# Suppose t is a positive random variable with pdf ft

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Unformatted text preview: ng function mapping the interval (a, b) onto the interval (A, B ). The situation is illustrated in Figure 3.21. The support of Y is the interval [A, B ]. So let A &lt; c &lt; B. There is a value g −1 (c) on the u axis such that g (g −1 (c)) = c, and: FY (c) = P {Y ≤ c} = P {X ≤ g −1 (c)} = FX (g −1 (c)). The derivative of the inverse of a function is one over the derivative of the function itself4 , to yield 1 g −1 (c) = g (g−1 (c)) . Thus, diﬀerentiating FY yields: fY (c) = 4 1 fX (g −1 (c)) g (g−1 (c)) 0 A&lt;c&lt;B else. Prove this by diﬀerentiating both sides of the identity g (g −1 (c)) = c (3.6) 108 CHAPTER 3. CONTINUOUS-TYPE RANDOM VARIABLES B c g(u) A f (u) X u a !1 g (c) b Figure 3.21: Monotone function of a continuous-type random variable. The expression for fY in (3.6) has an appealing form. Figure 3.21 shows a small interval with one endpoint c on the vertical axis, and the inverse image of the interval on the u axis, which is a small interval with one endpoint g −1 (c). The probability Y falls into the...
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## This note was uploaded on 02/09/2014 for the course ISYE 2027 taught by Professor Zahrn during the Spring '08 term at Georgia Institute of Technology.

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