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Step 2: Showing\({{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\)

(a):

The \({\rm{cdf}}\) of \({\rm{Z = }}{{\rm{X}}_{\rm{1}}}{\rm{ + }}{{\rm{X}}_{\rm{2}}}\) is

\({{\rm{F}}_{\rm{Z}}}{\rm{(z) = }}\int_{\rm{0}}^{\rm{z}} {\left( {\int_{\rm{0}}^{{\rm{z - }}{{\rm{x}}_{\rm{1}}}} {\frac{{\rm{1}}}{{{{\rm{2}}^{{{\rm{v}}_{\rm{1}}}{\rm{/2}}}}{\rm{\Gamma }}\left( {{{\rm{v}}_{\rm{1}}}{\rm{/2}}} \right)}}} \frac{{\rm{1}}}{{{{\rm{2}}^{{{\rm{v}}_{\rm{2}}}{\rm{/2}}}}{\rm{\Gamma }}\left( {{{\rm{v}}_{\rm{2}}}{\rm{/2}}} \right)}}{\rm{ \times x}}_{\rm{1}}^{{{\rm{v}}_{\rm{1}}}{\rm{/2 - 1}}}{\rm{ \times x}}_{\rm{2}}^{{{\rm{v}}_{\rm{2}}}{\rm{/2 - 1}}}{{\rm{e}}^{{\rm{ - }}\left( {{{\rm{x}}_{\rm{1}}}{\rm{ + }}{{\rm{x}}_{\rm{2}}}} \right){\rm{/2}}}}{\rm{d}}{{\rm{x}}_{\rm{2}}}} \right)} {\rm{d}}{{\rm{x}}_{\rm{1}}}\)

Although the integral is difficult, it can be demonstrated that dividing it yields, which is really a pdf of chi squared random variable with degrees of freedom \({{\rm{v}}_{\rm{1}}}{\rm{ + }}{{\rm{v}}_{\rm{2}}}\).

(b):

The sum of two chi-squared random variables is chi-squared random variable when \({\rm{(a)}}\) is used. Because \({{\rm{Z}}^{\rm{2}}}\)has a chi-squared distribution with one degree of freedom, the total will have a chi-squared distribution with two degrees of freedom \({\rm{2}}{\rm{.}}\).

Similarly, for three random variables

\({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}{\rm{ + Z}}_{\rm{3}}^{\rm{2}}{\rm{ = }}\left( {{\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}} \right){\rm{ + Z}}_{\rm{3}}^{\rm{2}}\)

has a chi-squared distribution with degrees of freedom 3 , because is a chi-squared random variable with degrees of freedom 2, and is a chi-squared random variable with degrees of freedom 1, which indicated that the sum is 3 .

Because \({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}\)is a chi-squared random variable with degrees of freedom\({\rm{2}}\), and \({\rm{Z}}_{\rm{3}}^{\rm{2}}\) is a chi-squared random variable with degrees of freedom \({\rm{3}}\), it has a chi-squared distribution with degrees of freedom\({\rm{3}}\). The total is \({\rm{3}}\) since is a chi-squared random variable with one degree of freedom.

By continuing this process, it is follows that

\({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}{\rm{ + \ldots + Z}}_{{\rm{n - 1}}}^{\rm{2}}{\rm{ + }}{{\rm{Z}}_{\rm{n}}}{\rm{ = }}\left( {{\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}{\rm{ + \ldots + Z}}_{{\rm{n - 1}}}^{\rm{2}}} \right){\rm{ + }}{{\rm{Z}}_{\rm{n}}}\)

Because \({\rm{Z}}_{\rm{1}}^{\rm{2}}{\rm{ + Z}}_{\rm{2}}^{\rm{2}}{\rm{ + \ldots + Z}}_{{\rm{n - 1}}}^{\rm{2}}\) (induction) is a chi-squared random variable with degrees of freedom \({\rm{n - 1}}\), and \({\rm{Z}}_{\rm{n}}^{\rm{2}}\)is a chi-squared random variable with degrees of freedom\({\rm{1}}\), the sum has a chi-squared distribution with degrees of freedom\({\rm{n}}\).

(c):

Here the \({\rm{(b)}}\) can be used because random variables

\({{\rm{I}}_{\rm{i}}}{\rm{ = }}\frac{{{{\rm{X}}_{\rm{i}}}{\rm{ - \mu }}}}{{\rm{\sigma }}}\)

have a standard normal distribution (this is the traditional way for deriving a standard normal distribution from a normal distribution), thus the sum

\(\sum\limits_{{\rm{i = 1}}}^{\rm{n}} {{\rm{U}}_{\rm{i}}^{\rm{2}}} \)

has a chi-squared distribution with degrees of freedom \({\rm{n}}\).