Protective index

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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,
fory<0,P(Y<y)=0andfory0,P(Y<y)=P(X2<y)=P(|X|<y)==FX(y)FX(y)=FX(y)(1FX(y))=2FX(y)1

fY(y)=2ddyFX(y)0=2ddy(y12πet22dt)=212πey2(y)'y=212πey2(12y12)=1212Γ(12)y121ey2

Where F and f are the cdf and pdf of the corresponding random variables.

Then Y=X2χ12.

Derivation of the pdf for two degrees of freedom

To derive the chi-squared distribution with 2 degrees of freedom, there could be several methods. Here presented is one of them which is based on the distribution with 1 degree of freedom.

Suppose that x and y are two independent variables satisfying xχ12 and yχ12, so that the probability density functions of x and y are respectively:

f(x)=1212Γ(12)x12ex2

and

f(y)=1212Γ(12)y12ey2

Simply, we can derive the joint distribution of x and y:

f(x,y)=12π(xy)12ex+y2

where Γ(12)2 is replaced by π. Further, let A=xy and B=x+y, we can get that:

x=B+B24A2

and

y=BB24A2

or, inversely

x=BB24A2

and

y=B+B24A2

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,yA,B)=|(B24A)121+B(B24A)122(B24A)121B(B24A)122|=(B24A)12

Now we can change f(x,y) to f(A,B):

f(A,B)=2×12πA12eB2(B24A)12

where the leading constant 2 is to take both the two variable change policies into account. Finally, we integrate out A to get the distribution of B, i.e. x+y:

f(B)=2×eB22π0B24A12(B24A)12dA

Let A=B24sin2(t), the equation can be changed to:

f(B)=2×eB22π0π2dt

So the result is:

f(B)=eB22

Derivation of the pdf for k degrees of freedom

Consider the k samples xi 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)dQ=𝒱i=1k(N(xi)dxi)=𝒱e(x12+x22++xk2)/2(2π)k/2dx1dx2dxk

where N(x) is the standard normal distribution and 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=1kxi2

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, 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)dQ=eQ/2(2π)k/2𝒱dx1dx2dxk

The integral is now simply the surface area A of the (k − 1)-sphere times the infinitesimal thickness of the sphere which is

dR=dQ2Q1/2.

The area of a (k − 1)-sphere is:

A=kRk1πk/2Γ(k/2+1)

Substituting, realizing that Γ(z+1)=zΓ(z), and cancelling terms yields:

P(Q)dQ=eQ/2(2π)k/2AdR=12k/2Γ(k/2)Qk/21eQ/2dQ