my problem, I have:

[itex] \sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_iA_j|X)={n \choose 2}P(A_1 A_2│X) [/itex]
 
  • #7
So, you mean that it is OK? Am I right?
thanks
 
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  • #8
sabbagh80 said:
So, you mean that it is OK? Am I right?
thanks

It depends on what you mean when you say that the [itex] A_i [/itex] are events "of the same type".

Must go to an appoinment now. Back later.
 
  • #9
Your idea is basically to take the formula
[tex] P( \cup_{i=1}^n A_i) = \sum_{i=1}^n P(A_i) - ...[/tex]
and solve for the last term on the right hand side, which is either plus or minus [itex] P(\cap_{i=1}^n A_i) [/itex].

The formula also applies when you restrict each of the [itex] A_i [/itex] by saying "given X", so that's OK.

What you aren't making clear (I think) is that you are assuming a large set of equalities such as:

[tex] P(A_1 A_2 | X) = P(A_3 A_4)| X) = P(A_3 A_5| X) [/tex] etc.
[tex] P(A_1 A_2 A_3| X) = P(A_2 A_3 A_4| X) = P(A_3 A_4 A_5 | X) [/tex] etc.
[tex] P(A_1 A_2 A_3 A_4| X) = P(A_2 A_3 A_4 A_5| X) [/tex] etc.


You should state this explicitly. You haven't revealed the application you are working on. It may be that you are dealing with independent events. Just remember that the fact that [itex] A1 [/itex] and [itex] A_2 [/itex] are independent does not mean that [itex] A_1|X [/itex] and [itex] A_2|X [/itex] must be independent.
 

FAQ: Is this conditional probability has derived correctly?

1. What is conditional probability?

Conditional probability is a measure of the likelihood of an event occurring given that another event has already occurred. It is expressed as P(A|B), where A is the event of interest and B is the condition.

2. How do you calculate conditional probability?

Conditional probability is calculated by dividing the probability of the joint event (both A and B occurring) by the probability of the condition (B) occurring. This can be represented as P(A|B) = P(A∩B) / P(B).

3. What is the difference between conditional and unconditional probability?

Unconditional probability is the likelihood of an event occurring without any prior knowledge or condition. Conditional probability, on the other hand, takes into account a specific condition that has already occurred.

4. What are some real-life examples of conditional probability?

One example of conditional probability is the probability of a person having a certain disease given that they have a specific genetic mutation. Another example is the probability of a flight being delayed given that there is a snowstorm.

5. Can conditional probability be derived incorrectly?

Yes, conditional probability can be derived incorrectly if the events are not independent or if the sample size is too small. It is important to carefully consider the conditions and ensure that the events are truly independent before calculating conditional probability.

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