How to approximate non-central chisquare distribution to Poisson weighted sum of central chi-square distribution in case of non-unit variances?

Question

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}

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 $\sigma^2$.