I have found that the component PDFs of a given vectorial quantity have the same exponential distribution. I am also interested in the PDF of the modulus (please see the picture enclosed). Is it possible to derive analytically the PDF distribution of such a quantity? Thank you!
Unless it's known that the components $X$, $Y$ and $Z$ are independent, the joint distribution for $(X,Y,Z)$ is not fixed by the marginal distributions for $X$, $Y$ and $Z$ separately. To determine the distribution of $R=\sqrt{X^2+Y^2+Z^2}$, the joint distribution is needed.
In the case of independent components, the integrals are a bit of work.
Richard, in fact, I don't know if I can say that the variables are statistically independent or not. To be more precise, the vector here is the angular velocity vector of particle pairs in isotropic homogeneous turbulence. The numerical results over a huge number of tracked pairs showed that the three components behave similarly, that indeed is an isotropy signature. If you think that the calculations would be analytically accessible if we consider that the variables are independent, please then assume that they are independent.
Dear all,
I think that with the Euclidean norm there is no hope. I'd try with the L1 norm:
||x|| = |x1| + |x2| + ... +|xn|, that is topologically equivalent to the Euclidean one and promisis tractable computations.
For the sake of clarity, the probability distribution you are working on is known as "Pascal distribution", while the "exponential" adjective is used to identify the one with density:
f(x) = 0 for x < 0
= a exp(-ax) for non-negative x.
Best
Lorenzo
Thank you Lorenzo for your clarifications. Let's go ahead with the L1 norm. What is the approach to use in order to derive this probability? Could you please tell us why do the Euclidean norm couldn't do the job.
Subrata, the support is all the axis of reals, since I included an absolute value. Please any help would be appreciated!! Thanks.
OK so kind of reflected exponential distribution not exactly exponential as we know ti. let us see !
I think Lorenzo was referring to the fact that while the PDF of x^2, y^2, and z^2 are easily derivable, their sum is hard due to the difficult integrals involved. Do you need it analytically solved? You could do it numerically...
If you choose the L1 norm, |x|, |y|, |z| are just exponential random variables. Their sum r^2 (assuming they are independent) is then simply a gamma(3,lambda) distribution (where lambda is a rate parameter). Do a random variable transformation from r^2 to r, and you've got the answer.
The thanformations required by the Euclidean norm induce "difficult" densities for the squares of the coimponents, and when you try to compute the convolution of densities you find integrals which cannot be computed in closed form. If you have Pascal components and you consider their absolute value, the distribution of the absolute value is (negative) exponential. Th convolution of exponential densities can be computed in closed form. In the (very) special case of equal parameters of the exponential distributions, the sum is distributed according to the Erlang/Gamma law.
That's all:
Ciao
Lorenzo
Thank you for your answers, Richard could you please explain why do we have a gamma distribution if we assume that the variables are stochastically independent? Thanks!!
For instance a/(a-t) is the mgf of an exponential. The sum of n indep exponential r.v. has mgf (a/(a-t))^n (owing to Laplace transform. The latter expression is the mgf of a Gamma distr.
Ciao
L
The square of an exponential is a Weibull distribution with shape parameter 1/2. I am totally out of my depth here but, according to a quick Google search, it is widely believed that no closed form for sums of independent Weibull variables is known. However I found an approximation in the link below which might potentially be of use to you.
Then you would only need to make a square-root transform to recover the vector modulus.
http://ieeexplore.ieee.org/stamp/stamp.jsp?arnumber=01665128
Thank you Pedro for your answer and helpful information. But how can I use IEEE Xplore to get the desired info? Thanks!
You may not be able to access the paper if your institution does not have a subscription to IEEE journals. I am at home now and, suprisingly, the IEEE website redirects you to a generic error page if you have no institutional access.
The paper is the following:
Article Simple precise approximations to Weibull sums
Unless it's known that the components $X$, $Y$ and $Z$ are independent, the joint distribution for $(X,Y,Z)$ is not fixed by the marginal distributions for $X$, $Y$ and $Z$ separately. To determine the distribution of $R=\sqrt{X^2+Y^2+Z^2}$, the joint distribution is needed.
In the case of independent components, the integrals are a bit of work.
When you show modulus equation : does it contain the means of probabilistic variables or probabilistic variables themselves. If variables, not their means, then you could use the same closed form (analytical) approach as I did for the vector modulus in "Iteration free vector orientation using maximum relative entropy with observational priors".
Since nothing is specied about the joint distribution, there are two most probable choices: independency of $(X,Y,Z)$ or invariance by rotation. This second choice is attractive since $X,Y,Z$ are identically distributed. Without loss of generality we take $\lambda=1$ and it is easy to prove that if $(X,Y,Z)$ is invariant by rotation then the density of $R$ is $re^{-r}$ on the positive line.
If the marginal distributions are indipendent, a partially explicit expression for P(r) is the following (Latex notations)
\begin{align*}
&P(r)=\frac{\lambda^3}{8}\frac{d}{dr}\int_0^{r}\left\{\int_{4\pi}\exp{\left[-\lambda s\left(\sin{\theta}\cos{\phi}+\sin{\theta}\sin{\phi}+\cos{\theta}\right)\right]}s^2\sin{\theta} d\theta d\phi \right\}ds=\\
&\frac{r^2\lambda^3}{8}\int_{4\pi}\exp{\left(-\lambda r\left(\sin{\theta}\cos{\phi}+\sin{\theta}\sin{\phi}+\cos{\theta}\right)}\sin{\theta} d\theta d\phi
\end{align*}
where $\theta$ and $\phi$ are polar and azimuthal angles respectively.
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