Stochastic

A little more calculus gives ex 2 z b a x2 b a dx b3

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Unformatted text preview: t line of the answer is easy to see. Since P (X > 0) = 1, we have P (X x) = 0 for x 0. For x 0 we compute Zx x P (X x) = e y dy = e y 0 = 1 e x 0 In many situations we need to know the relationship between several random variables X1 , . . . , Xn . If the Xi are discrete random variables then this is easy, we simply give the probability function that specifies the value of P (X1 = x1 , . . . , Xn = xn ) whenever this is positive. When the individual random variables have continuous distributions this is described by giving the joint density function which has the interpretation that Z Z P ((X1 , . . . , Xn ) 2 A) = · · · f (x1 , . . . , xn ) dx1 . . . dxn A By analogy with (A.9) we must require that f (x1 , . . . , xn ) Z Z · · · f (x1 , . . . , xn ) dx1 . . . dxn = 1 0 and Having introduced the joint distribution of n random variables, we will for simplicity restrict our attention for the rest of the section to n = 2. The first question we will confront is: “Given the joint distribution of (X, Y ), how do we recover the distri...
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This document was uploaded on 03/06/2014 for the course MATH 4740 at Cornell.

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