Continuous Probabilistic Models

Paulo Fagandini

Lisbon Accounting and Business School – Polytechnic University of Lisbon

Disclaimer

These slides are a free translation and adaptation from the slide deck for Estatística I by Prof. Sandra Custódio and Prof. Teresa Ferreira from the Lisbon Accounting and Business School - Polytechnical University of Lisbon.

Uniform Distribution

Continuous Uniform Distribution

A r.v. \(X\) follows a uniform distribution in \([a,b]\subset\mathbb{R}\) with \(-\infty<a<b<\infty\), if its probability density is given by:

\[ f_X(x)=\begin{cases} \frac{1}{b-a} & a\leq x \leq b\\ 0 & otherwise \end{cases} \]

We write \(X\sim U[a,b]\)

Continuous Uniform Distribution

Continuous Uniform Distribution

The cumulative distribution function is given by:

\[ F_X(x)=\begin{cases} 0 & x< a\\ \frac{x-a}{b-a} & a\leq x < b\\ 1 & x\geq b \end{cases} \]

Continuous Uniform Distribution

Note, this distribution is symmetric, and its first two moments are:

  • \(E[X]=\frac{a+b}{2}\)
  • \(V[X]=\frac{(b-a)^2}{12}\)

Continuous Uniform Distribution - Example

The length of small spots in a TV network is a r.v. \(X\) distributed \(U[5,12]\).

  1. Find its distribution function.
  2. What is the probability of a small spot last at least 7 seconds?
  3. The probability that a small spot lasts more than 6 seconds, given it never lasts more than 10 seconds is?
  4. Find \(E[X]\) and \(V[X]\)

Solution

Continuous Uniform Distribution - Example

Let the r.v. \(X\) be distributed \(U[2,b]\) with \(b>2\). What value must \(b\) take to make \(P(3\leq X\leq 5)=0.4\)?

Solution

Exponential Distribution

Exponential Distribution

The exponential distribution is rooted in the Poisson distribution, reflecting the waiting time between events originated according to a Poisson process.

Nevertheless, we can apply the exponential distribution to many other phenomena.

Exponential Distribution

A r.v \(X\) is distributed exponentially, with parameter \(\lambda>0\), \(X\sim Exp(\lambda)\), if its probability density function is given by:

\[ f_X(x)=\begin{cases} \lambda e^{-\lambda x} & x\geq 0\\ 0 & x< 0 \end{cases} \]

\(\lambda\) can be interpreted as the expected waiting time (or space) between events.

Exponential Distribution

Exponential Distribution

The cumulative probability function is:

\[ F_X(x)=\begin{cases} 0 & x< 0\\ 1-e^{-\lambda x} & x\geq 0 \end{cases} \]

And its moments are:

  • \(E[X]=\frac{1}{\lambda}\)
  • \(V[X]=\frac{1}{\lambda^2}\)

Normal Distribution

Appendix

Uniform Exercise

  1. $$ F_X(x)= \[\begin{cases} 0 & x<5\\ \frac{x-5}{12-5} & 5\leq x < 12\\ 1 & x\geq 12 \end{cases}\]
  2. \(F_X(12)-F_X(7)=1-\frac{2}{7}=\frac{5}{7}\)
  3. \(\frac{F_X(10)-F_X(6)}{F_X(10)}=\frac{5/7-1/7}{5/7}=4/5\)
  4. \(E[X]=\frac{5+12}{2}=8.5\), \(V[X]=\frac{(12-5)^2}{12}=\frac{49}{12}\approx 4 \Rightarrow \sigma_X\approx\sqrt{4}\approx 2\)

Back to the exercise

Uniform Exercise

  1. \(P(3\leq X\leq 5)=0.4\ \Rightarrow\ F_X(5)-F_X(3)=0.4\)
  2. \(\frac{5-2}{b-2}-\frac{3-2}{b-2}=0.4\ \Rightarrow\ \frac{3}{b-2}-\frac{1}{b-2}=0.4\)
  3. \(\frac{2}{b-2}=0.4\ \Rightarrow\ \frac{1}{b-2}=0.2\), but \(0.2 = \frac{2}{10} = \frac{1}{5}\)
  4. \(\frac{1}{b-2}=\frac{1}{5}\ \Rightarrow\ 5=b-2\ \Rightarrow\ b=7\)

Back to the exercise