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

Practice

The exercises go from easy to hard:

• 🟢 warm-up: one formula, one step
• 🟡 standard: two or three steps
• 🟠 exam level: this is what you will find in the midterm
• 🔴 stretch: you need to combine several things

🟢 Practice · Exercise 1

Let \(X\sim Exp(0.5)\). Find \(P(X>2)\).

🟢 Practice · Exercise 2

Let \(Z\sim N(0,1)\). Find:

(a) \(P(0<Z\leq 2.05)\) (b) \(P(-1.22\leq Z<1.05)\) (c) \(P(Z\geq -2.05)\)

(d) \(k\) such that \(P(|Z|>k)=0.05\) (e) \(k\) such that \(P(|Z|<k)=0.90\)

🟡 Practice · Exercise 3

The number of cars 🚗 checked in an inspection center, per hour, is distributed Poisson with an expected value of 4.

(a) What is the expected time between two cars?

(b) Find the probability of waiting more than 30 minutes for the next car.

🟡 Practice · Exercise 4

The bus 🚌 arrives at your stop at a random moment between 0 and 20 minutes from now, all moments equally likely. Let \(W\) be your waiting time.

(a) Find \(P(W>15)\).

(b) You have waited 10 minutes already. Find \(P(W>15|W>10)\).

(c) Find \(E[W]\) and \(V[W]\).

🟠 Practice · Exercise 5

The lifetime of a light bulb 💡, in hours, is distributed exponentially with mean 1000 hours.

(a) Find the probability that a bulb lasts more than 1500 hours.

(b) Find the median lifetime. Is it above or below the mean?

(c) A box has 4 bulbs, independent. Find the probability that at least one of them lasts more than 1500 hours.

✅ Practice · Exercise 5 · Solution

🔴 Practice · Exercise 6

Go back to the bus 🚌 of Exercise 4. Now suppose the waiting time is instead \(W\sim Exp(0.1)\), so it has the same mean, 10 minutes.

(a) Find \(P(W>15)\) and \(P(W>15|W>10)\).

(b) Compare with Exercise 4. With which bus is it worth to keep waiting?

📝 Homework

Problem set 3.2, Questions 1 to 7.

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