Assume the amount of evidence against a defendant in a criminal trial is an exponential random variable $X$. If the defendant is innocent, then $X$ has mean $1$, and if the defendant is guilty, then $X$ has mean $2$. The defendant will be ruled guilty if $X>c$, where $c$ is a suitably chosen constant.

If the judge wants to be $95\%$ certain that an innocent man will not be convicted, what should the value of $c$ be? For this $c$ value, what is the probability that a guilty defendant will be convicted? Assume before the trial begins, you believe the defendant to be guilty with probability $10\%$. If the defendant is convicted, what is your updated belief about the probability of their guilt?

For this question, I am not sure as to where to begin for finding the initial $c$. I know that $\lambda_{\text{innocent}}=1$ and $\lambda_{\text{guilty}}=0.5$. my thought was to compute $0.95=\int(\lambda e^{-\lambda x})dx$ from $0$ to $c$ and solving for $c$ in both the guilty and innocent cases. I obtained negative values for both, which I assume are wrong. Am I on the right track or is there another approach I am not seeing?

After $c$ is known, finding the probability the guilty defendant will be convicted should be the straight forward integral for an exponential distribution from $0$ to $c$?

If the defendant is convicted, would be probability of their guilty be increased to $100\%$? or $0.1\times0.95$?

Thank you in advance for any help in understanding the problem!

  • $\begingroup$ $0.95 \leq \int_0^c \lambda e^{-\lambda x}\mathrm{x} \implies 0.95 \leq 1- e^{-\lambda c} \implies c \geq \frac{\ln(20)}{\lambda} $ $\endgroup$ – Graham Kemp Apr 30 '14 at 2:30
  • $\begingroup$ I do not see how this is correct.. where is the 20 coming from? I obtained the same integral, however when solving for c I got (-lambda*ln(0.05)) which equaled 1.497866137 for the guilty defendant which is incorrect. $\endgroup$ – user140624 Apr 30 '14 at 2:46
  • $\begingroup$ The negative of a log is the log of the reciprocal. $-\ln(0.05) = \ln(0.05^{-1}) = \ln(20)$ $\endgroup$ – Graham Kemp Apr 30 '14 at 2:48
  • $\begingroup$ ohh gotcha thanks! So I found c = ln(20) and the probability that a guilty defendant is convicted to be 0.223606. I am unsure about how to update the original believed guilty probability of 10%. Any ideas? $\endgroup$ – user140624 Apr 30 '14 at 3:00
  • $\begingroup$ Calculate the probabilities of conviction conditional on guilt and innocence, and update your belief given the conviction using Bayes theorem: $B(G\mid F) = \frac{B(G) P(F\mid G)}{P(F)}$ where $B(G)=0.10$ is your prior belief and $B(G\mid F)$ is the posterior belief. $\endgroup$ – Graham Kemp Apr 30 '14 at 3:48

Let $F$ represent the event: 'is found guilty' (also, 'is convicted').

Let $G$ represent the event: 'is guilty'. So $\lambda_G = 0.5, \lambda_{\neg G}=1$

We want $P(\neg F \mid \neg G) \geq 0.95$

Given exponentially distributed evidence $X$, the probability of being found innocent, determined by the cutoff $c$ and the lambda value $\lambda = E(X)^{-1}$ is: $$P(X \leq c) = \int_0^c \lambda e^{-\lambda x} \mathrm{d} x \\ = 1-e^{-\lambda c}$$

So we want a value of $c$ such that: $P(X \leq c \mid \lambda=1) \geq 0.95$.

$$0.95 \leq 1-e^{-c} \implies c \geq \ln(20)$$


$P(F \mid G) = P(X>\ln 20 \mid \lambda = 0.5) = 20^{-0.5} = \frac 1 {\sqrt{20}} \approx 0.2236\dotsc$

$P(F \mid \neg G) = P(X>\ln 20 \mid \lambda = 1) = 20^{-1} = \frac 1 {20} = 0.05$

Next, we wish you update our belief given conviction.

Let $B(G)$ be your prior belief that the defendant is guilty and $B(G\mid F)$ be your posterior belief of guilt given the conviction. Basically, your belief is an estimation of the probability of guilt.

By Bayes theorem: $$B(G\mid F) = \frac{B(G)\cdot P(F\mid G)}{B(G)\cdot P(F\mid G) + (1-B(G))\cdot P(F\mid \neg G)} \\ = \frac{\frac{1}{10}\times \frac{1}{\sqrt{20}}}{\frac{1}{10}\times \frac{1}{\sqrt{20}}+\frac{9}{10}\times\frac{1}{20}} \\ = \frac{20}{20+9\sqrt{20}} \\ \approx 0.33195\dotsc $$


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