# 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.

• What is the meaning of $I^\prime$? Is it $I^\text{c}$? Dec 27, 2015 at 15:25
• @Xi'an yes $I'$ is $I^{c}$ in another notation Dec 29, 2015 at 0:52

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 $M_1$: $$P(M_1\mid M_2\cap I')\geq P(M_1\mid I')$$
• 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. Dec 29, 2015 at 1:53