chisquare - Notes on the Chi-Squared Distribution October...

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Notes on the Chi-Squared Distribution October 19, 2005 1 Introduction Recall the defnition oF the chi-squared random variable with k degrees oF Freedom is given as χ 2 = X 2 1 + ··· + X 2 k , where the X i ’s are all independent and have N (0 , 1) distributions. Also recall that I claimed that χ 2 has a gamma distribution with parameters r = k/ 2 and α = 1 / 2: Let f ( x ) = ± 1 2 ² k/ 2 x k/ 2 - 1 e - αx Γ( k/ 2) , where Γ( t ) is the gamma Function, given by Γ( t ) = Z 0 x t - 1 e - x dx, For t > 0 . Then, we are saying that P ( χ 2 a ) = Z a ± 1 2 ² k/ 2 x k/ 2 - 1 e - αx Γ( k/ 2) dx. In this set oF notes we aim to do the Following two things: 1) Show that the chi-squared distribution with k degrees oF Freedom does indeed have a gamma distribution; 2) Discuss the chi-squared test, which is similar to the one you have already seen in class. 1
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2 Proof that chi-squared has a gamma distri- bution We frst recall here a standard Fact about moment generating Functions: Theorem 1. Suppose that X and Y are continuous random variables having moment generating Functions M X ( t ) = E ( e tX ) and M Y ( t ) = E ( e tY ), respec- tively. ±urther, suppose that these Functions exist For all t in a neighborhood oF 0, and that they are continuous at t = 0. Then, P ( X a ) = P ( Y a ) For all a ⇐⇒ M X ( t ) = M Y ( t ) . Note: Stronger version oF this theorem are possible, but this is good enough For our purposes. We will not prove this theorem here, as it is long and technical. However, let us now use it to determine the moment generating Function For the chi- square distribution; frst, we determine the m.g.F. For a gamma distribution: Suppose Z is a random variable having a gamma distribution with parameters r > 0 and α > 0. Then, it has pdF given by g ( x ) = α ( αx ) r - 1 e - αx Γ( r ) , where x 0 . Since it is a pdF, we know that Z 0 g ( x ) dx = 1 , regardless oF the parameters α and r . Now, then, we have that M Z ( t ) is given by M Z ( t ) = E ( e tZ ) = Z 0 e tz g ( z ) dz = Z 0 α r z r - 1 e - z ( α - t ) Γ( r ) dz = α r ( α - t ) r Z 0 ( α - t )(( α - t ) z ) r - 1 e - z ( α - t ) Γ( r ) dz 2
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= α r ( α - t ) r 1 = 1 (1 - α - 1 t ) r . Notice here that we used the fact that the integral
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This note was uploaded on 10/23/2011 for the course MATH 3225 taught by Professor Staff during the Spring '08 term at Georgia Institute of Technology.

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chisquare - Notes on the Chi-Squared Distribution October...

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