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?


2 Answers 2


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 $C$.

  • 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, 2017 at 21:59
  • 3
    $\begingroup$ That's a second question, I guess ;) $\endgroup$ Nov 28, 2017 at 22:09

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.)

  • $\begingroup$ Why is this answer not validated? $\endgroup$ Mar 9, 2018 at 20:14

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