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Chapter 4: Q4E (page 146)

Let \({\rm{X}}\) denote the vibratory stress (psi) on a wind turbine blade at a particular wind speed in a wind tunnel. The article 鈥淏lade Fatigue Life Assessment with Application to VAWTS鈥� (J. of Solar Energy Engr., \({\rm{1982: 107 - 111}}\)) proposes the Rayleigh distribution, with pdf

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2}}{{\rm{\theta }}^{\rm{2}}}{\rm{)}}}}}&{{\rm{x > 0}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

otherwise as a model for the \({\rm{X}}\) distribution.

a. Verify that \({\rm{f(x;\theta )}}\) is a legitimate pdf.

b. Suppose \({\rm{\theta = 100}}\) (a value suggested by a graph in the article). What is the probability that \({\rm{X}}\) is at most \({\rm{200}}\)? Less than \({\rm{200}}\)? At least \({\rm{200}}\)?

c. What is the probability that \({\rm{X}}\) is between \({\rm{100}}\) and \({\rm{200}}\) (again assuming \({\rm{\theta = 100}}\))?

d. Give an expression for \({\rm{P(X}} \le {\rm{x)}}\).

Short Answer

Expert verified

(a) It is verified that\({\rm{f(x;\theta )}}\)is a legitimate pdf.

(b) The probability that\({\rm{X}}\)is at most\(200\), less than\(200\)and at least\(200\)is\({\rm{0}}{\rm{.8647, 0}}{\rm{.8647}}\)and\({\rm{0}}{\rm{.1353}}\)respectively.

(c) The that\({\rm{X}}\)is between\(100\)and\(200\)is\({\rm{0}}{\rm{.2582}}\).

(d) The expression for\({\rm{P(X}} \le {\rm{x)}}\)is\({\rm{1 - }}{{\rm{e}}^{{\rm{ - (x}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\).

Step by step solution

01

Concept Introduction

Probability refers to the likelihood of a random event's outcome. This word refers to determining the likelihood of a given occurrence occurring.

02

Verifying the pdf

(a)

The pdf given is 鈥�

\({\rm{f(x;\theta ) = \{ }}\begin{array}{*{20}{c}}{\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/(2}}{{\rm{\theta }}^{\rm{2}}}{\rm{)}}}}}&{{\rm{x > 0}}}\\{\rm{0}}&{{\rm{otherwise}}}\end{array}\)

For a pdf to be a legitimate pdf, it must satisfy the following two conditions 鈥�

1. \({\rm{f(x)}} \ge {\rm{0}}\)for all\({\rm{x}}\)

2. \(\int_{{\rm{ - }}\infty }^\infty {{\rm{f(x)}} \cdot {\rm{dx = 1}}} \)

It can be seen that given pdf is positive for all\({\rm{x}}\), hence it satisfies condition\({\rm{(1)}}\)鈥�

\(\int_{{\rm{ - }}\infty }^\infty {\rm{f}} {\rm{(x)}} \cdot {\rm{dx = }}\int_{\rm{0}}^\infty {\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\)

Here a simple observation can be made that\(\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}{\rm{ = }}\frac{{{\rm{d}}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}}}{{{\rm{dx}}}}\). Hence, above integral can be written as 鈥�

\(\begin{aligned}\int_{{\rm{ - }}\infty }^\infty {\rm{f}} {\rm{(x)}} \cdot dx &= \int_{\rm{0}}^\infty {\frac{{{\rm{d}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right.}}{{{\rm{dx}}}}} \cdot {\rm{dx}}\\ &= - \left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right)_{\rm{0}}^\infty \\&= - \left( {{{\rm{e}}^{{\rm{ - (}}\infty {{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}{\rm{ - }}{{\rm{e}}^{{\rm{ - (0}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right)\\ &= - (0 - 1) = 1\end{aligned}\)

Therefore, the pdf is legitimate.

03

Finding the Probability

(b)

First, denote these probabilities into usual notations 鈥�

\(\begin{array}{c}{\rm{P(X is at most 200) = P(X}} \le {\rm{200)}}\\{\rm{P(X is less than 200) = P(X < 200)}}\\{\rm{P(X is atleast 200) = P(X}} \ge {\rm{200)}}\\{\rm{P(X}} \le {\rm{200) = }}\int_{\rm{0}}^{{\rm{200}}} {\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\\{\rm{ = }}\int_{\rm{0}}^{{\rm{200}}} {\frac{{{\rm{ - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}}}{{{\rm{dx}}}}} \cdot {\rm{dx}}\\{\rm{ = - }}\left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2(100}}{{\rm{)}}^{\rm{2}}}}}} \right)_{\rm{0}}^{{\rm{200}}}\end{array}\)

Since value of \({\rm{\theta }}\) is given: \({\rm{\theta = 100}}\)

\(\begin{aligned}{\rm{P(X}} \le 200) &= - \left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/2(100}}{{\rm{)}}^{\rm{2}}}}}} \right)_{\rm{0}}^{{\rm{200}}}\\ & = - \left( {{{\rm{e}}^{{\rm{ - (200}}{{\rm{)}}^{\rm{2}}}{\rm{/2(100}}{{\rm{)}}^{\rm{2}}}}}{\rm{ - }}{{\rm{e}}^{{\rm{ - (0}}{{\rm{)}}^{\rm{2}}}{\rm{/2(100}}{{\rm{)}}^{\rm{2}}}}}} \right)\\ &= - \left( {{{\rm{e}}^{{\rm{ - 2}}}}{\rm{ - 1}}} \right)\\ &= - (0{\rm{.1353 - 1)P(X}} \le {\rm{200)}}\;{\rm{ = 0}}{\rm{.8647}}\end{aligned}\)

Since the given pdf is continuous, hence 鈥�

\({\rm{P(X < 200) = P(X}} \le {\rm{200) = 0}}{\rm{.8647}}\)

Also using properties of pdf, it can be written 鈥�

\(\begin{aligned}{\rm{P(X}} \ge 200) &= 1 - P(X < 200) \\ &= 1 - 0 {\rm{.8647}}\\ &= 0{\rm{.1353}}\end{aligned}\)

Practical consequence for continuous random variable 鈥�

When\({\rm{X}}\)is continuous random variable, then the probability that\({\rm{X}}\)lies in some interval between\({\rm{a}}\)and\({\rm{b}}\)does not depend on whether the lower limit\({\rm{a}}\)or the upper limit\({\rm{b}}\)is included in the probability calculation 鈥�

\({\rm{P(a < X < b) = P(a}} \le {\rm{X < b) = P(a < X}} \le {\rm{b) = P(a}} \le {\rm{X}} \le {\rm{b)}}\)

Therefore, the values obtained are \({\rm{0}}{\rm{.8647, 0}}{\rm{.8647}}\) and \({\rm{0}}{\rm{.1353}}\).

04

Finding the Probability

(c)

Probability that\({\rm{X}}\)is between\({\rm{100}}\)and\({\rm{200}}\)is denoted by 鈥�

\({\rm{P(100 < X < 200) = P(X < 200) - P(X > 100)}}\)

\({\rm{P(X < 200)}}\)is obtained as\({\rm{P(X < 200) = 0}}{\rm{.8647}}\).

To calculate\({\rm{P(X > 100)}}\)write that 鈥�

\(\begin{array}{c}{\rm{P(X > 100) = 1 - P(X}} \le {\rm{100)}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{{\rm{100}}} {\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}} {\rm{ \times }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\\{\rm{ = 1 - }}\int_{\rm{0}}^{{\rm{100}}} {\frac{{{\rm{d}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right.}}{{{\rm{dx}}}}} {\rm{ \times dx}}\\{\rm{ = 1 + }}\left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right)_{\rm{0}}^{{\rm{100}}}\end{array}\)

Since value of \({\rm{\theta }}\) is given: \({\rm{\theta = 100}}\)

\(\begin{aligned} P(X > 100) &= 1 + \left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2(100}}{{\rm{)}}^{\rm{2}}}} \right)}}} \right)_{\rm{0}}^{{\rm{100}}}\\&= 1 + \left( {{{\rm{e}}^{{\rm{ - (100}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2(100}}{{\rm{)}}^{\rm{2}}}} \right)}}{\rm{ - }}{{\rm{e}}^{{\rm{ - (0}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2(100}}{{\rm{)}}^{\rm{2}}}} \right)}}} \right)_{\rm{0}}^{{\rm{100}}}\\ & = 1 + \left( {{{\rm{e}}^{{\rm{ - 1/2}}}}{\rm{ - 1}}} \right)\\ &= 1 + (0{\rm{.6065 - 1)}}\\P(X > 100) &= 0{\rm{.6065}}\end{aligned}\)

Finally calculate\({\rm{P(100 < X < 200)}}\)as 鈥�

\(\begin{aligned}P(100 < X < 200) &= P(X < 200) - P(X > 100)\\ &= 0 {\rm{.8647 - 0}}{\rm{.6065}}\\ &= 0 {\rm{.2582}}\end{aligned}\)

Therefore, the value obtained is\({\rm{0}}{\rm{.2582}}\).

05

Expression for \({\rm{P(X}} \le {\rm{x)}}\)

(d)

The expression can be obtained as 鈥�

\(\begin{aligned}{\rm{P(X}} \le x) &= \int_{\rm{0}}^{\rm{x}} {\frac{{\rm{x}}}{{{{\rm{\theta }}^{\rm{2}}}}}} \cdot {{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\\ &= \int_{\rm{0}}^{\rm{x}} {\frac{{{\rm{d}}\left( {{\rm{ - }}{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right.}}{{{\rm{dx}}}}} \cdot {\rm{dx}}\\ &= - \left( {{{\rm{e}}^{{\rm{ - }}{{\rm{x}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right)_{\rm{0}}^{\rm{x}}\\ &= - \left( {{{\rm{e}}^{{\rm{ - (x}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}{\rm{ - }}{{\rm{e}}^{{\rm{ - (0}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}} \right)\\&= - \left( {{{\rm{e}}^{{\rm{ - (x}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}{\rm{ - 1}}} \right)\\ &= 1 - {{\rm{e}}^{{\rm{ - (x}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\end{aligned}\)

Therefore, the expression obtained is\({\rm{1 - }}{{\rm{e}}^{{\rm{ - (x}}{{\rm{)}}^{\rm{2}}}{\rm{/}}\left( {{\rm{2}}{{\rm{\theta }}^{\rm{2}}}} \right)}}\).

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