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

Question

. 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

Answer 1

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.