Harman Patil (Editor)

Proofs related to chi squared distribution

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The following are proofs of several characteristics related to the chi-squared distribution.

Contents

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 ,     P ( Y < y ) = 0     and for   y ≥ 0 ,     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

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

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

Then Y = X 2 ∼ χ 1 2 .

Alternative proof using directly the change of variable formula

The change of variable formula (implicitly derived above), for a monotonic transformation y = g ( x ) , is:

f Y ( y ) = ∑ i f X ( g i − 1 ( y ) ) | d g i − 1 ( y ) d y | .

In this case the change is not monotonic, because every value of Y has two corresponding values of 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 | .

In this case, the transformation is: x = g − 1 ( y ) = y , and its derivative is d g − 1 ( y ) d y = 1 2 y .

So here:

f Y ( y ) = 2 1 2 π e − y / 2 1 2 y = 1 2 π y e − y / 2 .

And one gets the chi-squared distribution, noting the property of the gamma function: Γ ( 1 2 ) = ( π )

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 and y are two independent variables satisfying x ∼ χ 1 2 and y ∼ χ 1 2 , so that the probability density functions of x and y are respectively:

f ( x ) = 1 2 1 2 Γ ( 1 2 ) x − 1 2 e − x 2

and

f ( y ) = 1 2 1 2 Γ ( 1 2 ) y − 1 2 e − y 2

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

f ( x , y ) = 1 2 π ( x y ) − 1 2 e − x + y 2

where Γ ( 1 2 ) 2 is replaced by π . Further, let A = x y and B = x + y , we can get that:

x = B + B 2 − 4 A 2

and

y = B − B 2 − 4 A 2

or, inversely

x = B − B 2 − 4 A 2

and

y = B + B 2 − 4 A 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

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

f ( A , B ) = 2 × 1 2 π A − 1 2 e − B 2 ( B 2 − 4 A ) − 1 2

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 × e − B 2 2 π ∫ 0 B 2 4 A − 1 2 ( B 2 − 4 A ) − 1 2 d A

Let A = B 2 4 sin 2 ⁡ ( t ) , the equation can be changed to:

f ( B ) = 2 × e − B 2 2 π ∫ 0 π 2 d t

So the result is:

f ( B ) = e − B 2 2

Derivation of the pdf for k degrees of freedom

Consider the k samples 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

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 = 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 , 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

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 .

The area of a (k − 1)-sphere is:

A = k R k − 1 π k / 2 Γ ( k / 2 + 1 )

Substituting, realizing that Γ ( z + 1 ) = z Γ ( 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

References

Proofs related to chi-squared distribution Wikipedia
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