According to Statistics libre texts Equation 5.9.20, a non-central chi square distribution can be approximated as sum of Poisson weighted central chi square distributions.
$\tag{1}g(y) = \sum_{k=0}^\infty e^{-\lambda/2} \frac{(\lambda/2)^k}{k!}f_{n+2k}(y)$
Here, the term $e^{-\lambda/2} \frac{(\lambda/2)^k}{k!}$ follows Poisson distribution,
$g(y)$ is the pdf of non-central chi-square distribution with non-centrality parameter $\lambda$ and $n$ degrees of freedom,
$f_{n+2k}(y)$ is the central chi square distribution with $n+2k$ degrees of freedom.
This holds true if the underlying gaussian distribution has unit variance according to Wikepedia non-central chi square distribution. In that case,
If $J\sim Poisson(1/\lambda)$, then then $\mathcal{X_{k+2J}^2} \sim \mathcal{X_{k}^{\prime2}}(\lambda)$
How do I approximate the non-central chi-square as Poisson distribution in case it has non-unit variances? I have been successful so far only with unit-variance. Should I also make changes to Poisson weights $e^{-\lambda/2} \frac{(\lambda/2)^k}{k!}$ to account for non-unit variance?

The textbook of digital communications by Prokais has evaluation for pdfs of both central chi-square distribution and non-central chi-square distributions with non-unit variances and I could successfully simulate them and they fit the distribution. In case it is of intrest, pdf of non-central chi-square can be written as, according to the text book Digital Communications, Prokais,
\begin{equation} \tag{2}p(x) = \begin{cases} \frac{1}{2\sigma^2}(\frac{x}{\lambda})^{\frac{n-2}{4}}e^{-\frac{\lambda+x}{2\sigma^2}\mathcal{I}_{\frac{n}{2}-1}}(\frac{\sqrt{\lambda}}{\sigma^2}\sqrt{x}), & \text{if } x > 0\\ 0, & \text{otherwise} \end{cases} \end{equation} Here, $\mathcal{I}_v(x)$ is the modified Bessel function of first kind of order $v$, n is the degrees of freedom, $\lambda$ is the non-centrality parameter given by $\lambda = \sum_{i=1}^{n}m_i^2$ where $m_i$ are the means of underlying Gaussian random variables with common variance $\sigma^2$.
The pdf of central chi-square for Gaussian variables of zero mean and common variance $\sigma^2$ is given by,
\begin{equation} \tag{3}p(x) = \begin{cases} \frac{1}{2^{(n/2)}\Gamma(\frac{n}{2})\sigma^n}x^{\frac{n}{2}-1}e^{-\frac{x}{2\sigma^2}}, & \text{if } x > 0\\ 0, & \text{otherwise} \end{cases} \end{equation}

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    $\begingroup$ Start by transforming the underlying Gaussian variables to give them unit variance. How does their mean change? What effect does this have on your non-centrality parameter? $\endgroup$ Commented Mar 17, 2023 at 11:50
  • $\begingroup$ I don't have a strong background in statistics, but I need this evaluation to go further in my work. As far as I understand, in case the Gaussian variables have non-unit variance, I can scale them to unit variance. Then I need to scale both the distribution and non-centrality parameter by $\sigma^2$ for the distribution to fit. This I have tried and it works. however, I would want to avoid scaling of the distribution itself, hence I used the formula in the textbook to account for variance while calculating the pdfs of the non-central chi-square distribution and chi-square distributions. $\endgroup$
    – amitha
    Commented Mar 17, 2023 at 12:27
  • $\begingroup$ I don't understand your question, because non-central chi-squared distributions usually do not have unit variance. Furthermore, because variance is determined by the scale of the variable, it seems you are merely asking how to change the units of measurement of your variable, which is as simple as it sounds. Could you clarify what you are trying to accomplish? $\endgroup$
    – whuber
    Commented Mar 17, 2023 at 12:38
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    $\begingroup$ I am trying to ask what you mean by "non-unit variance." $\endgroup$
    – whuber
    Commented Mar 17, 2023 at 13:50
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    $\begingroup$ Does this answer your question? Sum of squares of non-standard Gaussians $\endgroup$
    – dherrera
    Commented Nov 6, 2023 at 20:48

1 Answer 1


If I understand your question correctly, you are interested in the distribution of $S_X = \sum_i^n X_i^2$ where $X \sim \mathcal{N}(\mu, \sigma^2 I_n)$. Because the texts that you reference give you a pdf for sums of unit-variance variables, as mentioned in the comments, you need to transform your variable to use those expressions, i.e. define $Y=\frac{X}{\sigma}$, and now $Y \sim \mathcal{N}(\frac{\mu}{\sigma}, I_n)$. Then, $S_Y=\sum_i^n Y_i^2$ will have a non-centered chi-squared distribution with non-centrality parameter $\lambda = \frac{\mu'\mu}{\sigma^2}$.

Now, we know the distribution of $S_Y$, but we want the distribution of $S_X$. We note that $S_X = \sigma^2 S_Y$ from simple algebra. Because the variable we are interested in is just a scaled version of a variable with known distribution, we use this result. What the results says, is that if $S_X = \sigma^2 S_Y$, then $P_{S_X}\left(S_X \right) = \frac{1}{\sigma^2}P_{S_Y}\left( \frac{S_Y}{\sigma^2} \right)$.

So, in order to adapt the formulas of the unit-variance case to the case where variance is $\sigma^2$, you just take the formula for the unit variance case and: 1) Use as the non-centrality parameter $\lambda = \frac{\mu'\mu}{\sigma^2}$, 2) Divide the arguments in the formulas by $\sigma^2$, 2), and 3) Divide the whole formula by $\sigma^2$.


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