Convolution of X1 and X2

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If X1 and X2 are independent random variables with probability density functions fX1(x1) and fX2 (x2), respectively, the probability density function of Y X1 X2 is fY 1 y2 fX1 1 y x2 f X2 1x2 dx (S5-5)

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The probability density function of Y in Equation S5-5 is referred to as the convolution of the probability density functions for X1 and X2. This concept is commonly used for transformations (such as Fourier transformations) in mathematics. This integral may be evaluated numerically to obtain the probability density function of Y, even for complex probability density functions for X1 and X2. A similar result can be obtained for discrete random variables with the integral replaced with a sum. In some problems involving transformations, we need to nd the probability distribution of the random variable Y h(X) when X is a continuous random variable, but the transformation is not one to one. The following result is helpful.

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Suppose that X is a continuous random variable with probability distribution fX (x), and Y h(X) is a transformation that is not one to one. If the interval over which X is de ned can be partitioned into m mutually exclusive disjoint sets such that each of the inverse functions x1 u1( y), x2 u2( y), p , xm um( y) of y u(x) is one to one, the probability distribution of Y is fY 1 y2 where Ji u 1 y2 , i i a fX 3ui 1 y2 4 0 Ji 0

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(S5-6)

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1, 2, p , m and the absolute values are used.

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To illustrate how this equation is used, suppose that X is a normal random variable with mean and variance 2, and we wish to show that the distribution of Y 1X 2 2 2 is a

chi-squared distribution with one degree of freedom. Let Z probability distribution of Z is the standard normal; that is, f 1z2 1 e 12

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The inverse solutions of y z2 are z 1y, so the transformation is not one to one. De ne z1 1y and z2 1y so that J1 11 22 1y and J2 11 22 1y. Then by Equation S5-6, the probability distribution of Y is , z fY 1 y2 1 1 1 1 e y2` ` e y2` ` 12 21y 12 21y 1 y1 2 1 e y 2, y 0 12 2 1 11 22 , so we may write f( y) as 1 2

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Now it can be shown that 1 fY 1 y2

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which is the chi-squared distribution with 1 degree of freedom. EXERCISES FOR SECTION 5-8

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S5-1. Suppose that X is a random variable with probability distribution fX 1x2 1 4, x 1, 2, 3, 4 S5-5. A current of I amperes ows through a resistance of R ohms according to the probability distribution fI 1i2 2i, 0 i 1

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Find the probability distribution of the random Y 2X 1. S5-2. Let X be a binomial random variable with p 0.25 and n 3. Find the probability distribution of the random variable Y X 2. S5-3. Suppose that X is a continuous random variable with probability distribution fX 1x2 x , 18 0 x 6

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Suppose that the resistance is also a random variable with probability distribution fR 1r2 1, 0 r 1

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(a) Find the probability distribution of the random variable Y 2X 10. (b) Find the expected value of Y. S5-4. Suppose that X has a uniform probability distribution fX 1x2 1, 0 x 1

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Assume that I and R are independent. (a) Find the probability distribution for the power (in watts) P I 2R. (b) Find E(P). S5-6. A random variable X has the following probability distribution: fX 1x2 e x, x 0 X 2. X1 2. ln X.

(a) Find the probability distribution for Y (b) Find the probability distribution for Y (c) Find the probability distribution for Y

Show that the probability distribution of the random variable Y 2 ln X is chi-squared with two degrees of freedom.

S5-7. The velocity of a particle in a gas is a random variable V with probability distribution fV 1v2 av2e