Noncentral chi-square distribution

Probability distribution
name =Noncentral chi-square
type =density
pdf_

cdf_

parameters =$k\; >\; 0,$ degrees of freedom
$lambda\; >\; 0,$ non-centrality parameter

support =$x\; in\; [0;\; +infty),$
pdf =$frac\{1\}\{2\}e^\{-(x+lambda)/2\}left\; (frac\{x\}\{lambda\}\; ight)^\{k/4-1/2\}\; I\_\{k/2-1\}(sqrt\{lambda\; x\})$
cdf =:$sum\_\{j=0\}^infty\; e^\{-lambda/2\}\; frac\{(lambda/2)^j\}\{j!\}\; frac\{gamma(j+k/2,x/2)\}\{Gamma(j+k/2)\},$| mean =$k+lambda,$
median =
mode =
variance =$2(k+2lambda),$
skewness =$frac\{2^\{3/2\}(k+3lambda)\}\{(k+2lambda)^\{3/2$
kurtosis =$frac\{12(k+4lambda)\}\{(k+2lambda)^2\}$
entropy =
mgf =$frac\{expleft(frac\{\; lambda\; t\}\{1-2t\; \}\; ight)\}\{(1-2\; t)^\{k/2$ for
char =$frac\{expleft(frac\{ilambda\; t\}\{1-2it\}\; ight)\}\{(1-2it)^\{k/2$

In probability theory and statistics, the noncentral chi-square or noncentral $chi^2$ distribution is a generalization of the chi-square distribution. If $X\_i$ are "k" independent, normally distributed random variables with means $mu\_i$ and variances $sigma\_i^2$, then the random variable

:$sum\_\{i=1\}^k\; left(frac\{X\_i\}\{sigma\_i\}\; ight)^2$

is distributed according to the noncentral chi-square distribution. The noncentral chi-square distribution has two parameters: $k$ which specifies the number of degrees of freedom (i.e. the number of $X\_i$), and $lambda$ which is related to the mean of the random variables $X\_i$ by:

:$lambda=sum\_\{i=1\}^k\; left(frac\{mu\_i\}\{sigma\_i\}\; ight)^2.$

Note that some references define $lambda$ as one half of the above sum.

Properties

The probability density function is given by:$f\_X(x;\; k,lambda)\; =\; sum\_\{i=0\}^infty\; frac\{e^\{-lambda/2\}\; (lambda/2)^i\}\{i!\}\; f\_\{Y\_\{k+2i(x),$where $Y\_q$ is distributed as chi-square with $q$ degrees of freedom.

From this representation, the noncentral chi-squared distribution is seen to be a Poisson-weighted mixture of central chi-squared distributions. Suppose that a random variable "J" has a Poisson distribution with mean $lambda/2$, and the conditional distribution of "Z" given $J=j$ is chi-squared with "k+2i" degrees of freedom. Then the unconditional distribution of "Z" is non-central chi-squared with "k" degrees of freedom, and non-centrality parameter $lambda$.

Alternatively, the pdf can be written as:$f\_X(x;k,lambda)=frac\{1\}\{2\}\; e^\{-(x+lambda)/2\}\; left\; (frac\{x\}\{lambda\}\; ight)^\{k/4-1/2\}\; I\_\{k/2-1\}(sqrt\{lambda\; x\})$

where $I\_\; u(z)$ is a modified Bessel function of the first kind given by

:$I\_a(y)\; :=\; (y/2)^a\; sum\_\{j=0\}^infty\; frac\{\; (y^2/4)^j\}\{j!\; Gamma(a+j+1)\}$

The moment generating function is given by

:$M(t;k,lambda)=frac\{expleft(frac\{\; lambda\; t\}\{1-2t\; \}\; ight)\}\{(1-2\; t)^\{k/2.$

The first few raw moments are:

:$mu^\text{'}\_1=k+lambda$:$mu^\text{'}\_2=(k+lambda)^2\; +\; 2(k\; +\; 2lambda)$:$mu^\text{'}\_3=(k+lambda)^3\; +\; 6(k+lambda)(k+2lambda)+8(k+3lambda)$:$mu^\text{'}\_4=(k+lambda)^4+12(k+lambda)^2(k+2lambda)+4(11k^2+44klambda+36lambda^2)+48(k+4lambda)$

The first few central moments are:

:$mu\_2=2(k+2lambda),$:$mu\_3=8(k+3lambda),$:$mu\_4=12(k+2lambda)^2+48(k+4lambda),$

The "n"th cumulant is

:$K\_n=2^\{n-1\}(n-1)!(k+nlambda).,$

Hence:$mu^\text{'}\_n\; =\; 2^\{n-1\}(n-1)!(k+nlambda)+sum\_\{j=1\}^\{n-1\}\; frac\{(n-1)!2^\{j-1\{(n-j)!\}(k+jlambda\; )mu^\text{'}\_\{n-j\}.$

Again using the relation between the central and noncentral chi-square distributions, the cumulative distribution function (cdf) can be written as

:$P(x;\; k,\; lambda\; )\; =\; sum\_\{j=0\}^infty\; e^\{-lambda/2\}\; frac\{(lambda/2)^j\}\{j!\}\; Q(x;\; k+2j)$

where $Q(x;\; k)$ is the cumulative distribution function of the central chi-squared distribution which is given by

:$Q(x;k)=frac\{gamma(k/2,x/2)\}\{Gamma(k/2)\},$

where $gamma(k,z)$ is the lower incomplete Gamma function.

Derivation of the pdf

The derivation of the probability density function is most easily done by performing the following steps:

# First, assume without loss of generality that $sigma\_1=ldots=sigma\_k=1$. Then the joint distribution of $X\_1,ldots,X\_k$ is spherically symmetric, up to a location shift.
# The spherical symmetry then implies that the distribution of $X=X\_1^2+ldots+X\_k^2$ depends on the means only through the squared length, $lambda=mu\_1^2+ldots+mu\_k^2$. Without loss of generality, we can therefore take $mu\_1=sqrt\{lambda\}$ and $mu\_2=dots=mu\_k=0$.
# Now derive the density of $X=X\_1^2$ (i.e. "k=1" case). Simple transformation of random variables shows that :
where $phi(cdot)$ is the standard normal density.
# Expand the cosh term in a Taylor series. This gives the Poisson-weighted mixture representation of the density, still for "k=1". The indices on the chi-squared random variables in the series above are "1+2i" in this case.
# Finally, for the general case. We've assumed, wlog, that $X\_2,ldots,X\_k$ are standard normal, and so $X\_2^2+ldots+X\_k^2$ has a "central" chi-squared distribution with "(k-1)" degrees of freedom, independent of $X\_1^2$. Using the poisson-weighted mixture representation for $X\_1^2$, and the fact that the sum of chi-squared random variables is also chi-squared, completes the result. The indices in the series are "(1+2i)+(k-1) = k+2i" as required.

Related distributions

*If $Z$ is chi-square distributed $Z\; sim\; chi\_k^2$ then $Z$ is also non-central chi-square distributed: $Z\; sim\; \{chi\text{'}\}^2\_k(0)$

*If $J\; sim\; Poisson(lambda/2)$, then $\{chi\text{'}\}\_k^2(lambda)\; sim\; chi\_\{k+2J\}^2$

Software

Many statistical software packages and libraries include functions for computing noncentral chisquare densities and probabilities. The table below gives commands for the following example problems:
# Density $f\_X(x;k,lambda)$ with "x=5.0", "k=3", $lambda=1.5$ (Answer: 0.097257)
# Cumulative probability $P(x;k,lambda)$ with "x=5.0", "k=3", $lambda=1.5$ (Answer: 0.649285)
# Quantile: Find "x" in $P(x;k,lambda)=a$ with "k=3", $lambda=1.5$, "a=0.5" (Answer: 3.668745)
# Random numbers: Generate 100 random observations from the distribution with "k=3", $lambda=0.5$

Note: Any software that produces the answers 0.101384, 0.490071, 5.09848 for the first three problems is including a factor of 0.5 in the definition of the noncentrality parameter. This is standard in statistics texts (e.g. [R. Christensen, "Plane Answers to Complex Questions" (3rd edition, 2002), Springer, NY, p.424.] ), but apparently not among programmers who don't read before writing their code.

References

* Abramowitz, M. and Stegun, I.A. (1972), "Handbook of Mathematical Functions", Dover. Section 26.4.25.
* Johnson, N. L. and Kotz, S., (1970), "Continuous Univariate Distributions", vol. 2, Houghton-Mifflin.