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LectureNotes13 - Functions of Random Variables Cyr Emile...

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Functions of Random Variables Cyr Emile M’LAN, Ph.D. [email protected] Functions of Random Variables – p. 1/31
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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
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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
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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
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The Distribution Function Method Example 7.1 : Suppose that a random variable Y has probability density function f ( x ) = 6 x (1 - x ) , 0 < x < 1 . What is the probability density function of U = 2 X + 1 ? Solution : F X ( y ) = integraldisplay x 0 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
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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 > 0 and b are constants? Solution : F U ( u ) = P ( U u ) = P ( aX + b u ) = P parenleftbigg X u - b a parenrightbigg = F X parenleftbigg u - b a parenrightbigg f U ( u ) = F prime U ( u ) = 1 a f X parenleftbigg u - b a parenrightbigg Remember to calculate the two endpoints where f U ( u ) integrate to 1.
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