Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I have a fundamental question regarding conditional probability. Lets say I have $n$ independent random variables $X_1, X_2, \ldots, X_n$. Another random variable, $W$ is conditioned on the conjunction of these random variables: $P(W | X_1, X_2, X_3, \ldots, X_n)$. Given that $X_1, \ldots, X_n$ are independent, is it possible to write

$$P(W | X_1, X_2, X_3, \ldots, X_n) = P(W | X_1) \cdot P(W | X_2) \cdot \cdots \cdot P(W | X_n).$$

Or is the only way to rewrite is the Bayes theorem?

share|improve this question
    
You can use $\TeX$ on this site by enclosing formulas in dollar signs; single dollar signs for inline formulas and double dollar signs for displayed equations. You can see the source code for any math formatting you see on this site by right-clicking on it and selecting "Show Math As:TeX Commands". Here's a basic tutorial and quick reference. There's an "edit" link under the question. –  joriki Sep 27 '12 at 2:43

2 Answers 2

That expression is usually false for $n \gt 1$.

For example if $W$ is independent of all the $X_i$ and if $P(W)=p$ then the left hand side is $p$ while the right hand side is $p^n$, which is rather smaller.

share|improve this answer
    
At Henry, I think you misunderstood the question. $X_i$ are independent, but not $\{ W\} \cup \{ X_i\}$. I am also interested in the answer. I guess you can due to the rule $P(X_1,X_2) = P(X_1)\cdot P(X_2)$ if $X_1, X_2$ are independent. –  Marie Hoffmann Mar 29 '13 at 19:34
    
@Marie: My counter-example was an example. The main point is that karthik A is multiplying a lot of fractions together on the right hand side, and this will usually (though not always) be smaller than the left hand side –  Henry Apr 3 '13 at 14:11

The answer is no, it is no possible in general. Here a counterexample:

Let the sample space be $\{(0,0),(0,1),(1,0),(1,1)\}$ and the events:

  • $A$: The first element is one, or $\{(1,0),(1,1)\}$.
  • $B$: The second element is one, or $\{(0,1),(1,1)\}$.

Then is easy to see that:

  • $P(A)=\frac{1}{2}$ (First element one)
  • $P(B)=\frac{1}{2}$ (Second element one)
  • $P(A,B)=\frac{1}{4}$ (Both ones)
  • $P(A)P(B)=\frac{1}{4}$

Hence, $A$ and $B$ are independents. Now, let:

$W$: Both ones, or $\{(1,1)\}$ (this is the same as A,B)

Then:

  • $P(W|A)=\frac{1}{2}$
  • $P(W|B)=\frac{1}{2}$

But

$$P(W|A,B)=1$$ Since W={A,B} (Both elements one). Also, $$ P(W|A)P(W|B)=\frac{1}{4}$$

so $$P(W|A,B) \neq P(W|A)P(W|B)$$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.