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=X/n1Y/n2=n2Xn1YV=\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 XiX_i has the chi-square distribution with nin_i degrees of freedom with probability density function


fX(x)=12ni/2Γ(ni/2)xni/21ex/2,i=1,2f_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 XiX_i is


MXi(t)=(12t)ni/2,t<12,i=1,2M_{X_i}(t)=(1-2t)^{-n_i/2}, t<\dfrac{1}{2}, i=1,2

The moment generating function of UU is


E[etU]=E[etXetY]=E[etX]E[etY]E[e^{tU}]=E[e^{tX}e^{tY}]=E[e^{tX}]\cdot E[e^{tY}]

=(12t)n1/2(12t)n2/2=(1-2t)^{-n_1/2}(1-2t)^{-n_2/2}

=(12t)(n1+n2)/2,t<12=(1-2t)^{-(n_1+n_2)/2}, t<\dfrac{1}{2}

which is the moment generating function of a chi-square random variable with n1+n2n_1+n_2 degrees of freedom.

Since XX and YY follow independently chi-square distribution with n1,n2,n_1,n_2, we have that UU has a chi-square distribution with n1+n2n_1+n_2 degrees of freedom.


UU has a chi-square distribution with n1+n2n_1+n_2 degrees of freedom.

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

UU and VV are independent.



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