Answer to Question #151709 in Statistics and Probability for Sudhanshu Yadav

Question #151709
. If X and Y are independent chi-square variates with n1 and n2 d.f. respectively. Show
that U = X + Y and V = n2X/n1Y are independently distributed. Also indentify their
distributions
1
Expert's answer
2020-12-17T19:25:53-0500

There is an important connection between an F variable and chi-squared variables. If X and Y are independent chi-squared rv’s with n1 and n2 df, respectively, then the rv


"V=\\dfrac{X\/n_1}{Y\/n_2}=\\dfrac{n_2X}{n_1Y}"

is said to have an F distribution with (n1, n2) degrees of freedom.

The random variable "X_i" has the chi-square distribution with "n_i" degrees of freedom with probability density function


"f_X(x)=\\dfrac{1}{2^{n_i\/2}\\Gamma(n_i\/2)}x^{n_i\/2-1}e^{-x\/2}, i=1,2"

 The moment generating function for "X_i" is


"M_{X_i}(t)=(1-2t)^{-n_i\/2}, t<\\dfrac{1}{2}, i=1,2"

The moment generating function of "U" is


"E[e^{tU}]=E[e^{tX}e^{tY}]=E[e^{tX}]\\cdot E[e^{tY}]"

"=(1-2t)^{-n_1\/2}(1-2t)^{-n_2\/2}"

"=(1-2t)^{-(n_1+n_2)\/2}, t<\\dfrac{1}{2}"

which is the moment generating function of a chi-square random variable with "n_1+n_2" degrees of freedom.

Since "X" and "Y" follow independently chi-square distribution with "n_1,n_2," we have that "U" has a chi-square distribution with "n_1+n_2" degrees of freedom.


"U" has a chi-square distribution with "n_1+n_2" degrees of freedom.

"V" has an F distribution with (n1, n2) degrees of freedom.

"U" and "V" are independent.



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