4
$\begingroup$

Let $R$, $C$, and $X$ be independent random variables defined on $(0,\infty)$ and $$Y=\underbrace{R\, X}_{Z}+C.$$ We are given the joint probability distribution of $X$ and $Y$, $P_{XY}(x,y)$ and are asked to calculate the probability distributions of $R$ and $C$.

This is kind of like a regression problem, except I want the full probability distributions for the slope and intercept, not just their mean.

Here is what I have so far $$ \begin{align} P_{XY}(x,y) &= P_X(x)P_Y(y|x)\\ &= P_X(x)\int_0^\infty P_C(c)P_Z(y-c|x)dc\\ &= P_X(x)\int_0^\infty P_C(c)\frac1xP_R\left(\frac{y-c}{x}\right)dc\\ &= \frac{P_X(x)}{x}\int_0^\infty P_C(c)P_R\left(\frac{y-c}{x}\right)dc \end{align}$$ Therefore, $$ \frac{x\, P_{XY}(x,y)}{P_X(x)} = \int_0^\infty P_C(c)\,P_R\left(\frac{y-c}{x}\right)dc.$$ The right hand side is something like a convolution (not quite), and its value is known for every pair of x and y. How do I find $P_C$ and $P_R$? Any hints for analytical or numerical solution will be appreciated.

I am reposting this from StackExchange: https://math.stackexchange.com/q/2541446/491395

Edit: As Bjørn's answer below shows, we need more assumptions for this to work. Here is what I'm trying to do: I have measured the joint probability distribution of $X$ and $Y$ and it looks like this

enter image description here

Assuming a linear model with random slope and intercept works on this data, I want to find the distribution of these slopes and intercepts. Not sure, exactly what the necessary and sufficient conditions are for this to be possible.

Edit 2: Here is a very inefficient way to do this:

Consider the conditional expected value of $Y$ given $X$ $$ \newcommand\mean[1]{\left\langle{#1}\right\rangle} \mean{Y|X=x} = \mean{RX+C|X=x} = \mean R x +\mean C. $$ Using two values for $x$ we can find the mean values of $R$ and $C$. Now consider the second conditional moment $$ \mean{Y^2|X=x} = \mean{(RX+C)^2|X=x} = \mean{R^2} x^2 +\mean{C^2}+2\mean R\mean C x. $$ Again using two values of $x$ and the previously measured values of $\mean C$ and $\mean R$, we can find the second moments of $C$ and $R$. Inductively, we can find all the moments of $C$ and $R$.

Now, is there a cleaner, more efficient way to do this?

$\endgroup$
3
$\begingroup$

It's not possible.

Let $X$ be constant equal to 1.

Let $B_1,B_2,B_3$ be independent Bernoullis.

Let $R_1=B_1+B_2$, $C_1=B_3$.

Let $R_2=B_1$, $C_2=B_2+B_3$.

Then $R_1X+C_1=R_2X+C_2$. So even if you know the distribution of $Y$, it does not determine the distributions of $R$ and $C$.

$\endgroup$
  • 1
    $\begingroup$ Is this only true for constant $X$? This is like trying to fit a line to one point. What if we require a nonzero density for $X$ on an interval? $\endgroup$ – stochastic Nov 28 '17 at 21:59
  • 3
    $\begingroup$ That's a second question, I guess ;) $\endgroup$ – Bjørn Kjos-Hanssen Nov 28 '17 at 22:09
1
$\begingroup$

Suppose that the set $S:=\{x>0\colon P_X(x)\ne0\}$ has a nonempty intersection with each right neighborhood of $0$.
In terms of the characteristic functions (ch.f.'s) $f_C$ and $f_R$ of $C$ and $R$, your convolution equation means that $g_x(t):=f_C(t)f_R(xt)$ is known for all $x\in S$ and all real $t$. By the continuity of the ch.f., $f_R(xt)\to f_R(0)=1$ for each $t$ as $x\downarrow0$; this simple observation, that $\lim_{t\to0}f(t)=1$ for any ch.f. $f$ (together with the above condition on $S$), is crucial for the recovery of the distributions of $C$ and $R$. Indeed, now we have $f_C(t)=\lim_{x\in S,\,x\downarrow0}g_x(t)$, and hence $f_R(t)=g_x(t/x)/f_C(t/x)$ -- for any one $x\in S$ and all real $t$. Using these recovered ch.f.'s $f_C$ and $f_R$, it will then remain to use the inverse Fourier transform to determine the distributions of $C$ and $R$. (Of course, this does not contradict the counterexample given by Bjørn Kjos-Hanssen, since in that example $X$ takes only one positive value and the condition on the set $S$ stated in the beginning of this answer does not hold.)

$\endgroup$
  • $\begingroup$ Why is this answer not validated? $\endgroup$ – Fabrice Pautot Mar 9 '18 at 20:14

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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