2

I am trying to find the probability distribution function (pdf) of the following

$$Y=\big|\sum_i X_i\big|^2=\big|\sum_{i\leq 2} h_i \, a_i\big|^2 $$

with $$h_i\sim \text{iid} \operatorname{Nakagami}(m=3,1/3)$$ $$a_i \,\text{positive constants} \,\,\, \forall \,\,\, i$$

How I solve the problem

First I try to find the pdf of $$X=a\,h\sim \frac{1}{a}f_{h} (x/a)\sim \operatorname{Also a Nakagami}$$

So then I need to find the distribution of $$\sum_{i}X_i=?$$ and then the magnitude squared..

I am looking for a tractable approach to find the new pdf, but I am out of ideas!

If you feel this is ugly to derive, do you have any ideas to approximate it? I am out of ideas !!

Thanks for any advice.

Stephan Kolassa
  • 95,027
  • 13
  • 197
  • 357
Tyrone
  • 205
  • 1
  • 8

1 Answers1

1

I am posting an answer for the two cases "square of sum of Nakagami" and "sum of squares of Nakagami"

Case 1: Square of sum

I would use the following approach here:

  • Find the pdf of $X = ah$, which is Nakagami.
  • Find the pdf of $Z = \sum_i X_i$ using the approximation presented in this reference:

J. C. S. S. Filho, M. D. Yacoub, "Nakagami-m approximation to the sum of M non-identical independent Nakagami-m variates." Electronics Letters 40.15 (2004): 951-952.

  • Finally, find the pdf of $Y = |Z|^2$ where $Z$ is a Nakagami random variable.

Drawback: This reference is behind a paywall, so I was not able to copy the approximation here.

Case 2: Sum of squares

I believe the paper by Karagiannidis, Sagias, Tsiftsis (2006) has the formula for this case. You can find it in page 2.

The formula is a bit complex. I am putting the screenshots of the section where it is presented.

Part 1

Part 2

Part 3

Part 4

tiagotvv
  • 729
  • 7
  • 11
  • 1
    Could you post the formula here? Links expire or become invalid in which case your answer will not be useful anymore to future readers. – Andy Feb 25 '15 at 20:20
  • Thanks I was not aware of that paper, but I am actually looking for the square of the sum and not the sum of the square, this is what the paper does right? @tiagotvv – Tyrone Feb 25 '15 at 20:52
  • 1
    The sum of squares would be the same as the sum of Gamma distributions. Full answers, with derivations and references (from 1984, considerably antedating Karagiannidis et al!), are provided at http://stats.stackexchange.com/questions/72479. – whuber Feb 25 '15 at 21:16
  • Yes, but I am looking for the square of sum @whuber – Tyrone Feb 25 '15 at 21:45
  • I know that: my comment was intended to clarify the nature of this answer, not the nature of the question, which deals with something considerably more difficult. That's why the comment appears below the answer and not below your question. – whuber Feb 25 '15 at 21:46
  • @whuber, in your opinion this is very difficult, do you know of any approximations that can come about solving this? – Tyrone Feb 25 '15 at 22:12
  • Sorry, @Tyrone, I overlooked the magnitude sign coming before the sum. I updated my answer which now deals with the two cases: square of sum and sum of squares. Too bad the reference is behind a paywall. – tiagotvv Feb 25 '15 at 22:15
  • thanks but why is $Z$ a Nakagami, does this paper say so ? – Tyrone Feb 25 '15 at 22:18
  • @Tyrone, Yes, the paper says so. Note that it is an approximation. – tiagotvv Feb 25 '15 at 22:24
  • The saddlepoint approximation method suggested by Kjetil Halvorsen at http://stats.stackexchange.com/a/137318 looks promising. – whuber Feb 25 '15 at 22:25
  • @tiagotvv so does this mean that $Y$ is Gamma since $Z$ is Nakagami now by approximation ? – Tyrone Feb 25 '15 at 22:29
  • @Tyrone Yes, since the square of a Nakagami rv is a Gamma rv. – tiagotvv Feb 25 '15 at 22:50
  • @tiagotvv thanks for the reference ! I searched alot but I havent found somthing like that. – Tyrone Feb 25 '15 at 22:52
  • Do you know how to expand this $$\mathbb{E}[R^n]=\sum_{n_1=0}^n\sum_{n_2=0}^{n_1}\cdots \sum_{n_{M-1}}^{n_{M-2}} {n\choose n_1}{n_1\choose n_2}\cdots {n_{M-2}\choose n_{M-1}}\mathbb{E}[R_1^{n-n_1}]\times \mathbb{E}[R_2^{n_1-n_2}]\cdots \mathbb{E}[R_M^{n_{M-1}}]$$ Assume that for my application $n=4$ and $M=2$, I got the equation from the paper you referenced @tiagotvv – Tyrone Feb 25 '15 at 23:26