3
$\begingroup$

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?

$\endgroup$
1
  • $\begingroup$ This makes sense as a mathematical question, where the solution is that the deconvolution does indeed exist, it can be computed, and corresponds to the singular distribution of $X.$ As a statistical question, it puts us into an unconstructive straitjacket because it proposes using an ineffective (impossible, really) estimation procedure. Consider, then, exploring the estimation procedures described in the statistical textbooks and their associated distributional models. $\endgroup$
    – whuber
    Feb 4, 2023 at 16:05

1 Answer 1

0
$\begingroup$

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 $f(X)$.

$\endgroup$
1
  • 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, 2018 at 22:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.