# Applying the Bayesian theorem to rape statistics

I'm studying the prosecutor's fallacy. Imagine that:

$$C = \text{Convicted of rape} \\ I = \text{Innocent of rape} \\ A = \text{Accused of rape}$$

If I want to know the probability of being innocent having been accused, I apply Bayes' theorem:

$$P(A|I) = \frac{P(I|A)\cdot P(A)}{P(I)}$$

So if $$P(I|A) = 4\%$$ (number of false rape accusations) and $$P(A) = \frac{26,3}{100.000}$$ (stats for Belgium) and $$P(I) = 1-\frac{26,3}{100.000}$$ then:

$$P(A|I) = 1.052277e^{-05}$$

So the probability of being accused while innocent for a random person is far under the probability of being innocent while accused ?

• I didn't verify your arithmetic, but the point of Bayes's theorem is that $P(A\vert B)$ and $P(B\vert A)$ need not be equal, and we cannot specify a direction of inequality without knowing $P(A)$ and $P(B)$. – Dave Mar 3 '20 at 20:24
• ok, thanks, my question was more about the interpretation of the result and if it is coherent – Jean de Léry Mar 3 '20 at 20:34
• Why would $P(I) = 1 - P(A)$? It should be $P(I) = 1 - P(C)$ along your reasoning? – Mark Verhagen Mar 4 '20 at 7:58
• You write "If I want to know the probability of being innocent having been accused", which is $P(I|A)$ in your notation, not $P(A|I)$. I would guess $P(A|I)$ is not a meaningful quantity to be inferring, because if you already know the person is innocent, then you also know that any accusations are baseless. Does that make sense, or am I missing something? – Maurits M Mar 4 '20 at 9:11
• @MauritsM What about the probability that we accuse the null hypothesis of being false when it is true? That seems to be a quantity that is deemed meaningful in many analyses! – CloseToC Mar 4 '20 at 11:19

At the moment your problem does not involve the event $$C$$ at all, and you have not yet specified sufficient information to obtain the probability of interest. Specifically, you have not specified $$\mathbb{P}(I|\bar{A})$$, which is the probability that a randomly chosen non-accused person is innocent of rape. Using the values you have stipulated, and assuming for simplicity that $$\mathbb{P}(I|\bar{A}) \approx 1$$ (it would actually be slightly lower than one, since some rapists have never been caught), you should obtain:
\begin{aligned} \mathbb{P}(A|I) &= \frac{\mathbb{P}(I|A) \mathbb{P}(A)}{\mathbb{P}(I|A) \mathbb{P}(A) + \mathbb{P}(I|\bar{A}) \mathbb{P}(\bar{A})} \\[6pt] &\approx \frac{0.04 \cdot 0.00263}{0.04 \cdot 0.00263 + 1 \cdot (1-0.00263)} \\[6pt] &= \frac{0.0001052}{0.0001052 + 0.99737} \\[6pt] &= \frac{0.0001052}{0.9974752} \\[6pt] &= 1.05466 \times 10^{-5}. \\[6pt] \end{aligned}
Not necessarily. You have incorrectly defined $$P(I) = 1 - P(A)$$, which does not logically hold to be necessarily true. Is the probability of being innocent equal to the compliment of being accused?
Instead you have defined a $$P(I|A)$$ and a $$P(A)$$.
I would think about conditional probability definitions to determine the correct formulation for $$P(I)$$ with respect to the two terms I have listed above, and then recheck your calculations.
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