0
$\begingroup$

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!

$\endgroup$
  • $\begingroup$ $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
0
$\begingroup$

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: $P(X \leq c \mid \lambda=1) \geq 0.95$.

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

Hence:

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

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


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

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.