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

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

❓ Uniform · Question 1

If \(X\sim U[a,b]\), then \(E[X]\) equals:

A. \(\frac{a+b}{2}\)

B. \(\frac{b-a}{2}\)

C. \(\frac{1}{b-a}\)

D. \(\frac{(b-a)^2}{12}\)

❓ Uniform · Question 2

For \(X\sim U[0,10]\), the probability \(P(X=5)\) is:

A. \(0.1\)

B. \(1\)

C. \(0.5\)

D. \(0\)

✏️ Uniform · Question 3

Let \(X\sim U[5,15]\).

Compute \(E[X]\), \(V[X]\), \(P(X\leq 8)\) and \(P(7<X\leq 12)\).

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:

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

\(\lambda\) is the rate: the expected number of events per unit of time (or space). The expected waiting time between events is \(\frac{1}{\lambda}\).

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:

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

Exponential Distribution

Considering the survival application of this distribution, this property states that, the time left to live is independent of what it already lived.

Exponential Distribution - Example

In a factory, the execution time of a piece is random variable distributed exponentially with expected value of 5 minutes.

1. For this distribution, what is the value of \(\lambda\)?
2. You know that the piece has been in execution by at least 2 minutes. What is the probability that 4 more minutes are necessary to complete the piece?
3. Picking randomly 5 pieces. What is the probability that 2 out of the 5 were produced in less than 4 minutes?

Solution

Exponential Distribution - Example

The time it takes until the first consultation, and between consultations, in the clinic of Dr. Shawn are independent and distributed exponentially with \(\lambda=0.1\).

What is the probability that no consultation occurs before the first 10 minutes?

Solution

❓ Exponential · Question 2

Memorylessness says that \(P(X>x+h\,|\,X>x)\) equals:

A. \(P(X>h)\)

B. \(P(X>x)\)

C. \(P(X>x+h)\)

D. \(0\)

✏️ Exponential · Question 3

Buses arrive with \(\lambda=0.05\) per minute.

Compute the expected waiting time, \(P(X>10)\), and \(P(X>25\,|\,X>15)\).

Normal Distribution

The Gaussian or Normal Distributions is one of the most used distributions, playing a key role in statistical inference.

The r.v. \(X\) is normally distributed, \(X\sim N(\mu,\sigma^2)\), if its density and cumulative probability distribution functions:

\[ f(x)=\frac{1}{\sigma\sqrt{2\pi}}e^{-\frac{(x-\mu)^2}{2\sigma^2}}\\ P(X\leq a)=F(a)=\int_{-\infty}^a f(t)dt \]

Where \(E[X]=\mu\) and \(Var[X]=\sigma^2\)

Normal Distribution

Standard Normal

If \(\mu=0\) and \(\sigma^2=1\) we call this the standard normal distribution, where \(Z\sim N(0,1)\).

Standard Normal

That is, we can standardize a r.v. distributed Normally. Its probability function is denoted \(\Phi\), and its density is \(\phi\).

\[ \phi(z)= \frac{1}{\sqrt{2\pi}}e^{-\frac{z^2}{2}} \\ \Phi(z)=P(Z\leq z) = \int_{-\infty}^z \frac{1}{\sqrt{2\pi}}e^{-\frac{t^2}{2}} dt \]

Standard Normal

How do we know if a r.v. follows a Normal distribution?

1. There is no rule, you need to look at the data (histogram) and decide.
2. This distribution is popular because of the Law of Large Numbers and the Central Limit Theorem (covered in Statistics II)
3. Even though it is by far the most popular, not always it is used correctly.

Example 1

Consider the r.v. \(Z\sim N(0,1)\). Find \(P(Z\leq 1.65)\).

\[P(Z\leq 1.65) = F(1.65)=\Phi(1.65)=0.9505\]

Example 1

Example 2

Consider the r.v. \(X\sim N(6,25)\). Find \(P(X\leq 12)\).

Example 2

Example 3

If \(X\sim N(6,25)\), find \(P(6<X\leq 12)\)

Example 3

Example 4

Let \(X\sim N(6,25)\). Find \(P(X\leq -4)\) and \(P(X\geq 16)\)

Example 4

Example 5

Check that \(P(X\leq -4)=P(X\geq 16)\)

Example 6

Consider the distribution \(X\sim N(6,25)\)

Find \(P(0\leq X\leq 8)\)

Example 7

Consider the r.v. \(X\sim N(6, 25)\). Find \(P(|X-6|>10)\)

Example 8

Consider the r.v. \(X\sim N(6,25)\), find \(k\) such that \(P(X>k)=0.9\)

Example 8

Using the table, \(P(Z\geq z)=0.1\) means \(\Phi(z)=0.9\), so \(z=\Phi^{-1}(0.9)\approx 1.282\), and by symmetry \(\Phi^{-1}(0.1)\approx -1.282\). Substituting we get:

\[\Phi(-1.282)=0.1\Leftrightarrow \frac{k-6}{5}=-1.282\] \[k=6-1.282\times 5 = -0.41\]

Example 8

Normal Distribution

To find \(\mu_T\) and \(\sigma_T^2\) remember the properties of the mean and variance.

Example 9

Let the r.v.s \(X\sim N(6,4)\) and \(Y\sim N(6,4)\), with \(T=0.5 X-Y\) Find \(\mu_T\) and \(\sigma_T^2\)

Example 9

Find \(P(T>0)\)

Example 9

Normal Distribution

Example 10

Let \(X_i\sim N(120, 64)\) be r.v.s representing the number of bank deposits made in a specific day. Then \(T=X_1+\dots+X_5\) are the weekly deposits.

Find the probability of the weekly deposits exceed 620.

Example 10

\[P(T>620)=P\left(Z>\frac{620-5\times 120}{\sqrt{5}\times 8}\right)=\\1-P(Z\leq 1.12)=1-\Phi(1.12)=0.1314\]

❓ Normal · Question 1

If \(X\sim N(\mu,\sigma^2)\), then \(Z=\frac{X-\mu}{\sigma}\) follows:

A. \(N(\mu,\sigma^2)\)

B. an Exponential

C. a Uniform

D. \(N(0,1)\)

❓ Normal · Question 2

For the standard Normal, \(\Phi(-z)\) equals:

A. \(1-\Phi(z)\)

B. \(\Phi(z)\)

C. \(-\Phi(z)\)

D. \(2\Phi(z)\)

References

• Murteira, B.; Silva Ribeiro, C.; Andrade e Silva, J. & Pimenta, C., Introdução à Estatística, Escolar Editora, McGraw-Hill, 2010
• Paulino, C. D. & Branco, J. A. (2005). Exercícios de Probabilidade e Estatística. Escolar Editora
• Pimenta, F., Andrade e Silva, J.; Silva Ribeiro, C. & Murteira, B., Introdução à Estatística, 3ª Edição, Escolar Editora, 2015
• Ferreira, T., Custódio, S.G., Modelos Probabilísticos, Síntese Teórica e Exercícios Resolvidos, Edições Sílabo (1ª Edição), 2023

Standard Normal Table

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

Exponential Exercise

In this case the r.v. \(X\) is the time until the first consultation, and \(X\sim Exp(0.1)\).

\[P(X>10)=1-P(X<10)=1-F(10)\]

Or

\[1-(1-e^{-0.1\times 10})=e^{-1}\approx 0.3679\]

Back to the exercise