# Sum of non-central chi-square using Laguerre expansion

I moved this question from my answer on this link. From Castano-Martinez and Lopez-Blazquez(2005), They explained pdf of sum of noncentral Chi-square on equation 3.2, that is:

$f(y)= \frac{e^{-\frac{y}{2\beta}}}{(2\beta)^{v/2}} \frac{y^{v/2-1}}{\Gamma(v/2)} \sum_{k\geq 0}{\frac{k! c_k}{(v/2)_k}}L_k^{v/2-1}(\frac{vy}{4\beta \mu_0}), \forall \mu_0 > 0$

They actually explain all of symbols here. But I still don't understand about what symbol $\beta$ is. I thought it was arbitrary. However, when I tried this in algorithm, the results are different significantly, depend on value of $\beta$. I also got the wrong result, and I think it has something to do with $\beta$ value.

I put additional details on it. I realized I did not put enough info in my question. Hopefully it can help:

The equation above is density function (PDF), expanded using Laguerre expansion of linear combination of non-central Chi-square RV. $v$ is the degree of freedom of each RV, actually I only need degree of freedom=1 in my case.

$L_k^{v/2-1}(\frac{vy}{4\beta \mu_0})$ is Laguerre polynomial.

The non-centrality factor and the weighting factor of linear combination are covered in $c_k$

$c_k$s can be expressed by:

$c_0 = (\frac{v}{2\mu_0})^{v/2} \exp{(-0.5*\sum_{i=1}^n{\frac{\delta_i \alpha_i (p-\mu_0)}{\beta \mu_0 + \alpha_i(p-\mu_0)})}} \prod_{i=1}^n{(1+\frac{\alpha_i}{\beta}(p/\mu_0-1))^{-v_i/2}}$

$c_k = \frac{1}{k} \sum_{j=0}^{k-1}c_j d_{k-j}, k \geq 1$

with $d_j$:

$d_j = \frac{-j \beta p}{2 \mu_0} \sum_{i=1}^n ({\delta_i \alpha_i (\beta -\alpha_i)^{j-1}(\frac{\mu_0}{(\beta \mu_0 + \alpha_i (p-\mu_0)) })^{j+1}} + \frac{v_i}{2}(\frac{1-\alpha_i/\beta}{1+(\alpha_i/\beta)(p/\mu_0-1)}))^j$

with $v_i$ degree of freedom of i-th RV, $\delta_i$ noncentrality, $\alpha_i$ weighting factor of linear combination, $v = \sum_{i=1}^{n}{v_i}$, $p=v/2$, and $\mu_0>0$ (arbitrary number, but influential in algorithm speed).

That's all I can only refer from the paper. Just let me know should you demand more information.

• Because you can write $f$ as a function of $y/\beta$, and this is the only way in which $\beta$ is involved, evidently $\beta$ establishes the units of measurement of $y$: it is a scale parameter. – whuber Dec 2 '16 at 16:12
• Yes, $\beta$ is scaling factor. But can $\beta$ be arbitrary, as long as $\beta > 0$ or dependent to another parameter? – enas Dec 3 '16 at 6:23
• Either one is possible, depending on the model. The only way to answer that question would be to refer to the paper. – whuber Dec 3 '16 at 17:05
• The paper said that $\beta>0$. Therefore I thought $\beta$ is arbitrary. However, when I tried it in a program, different $\beta$ produces very different results. Now I think $\beta$ should be determined with specific formula, but still figuring out on it. – enas Dec 4 '16 at 4:53
• Because the article is behind a paywall, we cannot expect readers to have access to it. That will make it difficult for us to answer unless you can provide the background information about $\beta$ explicitly. – whuber Dec 4 '16 at 19:46