Prove $P(G\mid E_1) > P(G)$ and $P(G\mid E_2) > P(G)$, but $P(G\mid E_1, E_2) < P(G)$

Question

I have a really simple question. The problem is as below.

Let $G$ be the event that a certain individual is guilty of a certain robbery. In gathering evidence, it is learned that an event $E_1$ occurred, and a little later it is also learned that another event $E_2$ also occurred. (a) Is it possible that individually, these pieces of evidence increase the chance of guilt (so $P (G\mid E_1 ) > P (G)$ and $P (G\mid E_2 ) > P (G))$, but together they decrease the chance of guilt (so $P(G\mid E_1,E_2) < P(G)$)?

The solution is:

Yes, this is possible. In fact, it is possible to have two events which separately provide evidence in favor of $G$, yet which together preclude $G$! For example, suppose that the crime was committed between $1$ pm and $3$ pm on a certain day. Let $E_1$ be the event that the suspect was at a nearby co↵eeshop from $1$ pm to $2$ pm that day, and let $E_2$ be the event that the suspect was at the nearby coffeeshop from $2$ pm to $3$ pm that day. Then $P(G\mid E_1) > P(G), P(G\mid E_2) > P(G)$ (assuming that being in the vicinity helps show that the suspect had the opportunity to commit the crime), yet $P(G\mid E_1,E_2) < P(G)$ (as being in the coffeehouse from $1$ pm to $3$ pm gives the suspect an alibi for the full time).

I can't understand why the intersection of $E_1$ and $E_2$ is being from $1$ pm to $3$ pm. Isn't it an empty one since there is no intersection between the event from $1$ pm to $2$ pm and the event from $2$ pm to $3$ pm?

Answer 1

The intersection of the two intervals of time is empty, or else consists of a single instant in time, at 2:00 pm.

But the intersection of the two events, that the suspect was there the entire first hour and that the suspect was there the entire second hour, just means that both events happened. Consider this set with three members: $$ \left\{ \begin{array}{l} \text{The suspect was there between 1:00 and 2:00 and} \\ \text{somewhere else between 2:00 and 3:00.} \\ {} \\ \text{The suspect was there between 1:00 and 3:00.} \\ {} \\ \text{The suspect was there between 2:00 and 3:00 and} \\ \text{somewhere else between 2:00 and 3:00.} \end{array} \right\} $$ This is a set of three epistemically possible mutually exclusive situations.

Here is the intersection: $$ \left\{ \begin{array}{l} \text{The suspect was there} \\ \text{between 1:00 and 2:00} \\ \text{and somewhere else} \\ \text{between 2:00 and 3:00.} \\ {} \\ \hline \\ \text{The suspect was there} \\ \text{between 1:00 and 3:00.} \end{array} \right\} \bigcap \left\{ \begin{array}{l} \text{The suspect was there} \\ \text{between 2:00 and 3:00} \\ \text{and somewhere else} \\ \text{between 1:00 and 2:00.} \\ {} \\ \hline \\ \text{The suspect was there} \\ \text{between 1:00 and 3:00.} \end{array} \right\} = \text{what?} $$ The first set above is the event that the suspect was there between 1:00 and 2:00 (and may or may not have been there during the next hour). The second set is the event that the suspect was there between 2:00 and 3:00 (and may or may not have been there during the previous hour).

The intersection of two "events" in probability theory is always the event that both of them are true.

Answer 2

Intersection of $E_1$ and $E_2$ is that the person was seen in the coffee shop between 1pm and 2pm again seen between 2pm and 3pm. These are not mutually exclusive as one does not preclude the other (so intersection is not empty).

An example of a mutually exclusive event would be that the person was seen at the coffee shop (say event A) and another house (say event B) at the same time. In this case $A\cap B = \varnothing$

$E_1,E_2$ together increase the likelihood of another event, say $E_3$ that the person stayed at coffee shop from 1pm to 3pm continuously. $\mathbb P[G\mid E_3]<\mathbb P[G]$.

Therefore $E_1,E_2$ together decrease the chance that the person is guilty.

PS: Too long for comment so posting as answer

Answer 3

Let $L(t)$ denote the suspect's Location at time $t$ pm.

• $E_1$ is the event that $L(t)$ is the nearby coffee shop for all $t$ such that $1 \leq t \leq 2$.
• $E_2$ is the event that $L(t)$ is the nearby coffee shop for all $t$ such that $2 \leq t \leq 3$.

and $E_1 \cap E_2$ is an event that $L(t)$ is the nearby coffee shop for all $t$ such that $1 \leq t \leq 2$ and $L(t)$ is the nearby coffee shop for all $t$ such that $2 \leq t \leq 3$. Together, this means $L(t)$ is the nearby coffee shop for all $t$ such that $1 \leq t \leq 3$.

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A bit more abstractly, we have $$E_1 = \bigcap_{1 \leq t \leq 2}\{L : L(t) \text{ is the nearby coffee shop}\} \\ E_2 = \bigcap_{2 \leq t \leq 3}\{L : L(t) \text{ is the nearby coffee shop}\} \\ E_1 \cap E_2 = \bigcap_{1 \leq t \leq 3}\{L : L(t) \text{ is the nearby coffee shop}\}$$ We are intersecting two intersections, so the time values over which we intersect is the union of the two sets of time values.