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I am working on a problem that is similar to the one discussed in this link. But in my case $X_i \sim \mathcal{N}(1, \sigma^2)$, i.e., $X_i$ is not a zero-mean Gaussian RV. Specifically, I want to find the distribution of $Z = \frac{\left(\sum_{i=1}^L X_i\right)^2} {L \sum_{i=1}^L X_i^2}$, where $L$ in my case is 11. I found the numerator is $\sim$ $\chi_1^{'^2}(L)$ and the denominator is $\sim$ $\chi_L^{'^2}(\frac{L}{\sigma^2})$; where $\chi_k^{'^2}(\lambda)$ denotes the non-central chi-square distribution. I am unable to proceed beyond this point. monte-carlo simulation of the R.V. Z results in the following CDF plot: enter image description here

As seen from the simulation, the value of the RV will lie between 0 and 1. The value that Z takes in my problem is a threshold between 0 and 1. Knowing the distribution (a closed-form equation) will help me determine other parameters like detection, false alarm, miss-detection probability etc.

Can someone kindly help how I can proceed. Is there a closed-form solution when $X_i$ is a non-zero mean Gaussian r.v.. Thanks.

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  • $\begingroup$ It appears that you're using "amplitude ratio" units for the values of $\sigma$ as $\sigma=0$dB seems to mean $\sigma=1$. If so please clarify that. Also what are your definitions of "parameters like detection, false alarm, miss-detection probability, etc." Given the 16 values of $\sigma$ and the simulations, you could fit a function of $\sigma$ for all of those parameters. $\endgroup$
    – JimB
    Commented Jul 17, 2022 at 18:16
  • $\begingroup$ Yes, $\sigma$ is in dB. For linear, $\sigma_{linear} = 10^{\frac{\sigma_{dB}}{10}}$. For e.g., prob of false detection will be the CCDF, P(Z1 > r) and prob of miss detection will be the CDF, P(Z2 < r). Z1 and Z2 are r.v. (defined as Z in the problem) observed at time t1 and t2. At t1, the signal is absent and ratio r determines the false detection rate. At t2, signal is present and ratio r determines the miss detection rate. Yes, I think I can fit a function for $\sigma$. Was wondering if there is a closed-form solution for this r.v. $\endgroup$ Commented Jul 19, 2022 at 18:13
  • $\begingroup$ If the moments of this distribution are enough, this other post may be useful stats.stackexchange.com/questions/605979/… $\endgroup$
    – dherrera
    Commented Mar 6, 2023 at 18:09

1 Answer 1

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I think that the article "Evaluating the density of ratios of noncentral quadratic forms in normal variables" may be what you are looking for, although there doesn't seem to be a nice tidy expression there.

In Proposition 1 in the paper, it gives the density for $R=\frac{X'AX}{X'BX}$ where $X\sim \mathcal{N}(\mu, I)$ and B is positive semi-definite. You need to use matrix $A$ with 1's in the diagonal and $\frac{1}{2}$ elsewhere, and $B=\mathbf{I}L$ in that formula, and you should get the desired density.

The formula given in the paper is quite complicated, but maybe it will simplify out when you substitute in the matrices $A$ and $B$ associated with your problem (that's what happened to me in this similar problem).

For completeness, the formulas in the paper are:

Extracted from S.Broad, M.S Paolella 2009

As a note, if what you want are the moments of the ratio, and not the density per se, then there are nicer formulas and approximations. In this other answer, I show an exact analytic formula for the first moment of quadratic forms of this type. To adapt it to your specific problem you'd just need to scale X so that instead of having L in the denominator, you have 1/L in the numerator, and do that in matrix form.

Edit: The previous version of this answer had a wrong matrix $A$, that didn't match the ratio in the question.

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