We will also state this result as follows. in .NET

Implementation QR Code 2d barcode in .NET We will also state this result as follows.
We will also state this result as follows.
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Suppose that X1 and X2 are discrete random variables with joint probability distribution fX1 X2 1x1, x2 2, and let Y1 h1(X1, X2) and Y2 h2(X1, X2) de ne one-to-one transformations between the points (x1, x2) and (y1, y2) so that the equations y1 h1(x1, x2) and y2 h2(x1, x2) can be solved uniquely for x1 and x2 in terms of y1 and y2. Let this solution be x1 u1(y1, y2) and x2 u2(y1, y2). Then the joint probability distribution of Y1 and Y2 is fY1Y2 1y1, y2 2 fX1 X2 3u1 1 y1, y2 2, u2 1 y1, y2 2 4 (S5-2)
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A very important application of Equation S5-2 is in nding the distribution of a random variable Y1 that is a function of two other random variables X1 and X2. That is, let Y1 h1(X1, X2) where X1 and X2 are discrete random variables with joint distribution fX1 X2 1x1, x2 2. We want to find the probability distribution of Y1, say, fY1 1 y1 2. To do this, we de ne a second function Y2 h2(X1, X2) so that the one-to-one correspondence between the points (x1, x2) and (y1, y2) is maintained, and we use the result in Equation S5-2 to nd the joint probability distribution of Y1 and Y2. Then the distribution of Y1 alone is found by summing over the y2 values in this joint distribution. That is, fY1 1 y1 2 is just the marginal probability distribution of Y1, or fY1 1 y1 2 EXAMPLE S5-2 a fY1Y2 1 y1, y2 2
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Consider the case where X1 and X2 are independent Poisson random variables with parameters X1 X2. 1 and 2, respectively. We will nd the distribution of the random variable Y1 The joint distribution of X1 and X2 is fX1X2 1x1, x2 2 fX1 1x1 2 fX2 1x2 2 e e
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because X1 and X2 are independent. Now to use Equation S5-2 we need to de ne a second function Y2 h2(X1, X2). Let this function be Y2 X2. Now the solutions for x1 and x2 are x1 y1 y2 and x2 y2. Thus, from Equation S5-2 the joint probability distribution of Y1 and Y2 is fY1Y2 1 y1, y2 2 e 1 y1
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Because x1 0, the transformation x1 y1 y2 requires that x2 y2 must always be less than or equal to y1. Thus, the values of y2 are 0, 1, . . . , y1, and the marginal probability distribution of Y1 is obtained as follows: fY1 1 y1 2 a fY1Y2 1 y1, y2 2
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The summation in this last expression is the binomial expansion of 1 fY1 1 y1 2 e
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y1! a 1y y2 0 1
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We recognize this as a Poisson distribution with parameter 1 2. Therefore, we have shown that the sum of two independent Poisson random variables with parameters 1 and 2 has a Poisson distribution with parameter 1 2. We now consider the situation where the random variables are continuous. Let Y with X continuous and the transformation is one to one. h(X),
Suppose that X is a continuous random variable with probability distribution fX (x). The function Y h(X) is a one-to-one transformation between the values of Y and X so that the equation y h(x) can be uniquely solved for x in terms of y. Let this solution be x u(y). The probability distribution of Y is fY 1 y2 fX 3u1 y2 4 0 J 0 (S5-3)
where J u ( y) is called the Jacobian of the transformation and the absolute value of J is used.
Equation S5-3 is shown as follows. Let the function y Now P1Y a2 P3X
u1a2
h(x) be an increasing function of x. u1a2 4
fX 1x2 dx If we change the variable of integration from x to y by using x and then
u( y), we obtain dx
u ( y) dy
fX 3u1 y2 4u 1 y2 dy
Since the integral gives the probability that Y a for all values of a contained in the feasible set of values for y, fX 3u1 y2 4u 1 y2 must be the probability density of Y. Therefore, the probability distribution of Y is fY 1 y2 If the function y fX 3u1 y2 4u 1 y2 fX 3u1 y2 4J