Is this conditional probability has derived correctly?

[sabbagh80]

Hi friends,

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

$P(X│A_1 A_2…A_n )= \frac{P(X)}{P(A_1 A_2…A_n )} P(A_1 A_2…A_n│X)=$

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

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

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

Is everything OK? thanks a lot in advance for your involvement!

 

[Stephen Tashi]

sabbagh80,

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

And I don't understand why the term $P(\cup_{i=1}^n A_i|X)$ would need a factor of $(-1)^{n-1}$ with it.

 

[sabbagh80]

Stephen Tashi,
By ${n \choose 2}P(A_1 A_2│X)$, I don't mean:
$\sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_i|X)p(A_j|X)$
I mean $\sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_iA_j|X)$

I just used the inclusion exclusion property of probability.
I want to write $P(A_1 A_2…A_n│X)$ according to the above property.

 

[Stephen Tashi]

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 $n \choose 2$ doesn't make sense.)

 

[Stephen Tashi]

I just used the inclusion exclusion property of probability.
I want to write $P(A_1 A_2…A_n│X)$ according to the above property.

That is the correct idea. Just get the notation fixed.

 

[sabbagh80]

As I mentioned in my problem, I have:

$\sum_{\{(i,j):1 \leq i \le n, 1 \leq j \leq n, i \neq j\}} p(A_iA_j|X)={n \choose 2}P(A_1 A_2│X)$

 

[sabbagh80]

So, you mean that it is OK? Am I right?
thanks

 

[Stephen Tashi]

So, you mean that it is OK? Am I right?
thanks

It depends on what you mean when you say that the $A_i$ are events "of the same type".

Must go to an appoinment now. Back later.

 

[Stephen Tashi]

Your idea is basically to take the formula
$$P( \cup_{i=1}^n A_i) = \sum_{i=1}^n P(A_i) - ...$$
and solve for the last term on the right hand side, which is either plus or minus $P(\cap_{i=1}^n A_i)$.

The formula also applies when you restrict each of the $A_i$ 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:

$$P(A_1 A_2 | X) = P(A_3 A_4)| X) = P(A_3 A_5| X)$$ etc.
$$P(A_1 A_2 A_3| X) = P(A_2 A_3 A_4| X) = P(A_3 A_4 A_5 | X)$$ etc.
$$P(A_1 A_2 A_3 A_4| X) = P(A_2 A_3 A_4 A_5| X)$$ 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 $A1$ and $A_2$ are independent does not mean that $A_1|X$ and $A_2|X$ must be independent.