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Can I compute pdf(z), if I known joint probability density pdf(X+z,Y+z)? X, Y and z are independent, zero mean, variables. pdf of X,Y,z are unknown, we can assume that pdf(X), pdf(Y) are gaussian. We can also assume that variance of X,Y is much larger than that of z. This last condition effectivly prevents measuring pdf(X,Y) by setting z=0 and than using deconvolution.

There is an indirect method, so I know that the problem is solvable. From pdf(X+z,Y+z) I can compute the cross-moments M(n,m). These are equal to M(n,m)=M_z(n+m) if n or m are odd, since all the terms having X,Y or z to an odd power will average to 0. Now from the moments of z we can compute it's pdf.

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    $\begingroup$ I'm confused. How do you know that the moments of $z$ determine its distribution? How do you know that the variance of $z$ exists? $\endgroup$ – Galton Jun 20 '18 at 14:59
  • $\begingroup$ Stricly speaking I do not. However, as I understand, for most distributions they do and variance exists. I am trying to solve a practical problem, not to obtain a mathematical result. $\endgroup$ – muh Jun 25 '18 at 12:04
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Let the pdfs of $X$, $Y$, and $Z$ be denoted by $f_X(x)$, $f_Y(y)$, and $f_Z(x)$, respectively.

I don't immediately see how this problem can be solved when $f_X(x)$ and $f_Y(y)$ are unknown. However, you say that $X$ and $Y$ are mean $0$ as well as Gaussian, so the only missing pieces of information are their variances. The following solution should work if $f_X(x)$ and $f_Y(y)$ are known.

Define new random variables $U = X + Z$ and $V = Y + Z$. The joint pdf of $U$ and $V$ is $$ f_{UV}(u, v) = \frac{\partial}{\partial u}\frac{\partial}{\partial v} F_{UV}(u, v)\, , $$ where $F_{UV}(u, v)$, the joint cdf of $u$ and $v$, is given by \begin{align} F_{UV}(u, v) &= P\left(x+z < u,\; y+z < v\right)\\[0.1in] &= P\left(x < u-z,\; y < v-z\right)\\[0.1in] &= \int_{-\infty}^{+\infty} dz\, f_Z(z) \int_{-\infty}^{u-z} dx\, f_X(x) \int_{-\infty}^{v-z} dy\, f_Y(y)\, . \end{align}

The joint pdf of $U$ and $V$ then becomes: \begin{align} f_{UV}(u, v) &= \frac{\partial}{\partial u}\frac{\partial}{\partial v} F_{UV}(u, v)\\[0.1in] &= \int_{-\infty}^{+\infty} dz\; f_Z(z)\; f_X(u-z)\; f_Y(v-z) \qquad\qquad (1) \end{align}

Define the two-dimensional Fourier transform of the pdf $f_{UV}(u, v)$ as \begin{equation} \hat{f}_{UV}(s, t) \;\equiv\; \int_{-\infty}^{+\infty}du\, e^{-isu} \int_{-\infty}^{+\infty}dv\, e^{-itv} f_{UV}(u, v)\, , \end{equation} where $s$ and $t$ are frequencies.

Using this definition, take a Fourier transform of both sides of Eq. (1) above to yield \begin{align} \hat{f}_{UV}(s, t) &= \int_{-\infty}^{+\infty} dz\, f_Z(z)\; \int_{-\infty}^{+\infty} du\, e^{-isu}\, f_X(u-z)\; \int_{-\infty}^{+\infty} dv\, e^{-itv}\, f_Y(v-z)\\[0.1in] &= \int_{-\infty}^{+\infty} dz\,e^{-i(s+t)z}\, f_Z(z)\; \hat{f}_{X}(s)\; \hat{f}_{Y}(t)\\[0.1in] &= \hat{f}_{Z}(s+t)\; \hat{f}_{X}(s)\; \hat{f}_{Y}(t)\, . \end{align} Between the first and second lines above, we have made a change of variables $u\rightarrow u+z$, $v\rightarrow v+z$.

From here, one can set either $s = 0$ or $t = 0$, to yield $$ \hat{f}_Z(t) = \frac{\hat{f}_{UV}(0, t)}{\hat{f}_Y(t)} $$ or $$ \hat{f}_Z(s) = \frac{\hat{f}_{UV}(s, 0)}{\hat{f}_X(s)}\, , $$ respectively. Note that we have used the fact that $\hat{f}_X(0) = \hat{f}_Y(0) = 1$, which must be true for the Fourier transform of any pdf.

Assuming that all of the relevant pdfs are analytically known originally, all that is left to do in principle is to take an inverse Fourier transform of one of the above expressions for $\hat{f}_Z$: \begin{align} f_Z(z) &= \frac{1}{2\pi}\int_{-\infty}^{+\infty}dt\, e^{+itz}\, \hat{f}_Z(t) \;=\; \frac{1}{2\pi}\int_{-\infty}^{+\infty}dt\, e^{+itz}\, \frac{\hat{f}_{UV}(0, t)}{\hat{f}_Y(t)}\\[0.1in] &= \frac{1}{2\pi}\int_{-\infty}^{+\infty}ds\, e^{+isz}\, \hat{f}_Z(s) \;=\; \frac{1}{2\pi}\int_{-\infty}^{+\infty}ds\, e^{+isz}\, \frac{\hat{f}_{UV}(s, 0)}{\hat{f}_X(s)} \end{align}

Edit:

It strikes me that $\hat{f}_{UV}(s, 0)$ and $\hat{f}_{UV}(0, t)$ are the Fourier transforms of the marginal pdfs $f_{U}(u)$ and $f_{V}(v)$ of $U$ and $V$, respectively. The final result above could therefore have been obtained in a much shorter way as follows: $$ f_{U}(u) = \int_{-\infty}^{+\infty}dz\, f_Z(z)\, f_X(u-z) \;\;\rightarrow\;\; \hat{f}_U(s) = \hat{f}_Z(s)\, \hat{f}_X(s) $$

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  • $\begingroup$ Thank you. Unfortunately this amounts to Fourier transform deconvolution of f_Z from known f_U or f_V and known or estimated f_X or f_Y. Which does not work very well, when variance of X and Y is >> variance of Z. Using marginal distributions is probably even worse, than directly deconvolving from multidimensional distribution since some information is lost in the process. $\endgroup$ – muh Jun 25 '18 at 12:11
  • $\begingroup$ I see what you mean. If $f_X(x)$ and $f_Y(y)$ are broad, then $\hat{f}_X(s)$ and $\hat{f}_Y(t)$ will be narrow, and you'll be dividing be something that is close to zero over much of the relevant domain. When trying to do this with actual data, you'd be losing the high-frequency components of $f_Z(z)$. Unfortunately, I suspect this problem would carry over in some form to any potential solution to the problem, e.g. the series of moments would have poor convergence. $\endgroup$ – John Barber Jun 25 '18 at 16:27
  • $\begingroup$ I think what you may need is some additional regularization or ansatz that allows you to treat the expressions above as constraints rather than as solutions. For example, what if you tried finding the minimum-entropy or minimum variation solution for $f_Z(z)$ that also satisfies $\hat{f}_U(s) = \hat{f}_Z(s)\, \hat{f}_X(s)$ on the values of $s$ for which $\hat{f}_X(s)$ is not too close to $0$? $\endgroup$ – John Barber Jun 25 '18 at 16:31
  • $\begingroup$ Yes, this seem to be a standard procedure for finding parametrized distribution. I was hoping for a non-parametrized approach. As a matter of fact, the original X, Y, Z are complex, so that my measured pdf is 4D. Thus in going via moment I through a lot of information away, from enormous histogram to several numbers. I was wondering if I can avoid that. To make matter even more complicated, X+Z and Y+Z and not really complex, but bandlimited real, from which complex signal is constructed with the Hilbert transform. So there is lots of steps and I wondered, whether I can take shortcuts. $\endgroup$ – muh Jun 26 '18 at 17:52

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