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In probability theory and statistics, the noncentral chi-square or noncentral ?2 distribution is a generalization of the chi-square distribution. If Xi are k independent, normally distributed random variables with means µi and variances , then the random variable

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 Xi), and ? which is related to the mean of the random variables Xi by:

Note that some references define ? as one half of the above sum.
Properties
The probability density function is given by

where Yq 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 ? / 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 ?.
Alternatively, the pdf can be written as

where I?(z) is a modified Bessel function of the first kind given by

The moment generating function is given by

The first few raw moments are:




The first few central moments are:



The nth cumulant is

Hence

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

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

where ?(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
. Then the joint distribution of is spherically symmetric, up to a location shift.
- The spherical symmetry then implies that the distribution of
depends on the means only through the squared length, . Without loss of generality, we can therefore take and .
- Now derive the density of
(i.e. k=1 case). Simple transformation of random variables shows that :
where 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
are standard normal, and so has a central chi-squared distribution with (k-1) degrees of freedom, independent of . Using the poisson-weighted mixture representation for , 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 V is chi-square distributed
then V is also non-central chi-square distributed: 
- If
, then 
Software and online calculator
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 fX(x;k,?) with x=5.0, k=3, ? = 1.5
- Cumulative probability P(x;k,?) with x=5.0, k=3, ? = 1.5
- Quantile: Find x in P(x;k,?) = q with k=3, ? = 1.5, q=0.5
- Critical noncentral parameter: Find ? in P(x;k,?) = q with x=5.0,k=3, and q=0.5
- Random numbers: Generate 100 random observations from the distribution with k=3, ?=1.5
| Software |
Density |
Cumulative Prob. |
Quantile |
Noncentral parameter |
Random numbers |
| Matlab |
ncx2pdf(5.0,3,1.5) |
ncx2cdf(5.0,3,1.5) |
ncx2inv(.5,3,1.5) |
fsolve(@(L)(ncx2cdf(5.0,3,L)-.5), 1) |
ncx2rnd(3,1.5,100,1) |
| R |
dchisq(5.0,3,1.5) |
pchisq(5.0,3,1.5) |
qchisq(.5,3,1.5) |
require(MBESS);
conf.limits.nc.chisq( 5,NULL,3,0,.5)
|
rchisq(100,3,1.5) |
| SAS |
PROBCHI(5.0,3,1.5) |
CDF('CHISQUARE',5.0,3,1.5) |
CINV(.5,3,1.5) |
CNONCT(5.0,3,.55) |
? |
| Stattab [1] |
NA |
5.0 3 1.5 ? . |
? 3 1.5 0.5 . |
? |
NA |
| Correct Answer |
0.097257 |
0.649285 |
3.668745 |
2.898530 |
Varies |
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. [2]), but apparently not among programmers who don't read before writing their code.
These parameters can also be calculated online.
References
- ^ MD Anderson Cancer Center [1]
- ^ R. Christensen, Plane Answers to Complex Questions (3rd edition, 2002), Springer, NY, p.424.
- 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.
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