# Bayes Theorem with multiple conditions

I don't understand how this equation was derived.

$P(I|M_{1}\cap M_{2}) \leq \frac{P(I)}{P(I')}\cdot \frac{P(M_{1}|I)P(M_{2}|I)}{P(M_{1}|I')P(M_{2}|I')}$

This equation was from the paper "Trial by Probability" where the case of OJ Simpson was given as an example problem. The defendant is under trial for double murder and two evidences are introduced against him.

$M_{1}$ is the event the defendant's blood matches a drop of blood found in a crime scene. $M_{2}$ is the event a victim's blood matches blood on a sock belonging to the defendant. Assuming guilt, the occurrence of one evidence increases the probability of the other. $I$ is the event a defendant is innocent while $I'$ is when he is guilty.

We are trying to get the CEILING of the probability that the defendant is innocent given the two evidences.

Values for some variables were given but what I'm interested in is how the equation was derived. I tried but got nowhere.

Yes, I've already checked the 'Questions that may already have your answer'.

• What is the meaning of $I^\prime$? Is it $I^\text{c}$? – Xi'an Dec 27 '15 at 15:25
• @Xi'an yes $I'$ is $I^{c}$ in another notation – Sakurabe Dec 29 '15 at 0:52

## 1 Answer

By Bayes' Theorem: \begin{align*}P(I\mid M_1\cap M_2)&=\frac{P(I)P(M_1\cap M_2\mid I)}{P(M_1\cap M_2)}\\&=\frac{P(I)P(M_1\cap M_2\mid I)}{P(I)P(M_1\cap M_2\mid I)+P(I')P(M_1\cap M_2\mid I')}.\end{align*} Now the paper you provided argues that

If $I$ is true, then $M_1$ and $M_2$ are independent. But assuming guilt, the occurrence of one would increase the probability of the other.

So $$P(M_1\cap M_2\mid I)=P(M_1\mid I)P(M_2\mid I),\label{eq1}\tag{1}$$ and $$P(M_1\cap M_2\mid I')=P(M_1\mid M_2\cap I')P(M_2\mid I')\geq P(M_1\mid I')P(M_2\mid I').\label{eq2}\tag{2}$$ Hence, \begin{align*}P(I\mid M_1\cap M_2)&=\frac{P(I)P(M_1\mid I)P(M_2\mid I)}{P(I)P(M_1\cap M_2\mid I)+P(I')P(M_1\cap M_2\mid I')}&& \text{(Substitute with \eqref{eq1})}\\&\leq\frac{P(I)P(M_1\mid I)P(M_2\mid I)}{P(I')P(M_1\cap M_2\mid I')}&& \text{(Lesser Denominator)}\\&\leq\frac{P(I)}{P(I')}\cdot\frac{P(M_1\mid I)P(M_2\mid I)}{P(M_1\mid I')P(M_2\mid I')}.&& \text{(Substitute with \eqref{eq2})} \end{align*}

To derive \eqref{eq2}, note \begin{align*}\frac{P(M_1\cap M_2\mid I')}{P(M_2\mid I')}&=\frac{P(M_1\cap M_2\cap I')/P(I')}{P(M_2\cap I')/P(I')}\\&=\frac{P(M_1\cap M_2\cap I')}{P(M_2\cap I')}\\&=P(M_1\mid M_2\cap I')\end{align*} and since the occurrence of $M_2$ would increase the probability of $M_1$: $$P(M_1\mid M_2\cap I')\geq P(M_1\mid I')$$

• I wanna first thank you for taking the time to help. But I'm still a bit confused. Could you please add equation numbers and indicate where you apply earlier equations in later substitutions? Things are starting to make sense but I still don't get the inequality after the 'and', and the part where you substitute in the denominator and the whole thing becomes an inequality. I'm guessing an explanation on how the quoted argument from the paper is translated mathematically would help. Thanks again! – Sakurabe Dec 27 '15 at 10:38
• @Sakurabe: better? – Francis Dec 27 '15 at 19:27
• Okay, now I got how the evidences reinforce each other. Last question, did we just drop $P(I)P(M_{1}\cap M_{2}|I)$ from the denominator? As in drop without a theorem or anything? I mean, it does make some sense since it wouldn't reverse the resulting inequality from (2) plus it is also what I assumed they did in an earlier example in the paper involving only one DNA evidence (with the +1 in the denominator). Thanks, I really appreciate your help. – Sakurabe Dec 29 '15 at 1:53
• @Sakurabe: Yes, because that term is non-negative, so dropping it will decrease the denominator. – Francis Dec 29 '15 at 1:58