Preliminary: Let's say we have $Y=X+Z$ ($Y$ is data, $X$ is latent variable and $Z$ is noise), where the random variables are all in $\mathbb{R}$. Then an inverse Fourier transform leads to

\begin{align} f_X(x)=\frac{1}{2\pi}\int e^{-itx}\varphi_X(t)dt=\frac{1}{2\pi}\int e^{-itx}\frac{\varphi_Y(t)}{\varphi_Z(t)}dt, \end{align}

where $\varphi$ are the characteristic functions. Now, suppose we know the noise distribution for $Z$ so we know the characteristic function $\varphi_Z$. If we observe data $\boldsymbol{y} = (y_1, ..., y_n)$ then we can estimate $f_Y$ and thereby estimate $\varphi_Y$ by subsitution. This gives us the estimator

$$\hat{f_X}(x) = \frac{1}{2\pi}\int e^{-itx} \frac{\hat{\varphi_Y}(t)}{\varphi_Z(t)}dt = \frac{1}{2\pi}\int e^{-itx} \frac{\int e^{ity}\hat{f}_Y(y)dy}{\varphi_Z(t)}dt.$$

My question: What if $X$ is one-dimensional but the noise and observation are two-dimensional?

For example, $X$ is a random variable that takes values on some deterministic line segment (in two dimensional space, but since it is on the line so essentially its one-dimensional), the error $Z$ is a two dimensional random variable so as $Y$. I understand that if we treat $X$ as a two dimensional object we still have $\varphi_X(t)=\varphi_Y(t)/\varphi_Z(t)$, but it says nothing about the fact that $f_X$ is a density on the one dimensional line segment. How should I do deconvolution in this kind of situations?


If the dimensions of $X$ and $Z$ are different, the concept of summing them up does not make much sense. If I got this correctly, $X$ is a one-dimensional deterministic value on a line, $Z$ is noise added to the $(x,y)$-coordinated of $X$ and $Y$ is the position of $X$ that is observed.

I think, you are actually in the situation, where $$Y = f(X) + Z,$$ where $f$ maps the random variable $X$ to the position $X$ refers to. If $X$ is on a straight line, $f$ would map something like $$x \mapsto \begin{pmatrix}a_1\\a_2 \end{pmatrix}x+\begin{pmatrix}b_1\\b_2 \end{pmatrix}.$$

Anyway. $f(X)$ is a random variable, since $X$ is a random variable. So, you just need to find $X$ and the characteristic function of $f(X)$.

  • 1
    $\begingroup$ Yes, I believe the question describes the second situation. Unfortunately, telling the OP that "you just need to find..." doesn't really provide any answers at all. Could you explain how? $\endgroup$ – whuber May 6 '18 at 22:20

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