Exercises#

Problem 1#

Question 11 of Chapter 3 in Introduction to Probability, Statistics, and Random Processes

The number of emails that I get in a weekday (Monday through Friday) can be modeled by a Poisson distribution with an average of 16 emails per minute. The number of emails that I receive on weekends (Saturday and Sunday) can be modeled by a Poisson distribution with an average of 130 emails per minute.

a) What is the probability that I get no emails in an interval of length 4 hours on a Sunday?

b) A random day is chosen (all days of the week are equally likely to be selected), and a random interval of length one hour is selected on the chosen day. It is observed that I did not receive any emails in that interval. What is the probability that the chosen day is a weekday?

a) Let X be the number of emails received on a weekend, in a time interval of length 1 minute. Then XPoisson(130) has a Poisson distribution with parameter λ=130.

In Poisson’s Assumptions, the linearity assumption states that the probability of an event occurring is proportional to the length of the time period. As a consequence, the value of λ is proportional to the length of the time period.

And since the problem asked for a time period of length 4 hours, we have that the λ is now,

λ=60×4×130=8

This should be intuitive because λ is the average number of occurences of an event in a time period T. Thus, if in 1 minute, there is 130 email, then in 240 minutes (4 hours), there should be 8 emails.

We can rephrase our initial statement as follows:

Let X be the number of emails received on a weekend, in a time interval of length 4 hours. Then XPoisson(8) has a Poisson distribution with parameter λ=8.

Subsequently, the probability of getting no emails in a time interval of length 4 hours is given by

P[X=0]=e8800!=e83.4×104

b) Let X be the number of emails received on a weekday, in a time interval of length 1 hour, and XPoisson(60×16)=Poisson(10).

Let Y be the number of emails received on a weekend, in a time interval of length 1 hour, and YPoisson(60×130)=Poisson(2).

Let Z be the disjoint union of X and Y, and Z=XY. Z is also a random variable with a Poisson distribution ZPoisson(12). This is a consequence of the the additivity property in Poisson’s Property 17.

Let W be a random chosen day in a week, and W is a random variable with a Uniform distribution WUniform(1,7).

Now we can further decompose W to W=W1W2, where W1 is the random variable that indicates whether the chosen day is a weekday, and W2 is the random variable that indicates whether the chosen day is a weekend. Note in particular that, W is a random variable, a function W:ΩR, where Ω is the sample space, and R is the range of W, as indicated in Definition 14.

The sample space Ω is just the set of days in a week, where we denote as Ω={1,2,3,4,5,6,7}.

Similarly, Z has a sample space that can be understood as the disjoint union of the sample spaces of X and Y. This is important as we want to invoke the Law of Total Probability.

Now we can formulate the problem as follows:

P[W1|Z=0]

which means what is the probability of a chosen day is a weekday given that I did not receive any emails in the time interval of 1 hour of a random chosen week.

By Bayes’ Theorem (Definition 13), we have that

(13)#P[W1|Z=0]=(a)P[Z=0|W1]P[W1]P[Z=0]=(b)e1057P[Z=0]

where =(b)’s e10 is derived from the Poisson formula when λ=10, because P[Z=0|W1]=P[X=0]=e10 since conditional probability shrinked the sample space of Z to X; the 57 is just the probability of a chosen day is a weekday.

The denominator P[Z=0] is the probability of not receiving any emails in a time interval of length 1 hour of a random chosen week. This is not straightforward and we have to use the Law of Total Probability (Theorem 3). That is,

P[Z=0]=P[Z=0|W1]P[W1]+P[Z=0|W2]P[W2]=P[X=0]P[W1]+P[Y=0]P[W2]=e1057+e227

With this, we have solved the problem,

P[W1|Z=0]=e1057e1057+e227