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LectureNotes13 - Functions of Random Variables Cyr Emile...

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Unformatted text preview: Functions of Random Variables Cyr Emile M’LAN, Ph.D. [email protected] Functions of Random Variables – p. 1/31 Introduction ♠ Text Reference : Introduction to Probability and Its Applications, Chapter 7. ♠ Reading Assignment : Sections 7.1-7.6, April 27 Often distributions are known for a random variable X 1 or a sequence of random variables X 1 ,X 2 , ··· ,X n , but what may be of interest is another random variable, U , which can be expressed as some function of the random variable X 1 , for e.g., U = h ( X 1 ) , or some function of the random variables X 1 ,X 2 , ··· ,X n , for e.g., U = h ( X 1 ,X 2 , ··· ,X n ) . Conversions and changes of scale (degrees Fahrenheit to degrees Celsius). The lifetime T of an electronic system depends on the sum X 1 + X 2 of two components. Functions of Random Variables – p. 2/31 Introduction Let X 1 ,X 2 , ··· ,X n denote a random sample. A statistic U = h ( X 1 ,X 2 , ··· ,X n ) is computed from the sample. Question : How to find the probability distribution of the random variable U ? The solution is centered around three basic ideas: The distribution function method The transformation method The moment generating function method Functions of Random Variables – p. 3/31 The Distribution Function Method The distribution function method is the most direct approach and it is typically used when dealing with continuous random variables X i ’s . 1. For each u , find the region U = u in the ( x 1 ,x 2 , ··· ,x n ) space. 2. Then find the region Γ u defined in terms of the x i ’s for which U ≤ u . 3. For each u , find F U ( u ) = P ( U ≤ u ) by integrating the probability density function f 1 ( x 1 ) or the joint probability density function f ( x 1 ,x 2 , ··· ,x n ) over the region Γ u . 4. The density of U is then obtained by differentiating the distribution F U ( u ) , that is, f U ( u ) = F prime U ( u ) . Functions of Random Variables – p. 4/31 The Distribution Function Method Example 7.1 : Suppose that a random variable Y has probability density function f ( x ) = 6 x (1- x ) , < x < 1 . What is the probability density function of U = 2 X + 1 ? Solution : F X ( y ) = integraldisplay x 6 t (1- t ) dt = 3 x 2- 2 x 3 F U ( u ) = P ( U ≤ u ) = P (2 X + 1 ≤ u ) = P parenleftbigg X ≤ u- 1 2 parenrightbigg = 3 parenleftbigg u- 1 2 parenrightbigg 2- 2 parenleftbigg u- 1 2 parenrightbigg 3 for 1 < u < 3 . f U ( u ) = 3 parenleftbigg u- 1 2 parenrightbigg- 3 parenleftbigg u- 1 2 parenrightbigg 2 = 3 4 ( u- 1)(3- u ) for 1 < u < 3 . Functions of Random Variables – p. 5/31 The Distribution Function Method Example 7.2 : Suppose that a continuous random variable X has probability density function f ( x ) . What is the probability density function of U = aX + b where a > and b are constants?...
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LectureNotes13 - Functions of Random Variables Cyr Emile...

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