The following are proofs of several characteristics related to the chi-squared distribution.
Derivations of the pdf
Derivation of the pdf for one degree of freedom Let random variable Y be defined as Y = X2 where X has normal distribution with mean 0 and variance 1 (that is X ~ N(0,1)). Then,
for y < 0 , F Y ( y ) = P ( Y < y ) = 0 and for y ≥ 0 , F Y ( y ) = P ( Y < y ) = P ( X 2 < y ) = P ( | X | < y ) = P ( − y < X < y ) = F X ( y ) − F X ( − y ) = F X ( y ) − ( 1 − F X ( y ) ) = 2 F X ( y ) − 1 {\displaystyle {\begin{alignedat}{2}{\text{for}}~y<0,&~~F_{Y}(y)=P(Y<y)=0~~{\text{and}}\\{\text{for}}~y\geq 0,&~~F_{Y}(y)=P(Y<y)=P(X^{2}<y)=P(|X|<{\sqrt {y}})=P(-{\sqrt {y}}<X<{\sqrt {y}})\\~~&=F_{X}({\sqrt {y}})-F_{X}(-{\sqrt {y}})=F_{X}({\sqrt {y}})-(1-F_{X}({\sqrt {y}}))=2F_{X}({\sqrt {y}})-1\end{alignedat}}}
f Y ( y ) = d d y F Y ( y ) = 2 d d y F X ( y ) − 0 = 2 d d y ( ∫ − ∞ y 1 2 π e − t 2 2 d t ) = 2 1 2 π e − y 2 ( y ) y ′ = 2 1 2 π e − y 2 ( 1 2 y − 1 2 ) = 1 2 1 2 Γ ( 1 2 ) y − 1 2 e − y 2 {\displaystyle {\begin{aligned}f_{Y}(y)&={\tfrac {d}{dy}}F_{Y}(y)=2{\tfrac {d}{dy}}F_{X}({\sqrt {y}})-0=2{\frac {d}{dy}}\left(\int _{-\infty }^{\sqrt {y}}{\frac {1}{\sqrt {2\pi }}}e^{\frac {-t^{2}}{2}}dt\right)\\&=2{\frac {1}{\sqrt {2\pi }}}e^{-{\frac {y}{2}}}({\sqrt {y}})'_{y}=2{\frac {1}{{\sqrt {2}}{\sqrt {\pi }}}}e^{-{\frac {y}{2}}}\left({\frac {1}{2}}y^{-{\frac {1}{2}}}\right)\\&={\frac {1}{2^{\frac {1}{2}}\Gamma ({\frac {1}{2}})}}y^{-{\frac {1}{2}}}e^{-{\frac {y}{2}}}\end{aligned}}}
Where F {\displaystyle F} and f {\displaystyle f} are the cdf and pdf of the corresponding random variables. Then Y = X 2 ∼ χ 1 2 . {\displaystyle Y=X^{2}\sim \chi _{1}^{2}.}
Alternative proof directly using the change of variable formula The change of variable formula (implicitly derived above), for a monotonic transformation y = g ( x ) {\displaystyle y=g(x)} , is:
f Y ( y ) = ∑ i f X ( g i − 1 ( y ) ) | d g i − 1 ( y ) d y | . {\displaystyle f_{Y}(y)=\sum _{i}f_{X}(g_{i}^{-1}(y))\left|{\frac {dg_{i}^{-1}(y)}{dy}}\right|.}
In this case the change is not monotonic, because every value of Y {\displaystyle \scriptstyle Y} has two corresponding values of X {\displaystyle \scriptstyle X} (one positive and negative). However, because of symmetry, both halves will transform identically, i.e.
f Y ( y ) = 2 f X ( g − 1 ( y ) ) | d g − 1 ( y ) d y | . {\displaystyle f_{Y}(y)=2f_{X}(g^{-1}(y))\left|{\frac {dg^{-1}(y)}{dy}}\right|.}
In this case, the transformation is: x = g − 1 ( y ) = y {\displaystyle x=g^{-1}(y)={\sqrt {y}}} , and its derivative is
d g − 1 ( y ) d y = 1 2 y . {\displaystyle {\frac {dg^{-1}(y)}{dy}}={\frac {1}{2{\sqrt {y}}}}.}
So here:
f Y ( y ) = 2 1 2 π e − y / 2 1 2 y = 1 2 π y e − y / 2 . {\displaystyle f_{Y}(y)=2{\frac {1}{\sqrt {2\pi }}}e^{-y/2}{\frac {1}{2{\sqrt {y}}}}={\frac {1}{\sqrt {2\pi y}}}e^{-y/2}.}
And one gets the chi-squared distribution, noting the property of the gamma function: Γ ( 1 / 2 ) = π {\displaystyle \Gamma (1/2)={\sqrt {\pi }}} .
Derivation of the pdf for two degrees of freedom There are several methods to derive chi-squared distribution with 2 degrees of freedom. Here is one based on the distribution with 1 degree of freedom. Suppose that X {\displaystyle X} and Y {\displaystyle Y} are two independent variables satisfying X ∼ χ 1 2 {\displaystyle X\sim \chi _{1}^{2}} and Y ∼ χ 1 2 {\displaystyle Y\sim \chi _{1}^{2}} , so that the probability density functions of X {\displaystyle X} and Y {\displaystyle Y} are respectively:
f X ( x ) = 1 2 1 2 Γ ( 1 2 ) x − 1 2 e − x 2 {\displaystyle f_{X}(x)={\frac {1}{2^{\frac {1}{2}}\Gamma ({\frac {1}{2}})}}x^{-{\frac {1}{2}}}e^{-{\frac {x}{2}}}}
and of course f Y ( y ) = f X ( y ) {\displaystyle f_{Y}(y)=f_{X}(y)} . Then, we can derive the joint distribution of ( X , Y ) {\displaystyle (X,Y)} :
f ( x , y ) = f X ( x ) f Y ( y ) = 1 2 π ( x y ) − 1 2 e − x + y 2 {\displaystyle f(x,y)=f_{X}(x)\,f_{Y}(y)={\frac {1}{2\pi }}(xy)^{-{\frac {1}{2}}}e^{-{\frac {x+y}{2}}}}
where Γ ( 1 2 ) 2 = π {\displaystyle \Gamma ({\tfrac {1}{2}})^{2}=\pi } . Further, let A = x y {\displaystyle A=xy} and B = x + y {\displaystyle B=x+y} , we can get that:
x = B + B 2 − 4 A 2 {\displaystyle x={\frac {B+{\sqrt {B^{2}-4A}}}{2}}}
and
y = B − B 2 − 4 A 2 {\displaystyle y={\frac {B-{\sqrt {B^{2}-4A}}}{2}}}
or, inversely
x = B − B 2 − 4 A 2 {\displaystyle x={\frac {B-{\sqrt {B^{2}-4A}}}{2}}}
and
y = B + B 2 − 4 A 2 {\displaystyle y={\frac {B+{\sqrt {B^{2}-4A}}}{2}}}
Since the two variable change policies are symmetric, we take the upper one and multiply the result by 2. The Jacobian determinant can be calculated as:
Jacobian ( x , y A , B ) = | − ( B 2 − 4 A ) − 1 2 1 + B ( B 2 − 4 A ) − 1 2 2 ( B 2 − 4 A ) − 1 2 1 − B ( B 2 − 4 A ) − 1 2 2 | = − ( B 2 − 4 A ) − 1 2 {\displaystyle \operatorname {Jacobian} \left({\frac {x,y}{A,B}}\right)={\begin{vmatrix}-(B^{2}-4A)^{-{\frac {1}{2}}}&{\frac {1+B(B^{2}-4A)^{-{\frac {1}{2}}}}{2}}\\(B^{2}-4A)^{-{\frac {1}{2}}}&{\frac {1-B(B^{2}-4A)^{-{\frac {1}{2}}}}{2}}\\\end{vmatrix}}=-(B^{2}-4A)^{-{\frac {1}{2}}}}
Now we can change f ( x , y ) {\displaystyle f(x,y)} to f ( A , B ) {\displaystyle f(A,B)} :
f ( A , B ) = 2 × 1 2 π A − 1 2 e − B 2 ( B 2 − 4 A ) − 1 2 {\displaystyle f(A,B)=2\times {\frac {1}{2\pi }}A^{-{\frac {1}{2}}}e^{-{\frac {B}{2}}}(B^{2}-4A)^{-{\frac {1}{2}}}}
where the leading constant 2 is to take both the two variable change policies into account. Finally, we integrate out A {\displaystyle A} to get the distribution of B {\displaystyle B} , i.e. x + y {\displaystyle x+y} :
f ( B ) = 2 × e − B 2 2 π ∫ 0 B 2 4 A − 1 2 ( B 2 − 4 A ) − 1 2 d A {\displaystyle f(B)=2\times {\frac {e^{-{\frac {B}{2}}}}{2\pi }}\int _{0}^{\frac {B^{2}}{4}}A^{-{\frac {1}{2}}}(B^{2}-4A)^{-{\frac {1}{2}}}dA}
Substituting A = B 2 4 sin 2 ( t ) {\displaystyle A={\frac {B^{2}}{4}}\sin ^{2}(t)} gives:
f ( B ) = 2 × e − B 2 2 π ∫ 0 π 2 d t {\displaystyle f(B)=2\times {\frac {e^{-{\frac {B}{2}}}}{2\pi }}\int _{0}^{\frac {\pi }{2}}\,dt}
So, the result is:
f ( B ) = e − B 2 2 {\displaystyle f(B)={\frac {e^{-{\frac {B}{2}}}}{2}}}
Derivation of the pdf for k degrees of freedom Consider the k samples x i {\displaystyle x_{i}} to represent a single point in a k-dimensional space. The chi square distribution for k degrees of freedom will then be given by:
P ( Q ) d Q = ∫ V ∏ i = 1 k ( N ( x i ) d x i ) = ∫ V e − ( x 1 2 + x 2 2 + ⋯ + x k 2 ) / 2 ( 2 π ) k / 2 d x 1 d x 2 ⋯ d x k {\displaystyle P(Q)\,dQ=\int _{\mathcal {V}}\prod _{i=1}^{k}(N(x_{i})\,dx_{i})=\int _{\mathcal {V}}{\frac {e^{-(x_{1}^{2}+x_{2}^{2}+\cdots +x_{k}^{2})/2}}{(2\pi )^{k/2}}}\,dx_{1}\,dx_{2}\cdots dx_{k}}
where N ( x ) {\displaystyle N(x)} is the standard normal distribution and V {\displaystyle {\mathcal {V}}} is that elemental shell volume at Q(x), which is proportional to the (k − 1)-dimensional surface in k-space for which
Q = ∑ i = 1 k x i 2 {\displaystyle Q=\sum _{i=1}^{k}x_{i}^{2}}
It can be seen that this surface is the surface of a k-dimensional ball or, alternatively, an n-sphere where n = k - 1 with radius R = Q {\displaystyle R={\sqrt {Q}}} , and that the term in the exponent is simply expressed in terms of Q. Since it is a constant, it may be removed from inside the integral.
P ( Q ) d Q = e − Q / 2 ( 2 π ) k / 2 ∫ V d x 1 d x 2 ⋯ d x k {\displaystyle P(Q)\,dQ={\frac {e^{-Q/2}}{(2\pi )^{k/2}}}\int _{\mathcal {V}}dx_{1}\,dx_{2}\cdots dx_{k}}
The integral is now simply the surface area A of the (k − 1)-sphere times the infinitesimal thickness of the sphere which is
d R = d Q 2 Q 1 / 2 . {\displaystyle dR={\frac {dQ}{2Q^{1/2}}}.}
The area of a (k − 1)-sphere is:
A = 2 R k − 1 π k / 2 Γ ( k / 2 ) {\displaystyle A={\frac {2R^{k-1}\pi ^{k/2}}{\Gamma (k/2)}}}
Substituting, realizing that Γ ( z + 1 ) = z Γ ( z ) {\displaystyle \Gamma (z+1)=z\Gamma (z)} , and cancelling terms yields:
P ( Q ) d Q = e − Q / 2 ( 2 π ) k / 2 A d R = 1 2 k / 2 Γ ( k / 2 ) Q k / 2 − 1 e − Q / 2 d Q {\displaystyle P(Q)\,dQ={\frac {e^{-Q/2}}{(2\pi )^{k/2}}}A\,dR={\frac {1}{2^{k/2}\Gamma (k/2)}}Q^{k/2-1}e^{-Q/2}\,dQ}