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?

  • $\begingroup$ 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. $\endgroup$ – joriki Sep 27 '12 at 2:43

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.

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  • $\begingroup$ 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. $\endgroup$ – Marie Hoffmann Mar 29 '13 at 19:34
  • $\begingroup$ @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 $\endgroup$ – 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)


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


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

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

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