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I want to analyze or approximate a random variable that is a sum of two scaled independent non central $\chi$-squared random variables with the same degrees of freedom.

For example, $$X = X_1 + a X_2$$ where $X_1 \sim \chi^2(k,\lambda_1)$ and $X_2 \sim \chi^2(k,\lambda_2)$, $k$ is the degree of freedom, $\lambda_1, \lambda_2$ are noncentral parameters.

Simulations show that pdf of $X$ has the form of non central $\chi$-squared, or Gamma, ... I did try using characteristic function $M_X(t) = M_{X_1}(t)M_{X_2}(at)$ but it does not seem that the inverse transform of $M_X(t)$ has analytic form.

I wonder if there exists well-known results for this kind of problems.

For example,

$k=7, l_1=2,l_2=5, a=3$ enter image description here

$k=7, l_1=2,l_2=5, a=1$ enter image description here

$k=7, l_1=2,l_2=5, a=-1$ enter image description here

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  • $\begingroup$ You have $X= X_1 + aX_2.$ There is a standard and well known answer in case $a=1.$ If $a\ne1,$ I suspect there is no closed form. $\endgroup$ – Michael Hardy Sep 3 '20 at 23:06
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There is one case which is well-known.

Lemma. Let $Q_i \sim \chi^2_{k_i}(\lambda_i)$ for $i=1,\dots,n$, be independent. Then, $Q = \sum_{i=1}^n Q_i$ is a noncentral-$\chi^2_{k}(\lambda)$, where $k=\sum_{i=1}^n k_i$ and $\lambda = \sum_{i=1}^n \lambda_i$.

Proof. The Moment Generating Function (MGF) of $Q_i$ is given by $$ \mathcal{M}_{Q_i}(t) = (1 -2t)^{-k_i/2} \exp \Big\{ \frac{\lambda_i t}{1 - 2t} \Big\} $$ The MGF of $Q$, using independence, is given by $$ \mathcal{M}_Q(t) = \prod_{i=1}^n \mathcal{M}_{Q_i}(t) = \prod_{i=1}^n (1 -2t)^{-k_i/2} \exp \Big\{ \frac{\lambda_i t}{1 - 2t} \Big\} = (1 - 2t)^{- \frac{1}{2} \sum_{i=1}^n k_i} \exp \Big\{ \frac{t}{1 - 2t} \sum_{i=1}^n \lambda_i \Big\} $$ Thus, $$ \mathcal{M}_Q(t) = (1 -2t)^{-k/2} \exp \Big\{ \frac{\lambda t}{1 - 2t} \Big\} $$ and the thesis immediately follows. $\blacksquare$

Note that, if $\lambda \neq 0$, the PDF (probability density function) of $Q$ does not admit a closed-form, as it involves Bessel functions.

The general case of a linear combination of independent $\chi^2_{k_i}(\lambda_i)$ $$ Q = \sum_{i=1}^n a_i Q_i $$ results in a so-called generalized chi-squared distribution. The probability density function (PDF) of the generalized $\chi^2$ is also complicated.

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  • $\begingroup$ Thanks for the info. However, this does not answer my question. $\endgroup$ – Sabrina Cantu Apr 19 '18 at 19:30
  • $\begingroup$ Can you please better explain your question? $\endgroup$ – PseudoRandom Apr 27 '18 at 9:23
  • $\begingroup$ If I follow your reasoning, the distribution I am looking at must be generalized chi squared distribution and its PDF is complicated. It is the definition. However, simulations show that the PDF looks like the one of scaled/non-central chi squared, etc. My question is whether we can approximate my interested PDF to a chi-squared-like PDF. $\endgroup$ – Sabrina Cantu Apr 29 '18 at 8:44

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