h(x) is a decreasing function of x, a similar argument holds. in .NET

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h(x) is a decreasing function of x, a similar argument holds.
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EXAMPLE S5-3 Let X be a continuous random variable with probability distribution fX 1x2 x , 8 0 x 4
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Find the probability distribution of Y h(X ) 2X 4. Note that y h(x) 2x 4 is an increasing function of x. The inverse solution is x u( y) (y 4) 2, and from this we nd the Jacobian to be J u 1 y2 dx dy 1 2. Therefore, from S5-3 the probability distribution of Y is fY 1 y2 1y 42 2 1 a b 2 y 32 4 , 4 y 12
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We now consider the case where X1 and X2 are continuous random variables and we wish to nd the joint probability distribution of Y1 h1(X1, X2) and Y2 h2(X1, X2) where the transformation is one to one. The application of this will typically be in nding the probability distribution of Y1 h1(X1, X2), analogous to the discrete case discussed above. We will need the following result.
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Suppose that X1 and X2 are continuous random variables with joint probability distribution fX1 X2 1x1, x2 2, and let Y1 h1(X1, X2) and Y2 h2(X1, X2) de ne a one-to-one transformation between the points (x1, x2) and (y1, y2). Let the equations y1 h1(x1, x2) and y2 h2(x1, x2) be uniquely solved for x1 and x2 in terms of y1 and y2 as x1 u1(y1, y2) and x2 u2(y1, y2). Then the joint probability of Y1 and Y2 is fY1Y2 1 y1, y2 2 fX1X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 0 J 0 (S5-4)
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where J is the Jacobian and is given by the following determinant: J ` x1 y1, x1 y2 ` x2 y1, x2 y2
and the absolute value of the determinant is used.
This result can be used to nd fY1Y2 1 y1, y2 2, the joint probability distribution of Y1 and Y2. Then the probability distribution of Y1 is fY1 1 y1 2 fY1Y2 1 y1, y2 2 dy2
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That is, fY1 ( y1) is the marginal probability distribution of Y1. EXAMPLE S5-4 Suppose that X1 and X2 are independent exponential random variables with fX1 1x1 2 and fX2 1x2 2 2e 2x2. Find the probability distribution of Y X1 X2 . 2e
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The joint probability distribution of X1 and X2 is fX1X2 1x1, x2 2 4e
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because X1 and X2 are independent. Let Y1 h1(X1, X2) X1 X2. The inverse solutions of y1 x1 x2 and y2 x1 x2 y2 11 y1 2 , and it follows that x1 y1 x2 y1 Therefore y2 11 y1 2 2 y2 11 y1 2 2 11 11 y1 1 y1 2 y1 2 y2 c y2 c 11 11 1 y1 2 2 d, x1 y2 x2 y2 c c 11 11
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X1 X2 and Y2 h2(X1, X2) x2 are x1 y1 y2 11 y1 2 and y1 1 d d
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and from Equation S5-4 the joint probability distribution of Y1 and Y2 is fY1Y2 1 y1, y2 2 fX1X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 0 J 0 y2 4e 23 y1 y2 11 y12 y2 11 y124 ` ` 11 y1 2 2 4e
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for y1 0, y2 0. We need to nd the distribution of Y1 ability distribution of Y1, or fY1 1 y1 2 fY1Y2 1y1, y2 2 dy2 4e
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X1 X2 . This is the marginal prob-
3y2 11 y1
y1 2 2 4 dy2 0
y1 2 2
An important application of Equation S5-4 is to obtain the distribution of the sum of two independent random variables X1 and X2. Let Y1 X1 X2 and let Y2 X2. The inverse solutions are x1 y1 y2 and x2 y2. Therefore, x1 y1 x2 y1 1 0 x1 y2 x2 y2 1 1
and J 1. From Equation S5-4, the joint probability density function of Y1 and Y2 is fY1Y2 1 y1, y2 2 f X1 1 y1 y2 2 f X2 1 y2 2
Therefore, the marginal probability density function of Y1 is fY1 1 y1 2 fX1 1 y1 y2 2 fX2 1 y2 2 dy2
The notation is simpler if the variable of integration y2 is replaced with x and y1 is replaced with y. Then the following result is obtained.