A traffic control engineer believes that the cars passing through a particular intersection arrive at a constant mean rate λ equal to either 2 or 4 every 10 minutes. Prior to collecting any data, the engineer believes that it is much more likely that the rate is 2. In fact, the engineer's prior probabilities are $P(λ = 2) = 0.75$ and $P(λ = 4) = 0.25$. One day, during a randomly selected 10 minute interval, the engineer observes x = 6 cars pass through the intersection.
In light of the observation that x = 6, calculate the posterior probabilities that λ = 2 and λ = 4, and hence discuss how the traffic control engineer's beliefs have been altered by the observation.
My Solution: Using Baye's theorem for λ = 2 , I get the following:
$P(λ = 2 |x = 6) = \frac{P(x = 6 | λ = 2)(P(λ = 2)}{P(x = 6)}$
Using the probability density formula for the Poisson distribution, $P(X = x) = \frac{λ^{x}e^{-λ}}{x!}$ , I get:
$P(λ = 2 | x = 6) = \frac{\frac{2^{6}e^{-2}}{6!} (0.75)}{\frac{2^{6}e^{-2}}{6!} + \frac{4^{6}e^{-4}}{6!}}$
Finally, I get $P(λ = 2 | x = 6) = 0.077628034$ and when I repeat the method for λ = 4, I get $P( λ = 4|x = 6) = 0.2241239887$.
The part I don't understand is how to comment on the traffic engineer's beliefs in light of these findings. Initially, I thought that the prior probability for $P(λ = 4) = 0.25$ was very close to the observed value of $P(λ = 4|x = 6)$ and therefore his suspicions were correct. However the observed value of $P(λ = 2|x = 6)$ is roughly 10 times smaller than $P(λ = 2)$ which makes me think I have done something wrong as $P( λ = 2|x = 6)$ and $P(λ = 4|x = 6)$ don't sum to 1, like the prior probabilities do.
to flesh out the comments:
This is a routine application of Bayes Theorem.
Using the standard Poisson distribution, we see that the probability of observing $6$ cars given that $\lambda =2$ is $.0120298$ and that with $\lambda =4$ it is $.10419563$
Now, using the given prior probabilities, the re-estimated probability that $\lambda =2$ should be $$\frac {.75\times .0120298}{.75\times .0120298+.25\times .10419563}= .25725711$$
so that the estimated probabilities have effectively switched. Now the engineer sees a (roughly) $75\%$ chance that $\lambda = 4$ and only a $25\%$ chance that $\lambda = 2$.