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 $\frac{1}{6}$ 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 $\frac{1}{30}$ 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 $X\sim \text{Poisson}(\frac{1}{30})$ has a Poisson distribution with parameter $\lambda =\frac{1}{30}$.

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 $\lambda$ 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 $\lambda$ is now,

$$\lambda =60\times 4\times \frac{1}{30}=8$$

This should be intuitive because $\lambda$ is the average number of occurences of an event in a time period $T$. Thus, if in $1$ minute, there is $\frac{1}{30}$ 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 $X\sim \text{Poisson}(8)$ has a Poisson distribution with parameter $\lambda =8$.

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

$$\mathbb{P}[X=0]=\frac{e^{-8}{8}^0}{0!}=e^{-8}\approx 3.4\times {10}^{-4}$$

b) Let $X$ be the number of emails received on a weekday, in a time interval of length $1$ hour, and $X\sim \mathrm{Poisson}(60\times \frac{1}{6})=\mathrm{Poisson}(10)$.

Let $Y$ be the number of emails received on a weekend, in a time interval of length $1$ hour, and $Y\sim \mathrm{Poisson}(60\times \frac{1}{30})=\mathrm{Poisson}(2)$.

Let $Z$ be the disjoint union of $X$ and $Y$, and $Z=X\bigsqcup Y$. $Z$ is also a random variable with a Poisson distribution $Z\sim \mathrm{Poisson}(12)$. 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 $W\sim \mathrm{Uniform}(1,7)$.

Now we can further decompose $W$ to $W=W_1\bigsqcup W_2$, where $W_1$ is the random variable that indicates whether the chosen day is a weekday, and $W_2$ is the random variable that indicates whether the chosen day is a weekend. Note in particular that, $W$ is a random variable, a function $W:\mathrm{\Omega }\to \mathbb{R}$, where $\mathrm{\Omega }$ is the sample space, and $\mathbb{R}$ is the range of $W$, as indicated in Definition 14.

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

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:

$$\mathbb{P}[W_1|Z=0]$$

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)$$\begin{array}{r}\begin{array}{rl}\mathbb{P}[W_1|Z=0]& \stackrel{\text{(a)}}{=}{\displaystyle \frac{\mathbb{P}[Z=0|W_1]\mathbb{P}\left[W_1\right]}{\mathbb{P}[Z=0]}}\\ & \stackrel{\text{(b)}}{=}{\displaystyle \frac{e^{-10}\cdot \frac{5}{7}}{\mathbb{P}[Z=0]}}\end{array}\end{array}$$

where $\stackrel{\text{(b)}}{=}$’s $e^{-10}$ is derived from the Poisson formula when $\lambda =10$, because $\mathbb{P}[Z=0|W_1]=\mathbb{P}[X=0]=e^{-10}$ since conditional probability shrinked the sample space of $Z$ to $X$; the $\frac{5}{7}$ is just the probability of a chosen day is a weekday.

The denominator $\mathbb{P}[Z=0]$ 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,

$$\begin{array}{r}\begin{array}{rl}\mathbb{P}[Z=0]& =\mathbb{P}[Z=0|W_1]\mathbb{P}\left[W_1\right]+\mathbb{P}[Z=0|W_2]\mathbb{P}\left[W_2\right]\\ & =\mathbb{P}[X=0]\mathbb{P}\left[W_1\right]+\mathbb{P}[Y=0]\mathbb{P}\left[W_2\right]\\ & =e^{-10}\cdot \frac{5}{7}+e^{-2}\cdot \frac{2}{7}\end{array}\end{array}$$

With this, we have solved the problem,

$$\mathbb{P}[W_1|Z=0]={\displaystyle \frac{e^{-10}\cdot \frac{5}{7}}{e^{-10}\cdot \frac{5}{7}+e^{-2}\cdot \frac{2}{7}}}$$