Distribution Theory of the Non-Central Chi-Square

The following theorem is of considerable help in deriving results concerning the non-central chi-square distribution.

Theorem 6..1

A random variable $W \sim \chi^2_p(\lambda)$ can be represented as the sum of a non-central chi-square variable with one degree of freedom and non-centrality parameter $\lambda$ and a (central) chi-square variable with $p-1$ degrees of freedom where the two variables are independent.

Proof

Let $X_1, X_2, \dots , X_p$ be independently distributed, where $X_i\sim N(\mu_i,1)$ and write
${\bf X}' = (X_1,X_2, \dots , X_p)$. Define

$$W = \sum^p_{i=1} X^2_i \ .$$ (6.1)

Choose an orthogonal matrix B such that the elements in the first row are defined by

$$b_{1j} =\mu_j\lambda^{-\frac 12} \ \ \mbox{for} \ \ j=1,2,\dots ,p$$ (6.2)

where $\lambda = \sum^p_{j=1} \mu_j^2$.

Define ${\bf Y}'=(Y_1,Y_2,\dots ,Y_p)$ by the orthogonal transformation

$${\bf Y} = {\bf B}{\bf X} \ .$$ (6.3)

Then using the result of Assignment 2, Q. 6, we see that ${\bf Y}\sim N_p({\bf B}\mbox{\boldmath$\mu$}, {\bf BIB}')$. That is, since B is orthogonal

$${\bf Y}\sim N_p ({\bf B}\mbox{\boldmath$\mu$}, {\bf I})$$ (6.4)

But, the mean of the vector ${\bf Y}$ can be written as

$$E({\bf Y})={\bf B}\mbox{\boldmath$\mu$} = \left[ \begin{array... ... . \\ [-0.2cm] . \\ \sum b_{pj}\mu_j \end{array}\right]$$

where, using (6.2) we have

$$E(Y_1)=\sum b_{1j}\mu_j = \sum \mu^2_j \lambda^{-\frac12}=\lambda / \lambda^{\frac12} = \lambda^{\frac12} \ .$$

Further, since the rows of B are mutually orthogonal

$$E(Y_i) = \sum^p_{j=1} b_{ij} \mu_j = 0 \ \ \mbox{for} \ \ i=2,3,\dots ,p \ .$$

From (6.3), $Y_1,Y_2, \dots , Y_p$ is a set of independent normally distributed random variables. Also

$$W = {\bf X}'{\bf X} = ({\bf B}^{-1} {\bf Y})'{\bf B}^{-1}{\bf Y} = {\bf Y}'({\bf B}^{-1})'{\bf B}^{-1} {\bf Y} = {\bf Y}' {\bf Y}$$

$$= Y^2_1 + \sum^p_{i=2} Y^2_i$$

$$= V + U$$ (6.5)

where $V=Y^2_1$ and $U=\sum^p_{i=2} Y^2_i$.

Since $U$ depends only on $Y_2,\dots ,Y_p, \ \ U$ is independent of $V$. Furthermore,
$Y_1 \sim N(\lambda^{\frac12},1)$ so that $Y^2_1$ is distributed as a chi-square with one degree of freedom and non-centrality parameter $\lambda$. Also, $U$ is distributed as a chi-square with $(p-1)$ degrees of freedom since $Y_2, \dots , Y_p$ are independently and identically distributed $N(0,1)$. This completes the proof.

This theorem will now be used to derive the density function for a random variable with a non-central $\chi^2$ distribution.

Theorem 6..2

If $W\sim \chi^2_n(\lambda )$, then the probability density function of $W$ is

$$g_W(w) = \frac{e^{-\frac12 w} e^{-\frac{\lambda}{2}} w^{\frac 12 n- 1}}{2^... ...c{1}{n(n+2)} . \frac{1}{2!}\left(\frac{w\lambda }{2}\right)^2 + \dots \right] ,$$

$$~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~ 0 \leq w < \infty \ .$$ (6.6)

Proof: Write $V=Y^2_1$ and $U=\sum^n_{i=2} Y^2_i$, so that by Theorem 6.1, we can write $W$ as $W=V+U$.

Now $U\sim \chi^2_{n-1}(0)$, so that the probability density function of $U$ is

$$f_U(u) = \frac{e^{-\frac12 u} u^{\frac{n-1}{2}-1}}{2^{\frac... ...-1)} \Gamma [\frac 12 (n-1)]} \ , \ \ 0 \leq u < \infty \ .$$ (6.7)

In Example 3.1 of Chapter 3, the density function of $V$ was found to be

$$f_V(v)=\frac{v^{-\frac 12} e^{-\frac v2}e^{-\frac{ \lambda}... ...bda )^{\frac 12}} + e^{-(v\lambda )^{\frac 12}} \right] \ .$$ (6.8)

But $U$ and $V$ are independent so that the joint p.d.f. of $U$ and $V$ is

$$\begin{eqnarray*} f_{U,V}(u,v) & = & f_U(u)f_V(v)\\ [0.2cm] & = & \frac{u^{\f... ...mbda v)^{\frac 12}} + e^{-(\lambda v)^{\frac 12}} \right] \ . \end{eqnarray*}$$

Define random variables $W$ and $T$ by

$$\left\{ \begin{array}{l} W=U+V\\ T=U \ . \end{array}\right.$$

Then

$$\left\{ \begin{array}{l} V=W-T\\ U=T \ . \end{array} \right.$$

Clearly the Jacobian of the transformation is $1$ so that

$$f_{T,W}(t,w) = \frac{e^{-\frac{w}{2}} t^{\frac{n-3}{2}} (w-t)... ... (w-t)^{\frac 12}}\right], 0\leq t\leq w, \ 0 \leq w < \infty.$$

Now write $(w-t)^{\frac12} = w\left(1-\frac tw \right)^{\frac 12}$ and expand the terms in the brackets so that

$$\begin{eqnarray*} f_{T,W}(t,w) & = & \frac{e^{-\frac w2} e^{-\frac \lambda 2} w... ...{2!} \left(1-\frac tw \right)^{\frac 12} + \dots \right\} \ . \end{eqnarray*}$$

To obtain the marginal density function of $W$ we integrate with respect to $t$, $(0 \leq t < w)$. Notice we have a series of integrals of the form

$$\begin{eqnarray*} \int^w_0 \left(\frac tw \right)^{\frac{(n-3)}{2}} \left( 1-\f... ...m] & = & \frac 1w B\left( \frac{n-1}{2}, \ r-\frac 12 \right) \end{eqnarray*}$$

for $r = 0,1,2, \dots$ .

Thus

$$g_W(w) = \frac{e^{-\frac w2} e^{-\frac \lambda 2} w^{(\frac{n... ...a }{2!} B\left(\frac{n-1}{2},\frac 32\right) + \dots \right\}$$

and using the relationship $B(m,n) = \Gamma (m) \Gamma (n)/ \Gamma (m+n)$ we obtain (6.6).

Note: No generality is lost by assuming unit variances as the more general case (where the variance is $\sigma^2$, say) can easily be reduced to this case. That is, if $X\sim N(\mu , \sigma^2)$ then $X/\sigma \sim N(\mu /\sigma , 1)$.