A and B are independent witness in a case. The probablity that A speaks the truth is 'x' and that of B is 'y'.If A and B agree on a certain statement, how to find the probability that the statement is true ?
How to find the probability of truth?
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0@Gigili, please refer to answers below to see why. It should be trivial to *everyone* not just *me* I hope. – 2011-11-28
3 Answers
Say the fact is True or False (T/F), and independent statements A, B are binary.
We are given P(A=1|T)=x and P(B=1|T)=y. By convexity P(A=0|T)=1-x and P(B=0|T)=1-y.
We can surmise P(A=0|F)=x and P(B=0|F)=y. Also by convexity P(A=1|F)=1-x and P(B=1|F)=1-y.
We want P(T|A=1,B=1).
By Bayes rule P(T|A=1,B=1) = P(A=1,B=1|T)P(T)/P(A=1,B=1).
By independence P(T|A=1,B=1) = P(A=1|T)P(B=1|T)P(T)/(P(A=1)P(B=1)).
Plugging in, the numerator is xyP(T).
The denominator requires P(A=1) = P(A=1|T)P(T) + P(A=1|F)P(F) = xP(T) + (1-x)P(F), same for P(B=1).
So I don’t think we are identified without knowing the marginal probability of the truth P(T).
Of course, for a given x and y you can provide P(T|A=1,B=1) as a function of P(T) between 0 and 1.
$ P(A)=x $ and $ P(B)=y $,
$A$ and $B$ are independent, so $P(A \cap B)=P(A)P(B)$
Therefore, the probability that both speak the truth will be $P(A \cap B)=P(A)P(B)=xy$.
And then, the probability that they agree on a certain statement is, they both speak the truth or they both tell lie, which would be:
$xy + (1-x)(1-y) $
As a result, the probability that the statement is true is:
$ \frac{xy}{xy+(1-x)(1-y)} $.
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1No, that's probability that both speak truth. – 2011-11-28
Let:
$A_t$ stand for "A says statement is true." and $A_f$ for "A says statement is false" and
$B_t$ stand for "B says statement is true." and $B_f$ for "B says statement is false" and
$S_t$ stand for "Statement is true" and $S_f$ for "Statement is false" and
Then, we know that:
$\text{Prob}(A_t | S_t) = \text{Prob}(A_f | S_f) = x$, and
$\text{Prob}(A_t | S_f) = \text{Prob}(A_f | S_t) = 1-x$, and
$\text{Prob}(B_t | S_t) = \text{Prob}(B_f | S_f) = y$, and
$\text{Prob}(B_t | S_f) = \text{Prob}(B_f | S_t) = 1-y$, and
We want to know:
$\text{Prob}(S_t | A_t \cap B_t)$
Using Bayes theorem, we have:
$\text{Prob}(S_t | A_t \cap B_t) = \frac{\text{Prob}(A_t \cap B_t |S_t) \text{Prob}(S_t)}{\text{Prob}(A_t \cap B_t)}$
But, we know that,
$\text{Prob}(A_t \cap B_t |S_t) = xy$ and
$\text{Prob}(A_t \cap B_t) = \text{Prob}(A_t \cap B_t |S_t) \text{Prob}(S_t) + \text{Prob}(A_t \cap B_t |S_f) \text{Prob}(S_f)$
Thus,
$\text{Prob}(A_t \cap B_t) = xy \text{Prob}(S_t) + (1-x)(1-y) (1-\text{Prob}(S_t))$
Simplifying the above, we get:
$\text{Prob}(A_t \cap B_t) = (1+2xy-x-y) \text{Prob}(S_t)$
Thus, we have:
$\text{Prob}(S_t | A_t \cap B_t) = \frac{xy}{1+2xy-x-y}$