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Question: problem statement

$$z^d = \sum_i x_i^d $$

pdf of $X = \frac{d(\lambda x^d)^n}{x \Gamma(n)} \exp(-\lambda x^d)$ which is a generalized Gamma distribution and they represent euclidean distances. The pdf is presented in Eq(26) http://www.cs.tut.fi/~moltchan/pubs/distances2011.pdf

I am not sure, if each of the r.v for distance is independent. I obtained a histogram of the distances calculated between data points of a univariate time series model of Moving Avreage (FIR filter) in higher dimension $d$ for the 2 cases (a) when the time series is clean given below hist_distances_clean_signal

and when the time series signal is noisy with a signal to noise ratio of 1 db. The noise added is a zero mean white gaussian noise.

hist_distances_noisy_signal

Question1: I do not know how to check if they are indeed Generalized gamma.

Srishti M
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1 Answers1

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If$$X_i\sim f(x)=\dfrac{d(\lambda x^d)^n}{x \Gamma(n)} \exp(-\lambda x^d)\,,$$then $$Y_i=X_i^d\sim g(y)=\dfrac{d(\lambda y)^n}{y^{1/d}\Gamma(n)} \exp(-\lambda y)\times\frac{1}{d}y^{1/d-1}=\dfrac{\lambda^ny^{n-1}}{\Gamma(n)}\exp(-\lambda y)$$means that $X_i^d\sim\text{G}(n,\lambda)$, hence that$$\sum_{i=1}^k X_i^d\sim\text{G}(kn,\lambda)$$from which it is straigtforward to deduce that $Z$ has a generalised Gamma distribution.

Xi'an
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  • Thank you for your answer. Could you please elaborate on your second step why we multiply by $y^{1/d-1}/d$; where this term comes and should there be a $\lambda^n$ term for $Y_i$? The result is applicable on the assumption that each $X_i^d$ is an iid r.v. Can you please clarify all this as my background is very dwindling in probability theory. – Srishti M Apr 09 '15 at 19:34
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    I indeed forgot the $\lambda^n$ in the rhs, now corrected. As for the $y^{1/d-1}/d$ additional term, this is simply the Jacobian $\text{d}x/\text{Y}$ of the transform $y=x^d$. And yes indeed the $X_i$'s have to be independent for the sum to be a Gamma variate. – Xi'an Apr 09 '15 at 19:46
  • Thank you once again, but I do not know why to calculate the Jacobian? Is this different from convolution formula to find the pdf? Can you please elaborate this point with the details? I found a tutorial http://www.statlect.com/ucdexp1.htm which mentions that the sum of exponential r.v is a gamma. In this case it is coming out to be generalized gamma. Why? – Srishti M Apr 09 '15 at 19:50
  • Will really be helpful if you can elaborate on how you obtained of of $Y_i$. I do not know why you used Jacobian, multiplication etc – Srishti M Apr 09 '15 at 19:54
  • I certainly cannot provide you with the details of the theory of change of variables for random variables, but any basic probability or statistics textbook like Ross or Casella and Berger can help you with this. – Xi'an Apr 09 '15 at 20:31
  • Hello. Sorry to bother again, but I have a confusion & shall be immensely grateful if you can address it. In your last expression, you mentioned that the sum of $X_i$ is a Gamma distribution and then you say that $Z$ is a generalized Gamma. Why $Z$ is Generalized and not just Gamma? Is this also coming from change of variables that $Z$ becomes generalized Gamma? – Srishti M Apr 11 '15 at 21:11
  • You get that $Z^d$ is Gamma $G(nk,\lambda)$. Hence $Z$ is generalised Gamma. – Xi'an Apr 12 '15 at 05:35