The exponential distribution is the probability distribution of the time between events in a Poisson point process, i.e., a process in which events occur continuously and independently at a constant average rate. It is a particular case of the gamma distribution.

\(Y \sim Exp(\lambda)\) with conjugate \(Gamma\) function.

prior: \(\lambda \sim Gamma(\alpha,\beta), \space\),mean \(=1/\lambda = \alpha/\beta, \space\) stddev \(=1/\alpha\)

posterior: \(\lambda|y \sim Gamma(\alpha,\beta) = Gamma(a+n,b+\Sigma y_i),\space\) mean \(=(a+n)/(b+\Sigma y_i)\)

Consider the chocolate chip cookie example from the lesson. As in the lesson, we use the Poisson likelihood to model the number of chips per cookie, and a conjugate gamma prior on \(\lambda\), the expected number of chips per cookie. Suppose our prior expection is \(\lambda=8\).

  1. Recall that we used the conjugate \(Gamma\) prior for \(\lambda\), the arrival rate of busses per minute. Suppose our prior belief about this rate is that it should have mean 1/20 arrivals per minute with standard deviation of 1/5. Then the prior is \(Gamma(a,b)\) with \(a=1/16\). Find the balue of \(b\). Answer: mean = \(a/b = 1/20,\space b=20a = 20/16 = 1.25\)

  2. Suppose that we wish to use a prior with the same mean (1/20), but with an effective sample size of one arrival. Then the prior for \(\lambda\) is \(Gamma(1,20)\). In addition to the original $Y_1=12, we observe the waiting times for four additional buses: \(Y_2=15, Y_3=8, Y_4=13.5, Y_5=25\). Recall that with multiple (independent) observations, the posterior for \(\lambda\) is \(Gamma(\alpha,\beta)\) where \(\alpha=a+n\) and \(\beta=b+\Sigma y_i\) What is the posterior mean for \(\lambda\)?

(1+5)/(20+73.5)
## [1] 0.06417112
  1. Find the posterior probability that \(\lambda < 1/10\).
pgamma(0.1,1+5,20+73.5)
## [1] 0.9039699
  1. Suppose we decide on a prior \(\Gamma(8,1)\), which has a prior mean of 8 and sample size of 1 cookie. We collect data, sampling 5 cookies and counting the chips in each. We find 9, 12, 10, 15 and 13 chips. What is the posterior distribution for \(\lambda\)? Answer: posterior is \(\Gamma(\alpha+\Sigma y_i, \beta+n) = \Gamma(59+8, 1+5) = \Gamma(67,6)\)
chips_count <- list(c(9,12,10,15,13))
sapply(chips_count,sum)
## [1] 59
  1. What do the plots of the prior density (dotted line) and posterior density (solid line) look like?
lambda=seq(from=0,to=20,by=.01)
plot(lambda,dgamma(lambda,67,6),type="l",ylab='f(lambda | y)')
lines(lambda,dgamma(lambda,8,1),type="l",lty=2)

  1. What is the posterior mean for \(\lambda\)?

\(\bar{\lambda}_{posterior} = (\alpha+\Sigma y_i)/(\beta+n)= (\beta/(\beta+n))(\alpha/\beta)+(n/(\beta+n))(\Sigma y_i/n)\)

\(\bar{\lambda}_{posterior} = (59+8)/(1+5)\) 

(59+8)/(1+5)
## [1] 11.16667
  1. Find the lower end of a 90% equal-tailed credible interval for \(\lambda\).
qgamma(.05,67,6)
## [1] 9.021382
(1+5)/(20+61.5)
## [1] 0.07361963
qgamma(.975,9,390)
## [1] 0.04041843
y=seq(from=0,to=120,by=0.5)
plot(y,dgamma(y,9,390),type="l")