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Let G be the event that a certain individual is guilty of a certain robbery. In gathering evidence, it is learned that an event E1 occurred, and a little later it is also learned that another event E2 also occurred. (a) Is it possible that individually, these pieces of evidence increase the chance of guilt (so P (G\mid E1 ) > P (G) and P (G\mid E2 ) > P (G)), but together they decrease the chance of guilt (so P(G\mid E1,E2) < P(G))?

This question is taken from Joe Blitzstein's Homework 02.

The solution provided 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 E1 be the event that the suspect was at a nearby co↵eeshop from 1 pm to 2 pm that day, and let E2 be the event that the suspect was at the nearby co↵eeshop from 2 pm to 3 pm that day.

However I would like to prove this mathematically.

$$P(G \mid E_i) > P(G), i \in {1, 2}$$

using Bayes Law:

$$\frac{P(G) \cdot P(E_i \mid G)}{P(E_i)} > P(G)$$

divide by $P(G)$ and multiply with $P(E_i)$:

$$P(E_i \mid G) > P(E_i)$$

I believe this is correct but now comes the difficult part:

$$P(G \mid E_1, E_2) < P(G)$$

again using Bayes Law:

$$\frac{P(G) \cdot P(E_1 \mid G) \cdot P(E_2 \mid G)}{P(E_1) + P(E_2)} < P(G)$$

again divide by $P(G)$ and multiply by $P(E_1) + P(E_2)$:

$$P(E_1 \mid G) \cdot P(E_2 \mid G) < P(E_1) + P(E_2)$$

now it is getting hard. I believe that this is true as the product of $a, b \in [0, 1]$ is smaller than their sums.

But how would can I show the last step?

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Please note that we need to prove that this situation is possible and don't need to prove that this is always true.

In this equation, if we replace RHS with something smaller and then show that that could be greater than LHS than we are done. $$ P(E_1 \mid G) \cdot P(E_2 \mid G) < P(E_1) + P(E_2) $$ so we could show that this is possible: $$ P(E_1 \mid G) \cdot P(E_2 \mid G) < P(E_2 \mid G) + P(E_1 \mid G) $$ Now this is obviously true as the product of two numbers or rather probabilities could be smaller than their sum, we are done.

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