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Prediction problem

September 8, 2011 Leave a comment

In class, I described a simple Bayesian prediction problem.  One observes the number of fire calls in a small California city for 30 days — the total number of fire calls in this period is 158.  An administrator is interested in predicting the total number of calls in the next seven days.

We assume that y, the number of calls y_i in a day, is Poisson with mean \lambda.  Suppose our prior beliefs about \lambda are described by a normal density with mean 10 and standard deviation 2.

I first write a simple function compute.post that computes and plots the posterior (using the likelihood times prior recipe) on a grid between lo and hi.
compute.post=function(lo, hi,...){
lambda = seq(lo, hi, length=100)
prior = dnorm(lambda, 10, 2)
like = lambda^158 * exp(-30*lambda)
post = prior * like
plot(lambda, post,...)
list(L = lambda, P = post)
}


With some trial and error, the interval (3.3, 7.5) seems to bracket the posterior well.

S=compute.post(3.3, 7.5, type="l", col="red")

Next I take a simulated sample from the posterior by using the sample function.

sim.lambda = sample(S$L, 1000, replace=TRUE, prob=S$P)

Last, I simulate a sample from the posterior predictive density of z, the number of fire calls in seven days.  Since [z | \lambda] is Poisson(7 \lambda, we simulate z by (1) simulating \lambda from the posterior and (2) simulating z from a Poisson(7 \lambda) distribution.  This is easy to do by using the vector properties of the random simulation functions.

sim.z = rpois(1000, 7*sim.lambda)

Here is a graph of the predictive distribution (by tabulating and plotting the simulated values of z).


		
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Categories: Single parameter

Brute force Bayes for one parameter

September 6, 2011 Leave a comment

Although we talk a lot about conjugate analyses, one doesn’t need to restrict oneself to the use of conjugate priors.  Here we illustrate learning about a Poisson mean using a normal prior on the mean \lambda .

Suppose I observe the following number of fire calls for a particular community each week:  0, 0, 1, 3, 2, 2.  If we assume y_i, the number of fire calls in week i, is Poisson(\lambda), then the likelihood function is given by

L(\lambda) = \exp(- 6 \lambda) \lambda^8, \, \, \lambda>0

Suppose I represent my prior beliefs about \lambda by a normal curve with mean 3 and standard deviation 1.  Then the posterior is given by (up to an unknown proportionality constant) by

g(\lambda | data) = L(\lambda) \times \exp(-(\lambda - 3)^2)

Here is a simple brute-force method of summarizing this posterior.

1.  Choose a grid of \lambda values that covers the region where the posterior is concentrated (this might take some trial and error).

2.  On this grid, compute the prior, likelihood, and posterior.

3.  Using the R sample function, take a large sample from the grid where the probabilities of the points are proportional to the like x prior values.

4.  Summarize this posterior simulated sample to learn about the location of the posterior.

Here’s some R code for this example.  I use the plot function to make sure the grid does cover the posterior.  The vector L contains 1000 draws from the posterior.

lambda = seq(0, 5, by=0.1)
like = exp(-6*lambda)*lambda^8
prior = dnorm(lambda, 3, 1)
post = like * prior
plot(lambda, post)
L = sample(lambda, size=1000, prob=post, replace=TRUE)
plot(density(L))  # density graph of simulated draws
Categories: Single parameter

Poisson mean example, part II

September 20, 2009 Leave a comment

In the last post, I described the process of determining a prior for \lambda, my son’s average number of text messages per day.   I decided to model my beliefs with a gamma(a, b) density with a = 44.4 and b = 4.4.

Now I observe some data.  I looked at the online record of text messaging for the first seven days that’s been at school and observe the counts

Sat  Sun  Mon  Tue  Wed  Thu  Fri
19     4      26      17    15       0    17

If we assume these counts are Poisson with mean \lambda, then the likelihood function is given by

L(lambda) =  lambda^s  exp(-n lambda),

where n is the number of observations (7) and s is the sum of the observations ( 98).

The posterior density (using the prior times likelihood recipe) is given by

L(lambda) x g(lambda) = lambda^{a+s-1}  exp(-(b+n) lambda),

which we recognize as a gamma density with shape a_1 = a + s = 142.4 and rate b_1 = b + n = 11.4.

Using the following R commands, we construct a triplot (prior, likelihood, and posterior).

> a=44.4; b=4.4
> s=sum(y); n=length(y)

> a=44.4; b=4.4
> s=sum(y); n=length(y)
> curve(dgamma(x,shape=s+a,rate=n+b),col=”red”,xlab=”LAMBDA”,
+   ylab=”DENSITY”,lwd=3,from=3,to=25)
> curve(dgamma(x,shape=a,rate=b),col=”blue”,lwd=3,add=TRUE)
> curve(dgamma(x,shape=s+1,rate=n),col=”green”,lwd=3,add=TRUE)
> legend(“topright”,c(“PRIOR”,”LIKELIHOOD”,”POSTERIOR”),
+  col=c(“blue”,”green”,”red”),lty=1,lwd=3)

gammapostNote that (1) the posterior is a compromise between the data information (likelihood) and the prior  and (2) the posterior is more precise since we are combining two sources of information.
I would like to predict my son’s text message use in the next month.  Let y_1^*, ..., y_{30}^* denote the number of text messages in the next month.  The total number of messages s^* has, conditional on \lambda, a Poisson distribution with mean 30 \lambda.
I am interested in computing the posterior predictive  distribution for s^*.  One can simulate this distribution by (1) simulating a value of the parameter \lambda from the gamma(142.4, 11.4) posterior and then (2) simulating s^* from a Poisson distribution with mean 30 \lambda.
We can do this simulation in R in two lines.   In the last line, I tabulate the values of s^* and plot the counts.
> lambda=rgamma(1000,shape=142.4,rate=11.4)
> ys=rpois(1000,30*lambda)
> plot(table(ys),ylab=”FREQUENCY”)
barchartLooking at the graph, it is likely that my son will have between 310 and 450 text messages in the next 30 days.
Categories: Single parameter

Constructing a prior for a Poisson mean

September 19, 2009 Leave a comment

I am interested in learning about my son’s cell phone use.   Suppose the mean number of text messages that he sends and receives per day is equal to \lambda.   I will observe the number of text messages y_1, ..., y_n for n days — we’ll assume that conditional on \lambda, y_1, ...., y_n are independent Poisson(\lambda).

How do I construct a prior density on \lambda?   Suppose I model my beliefs by use of a gamma density with shape parameter a and rate parameter b.  I want to figure out the values of the prior parameters a and b.

There are two distinct ways of assessing your prior.  One can think about plausible values of the mean number of daily text messages \lambda.  Alternatively, it may be easier to think about the actual number of text messages y — if we assume a gamma prior, then one can show that the predictive density of y is given by

f(y) = \frac{\Gamma(a+y)}{\Gamma(a) y!} \frac{b^a}{(b+1)^{a+y}}, y= 0, 1, 2, ...

Personally, I think it is easier to think about the number of text messages y.  My best prediction at y is 10.  After some additional thought, I decide that my standard deviation for y is equal to 3.5.    One can show that the mean and standard deviation of the predictive density are given by

E(y) = \frac{a}{b}, SD(y) = \sqrt{\frac{a}{b} + \frac{a}{b^2}}

If one matches my guesses with these expressions, one can show a = 44.4, b = 4.4.

To see if these values make any sense, I plot my predictive density.  This density is a special case of a negative binomial density where (using the R notation)

size = a,  prob = b /(b + 1).

I graph using the following R commands.  One can compute P(5 <= y <= 15) = 0.89 which means that on 89% of the days, I believe Steven will send between 5 and 15 messages.

a = 44.4; b = 4.4
plot(0:30,dnbinom(0:30,size=a,prob=b/(b+1)),type=”h”,
xlab=”y”, ylab=”Predictive Prob”, col=”red”, lwd=2)

a = 44.4; b = 4.4
plot(0:30,dnbinom(0:30,size=a,prob=b/(b+1)),type=”h”,
xlab=”y”, ylab=”Predictive Prob”, col=”red”, lwd=2)

poisson

Categories: Priors, Single parameter

Modeling with Cauchy errors

September 18, 2009 Leave a comment

One attractive feature of Bayesian thinking is that one can consider alternative specifications for the sampling density and prior and these alternatives makes one think more carefully about the typical modeling assumptions.

Generally we observe a sample y_1, ..., y_n distributed from a sampling density f(y|\theta) depending on a parameter \theta.  If we assign \theta  a prior density g(\theta), then the posterior density is given (up to a proportionality constant) by

g(\theta | y) = g(\theta) \prod_{i=1}^n f(y_i | \theta)

Consider the following Cauchy/Cauchy model as an alternative to the usual Normal/Normal model.  We let y_i be distributed from a Cauchy density with location \theta and known scale parameter \sigma, and we assign \theta a Cauchy prior with location \mu_0 and scale \tau.

Of course, this is not going to be a conjugate analysis — the posterior density has a complicated form.  But it is easy to use R to plot and summarize the posterior density.

First we define the posterior density for \theta.  The arguments to the function are theta (which could be a vector), the vector of data values, the known scale parameter \sigma, and the prior parameters \mu_0 andn \tau.  The output is a vector of values of the (unnormalized) posterior density.

posterior=function(theta,data,scale,mu0,tau)
{
f=function(theta) prod(dcauchy(data,theta,scale))
likelihood=sapply(theta,f)
prior=dcauchy(theta, mu0, tau)
return(prior*likelihood)
}

posterior=function(theta,data,scale,mu0,tau)
{
f=function(theta) prod(dcauchy(data,theta,scale))
likelihood=sapply(theta,f)
prior=dcauchy(theta, mu0, tau)
return(prior*likelihood)
}

(This function is a little tricky to program since both theta and data are vectors.)
We can plot the posterior density using the curve function.   We first give the data vector, assume \sigma = 1 and assign a Cauchy(0, 1) prior for \theta.
data=c(2,5,4,3,10,11,10,9)
We then use the curve function to plot the posterior.  After some preliminary work, we know that the probability mass falls between 0 and 15, so we use from = 0, to = 15 in the function.
curve(posterior(x,data,1,0,1),from=0,to=15,xlab=”THETA”,ylab=”DENSITY”)
Here is the plot that is produced.  Note the interesting bimodal shape.  One can obtain different posterior shapes when we model using Cauchy densities.  (Question:  Why does the posterior have this particular shape?)
cauchyplot
Categories: Single parameter

How many more home runs will Ryan Howard hit?

September 13, 2009 Leave a comment

Here is a basic proportion problem.  We are approaching the end of the 2009 baseball season and Ryan Howard of the Philadelphia Phillies currently has 38 home runs in 540 at-bats (opportunities).  I project that he will have 85 more at-bats this season.  How many additional home runs will he hit?

Let p be the probability that Howard hits a home run in a single at-bat during the 2009 season.  We’ll address this prediction problem in three steps.  First, we’ll use past data from previous seasons to form a prior distribution for p.  Second, we’ll update our beliefs about p using the 2009 data.  Last, we’ll predict the number of future at-bats from the posterior predictive distribution.  In this process, we’ll illustrate the use of some functions from the LearnBayes package in R.

1.  My prior.

In the four previous seasons (2005 through 2008) his home runs/at-bats have been 22/312. 58/581. 47/529, and 48/610.  The home run rates are 0.071, 0.100, 0.089, 0.079.  Remember that p is the 2009 home run hitting probability.  Based on these data, I believe that

(1) P(p < 0.075) = 0.5

(2) P(p < 0.115) = 0.9

In other words, my best guess at p is 0.075 and I’m pretty confident that p is smaller than 0.115.  I find the beta(a, b) that matches this information by use of the beta.select function in LearnBayes.  I first define the two quantiles as lists — these quantiles are the arguments in beta.select.

> quantile1=list(p=.5,x=.075)
> quantile2=list(p=.9,x=.115)
> beta.select(quantile1, quantile2)
[1]  7.26 85.77

We see that a beta(7.26, 85.77) matches this information.
2.  Update my prior with the 2009 data.
In the 2009 season, Howard had 38 successes (home runs) and 502 failures (not home runs).  The posterior for p will be beta with updated parameters
a = 7.26 + 38 = 45.26, b = 85.77 + 502 = 587.77
3.  Predict.
The predictive distribution with a beta density has the beta/binomial form.  These probabilities are computed using the pbetap function in LearnBayes.  The arguments are the beta parameters (a, b), the future sample size, and the values of y of interest.  Here we use a = 45.66, b = 587.77, we use 85 at-bats, and we’re interested in the probabilities of y = 0, 1, …, 85.
> ab=c(45.66, 587.77)
> n=85
> y=0:85
> pred.probs=pbetap(ab,n,y)
The predictive probabilities are stored in the vector pred.probs.  To summarize these probs, we use the function disc.int.  The inputs are the probability distribution where the first column are the values of y and the second column are the probabilities, and a probability content of interest.  The output is an interval of values that contain y with a given probability.  We use this function twice — once to find a 50% interval and a second time to find a 90% interval.
> discint(cbind(y,pred.probs),.5)
$prob
[1] 0.575821
$set
[1] 4 5 6 7
> discint(cbind(y,pred.probs),.9)
$prob
[1] 0.9301308
$set
[1]  2  3  4  5  6  7  8  9 10
So P(4 <= y <= 7) = 0.58 and P(2 <= y <= 10) = 0.93.  This means that I’m somewhat confident that Howard will hit between 4 and 7 home runs, and very confident that Howard will hit between 2 and 10 home runs in the remainder of the season.
> quantile1=list(p=.5,x=.075)
> quantile2=list(p=.9,x=.115)
> beta.select(quantile1, quantile2)
[1]  7.26 85.77
Categories: Single parameter

Do graduate students prefer McDonalds?

September 2, 2009 Leave a comment

Today I did an illustration of discrete Bayes for a proportion.  I’m interested in the proportion p of graduate students who answer “McDonalds” when asked the question “McDonalds, Wendys, or Burger King?”

I believe p can be one of the five values 0.1, 0.2, 0.3, 0.4, 0.5 and I assign the respective prior weights 1, 2, 5, 10, 5.  I define this prior in R:

> p=c(.1,.2,.3,.4,.5)
> prior=c(1,2,5,10,5)
> prior=prior/sum(prior)
> names(prior)=p

> p=c(.1,.2,.3,.4,.5)
> prior=c(1,2,5,10,5)
> prior=prior/sum(prior)
> names(prior)=p

I collected data from my class.  Of the 25 students, 11 responded with “McDonalds”.
I update my probabilities using the function discrete.bayes.  You can read in the function and associated plot and print methods by typing in
The updating is done by discrete.bayes.  The arguments are the sampling density dbinom, the prior probabilities defined in prior, the number of yes’s (11) and the sample size (25).
> s=discrete.bayes(dbinom,prior,11,size=25)
I compare the prior and posterior probabilities using two bar graphs.
> par(mfrow=c(2,1))
> barplot(prior,ylim=c(0,.6),xlab=”p”,main=”PRIOR”)
> plot(s,xlab=”p”,main=”POSTERIOR”)
twoplotsNote that the posterior probs are more precise than the prior probabilities.  I am more confident that the proportion of McDonalds fans is equal to 0.4.
Categories: Single parameter