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Let $$X = Y + Z,$$ where $Y,Z$ are independent continuous random variables. Can it happen that
$$ X \text{ and }\frac{Y}{X} \text{ are independent}? $$

If yes (I believe it can happen, but only in some weird degenerative case), can you somehow characterize such pairs of variables?

I was trying to work it out by conditioning. Write $W(a):= [Y\mid X=a]$ which is a random variable $Y$ conditioned on $X=a$. Then, $ W(a)/a \overset{d}{=} W(b)/b$ for all $a,b$ in the support of $X$. However, explicitly writing down a distribution of $W(a)$ is typically non-trivial. Some other ideas on how to proceed?

Albert Paradek
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  • I haven't gotten anywhere with this, but perhaps you can make use of it: If you introduce $C=Y/X$ , you can express $Y$ and $Z$ in terms of the independent variables $X$ and $C$, namely $Y=CX$ and $Z=X-CX$. Thus $CX\perp X-CX$, and also $CX\perp(C-1)X$. That looks hard to achieve (it's impossible if $C$ and $X$ are positive), but I haven't been able to derive a contradiction. – joriki Jan 17 '23 at 17:00
  • I also tried to work with the expectation values, e.g. $E[YZ]=E[Y]E[Z]$ yields $E[CX(X-CX)]=E[CX]E[X-CX]$, and those all factor because $C\perp X$, but I didn't get anything out of that, either. – joriki Jan 17 '23 at 17:04
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    This is known to hold for $Y,Z$ independent Gamma with a common rate/scale parameter: https://math.stackexchange.com/a/190695/321264, https://math.stackexchange.com/q/261337/321264 – StubbornAtom Jan 17 '23 at 17:34
  • The statement in my comment above that this is impossible if $C$ and $X$ are positive is obviously wrong – they're positive in the example given by@StubbornAtom. – joriki Jan 17 '23 at 17:58
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    The simplest example is $Y$ and $Z$ iid exponential distributions where $\frac{Y}{Y+Z} \sim U[0,1]$ both unconditionally and conditioned on $Y+Z$ – Henry Jan 20 '23 at 13:55

1 Answers1

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Some thoughts too long for a comment:

Let $X=Y+Z$ and $U=Y/(Y+Z)$.

The Jacobian for the transformation $(x, u) \mapsto (xu, x(1-u))$ is $\left|\det \begin{bmatrix} u & x \\ 1-u & -x \end{bmatrix}\right| = |x|$, so

$$f_{Y, Z}(y, z) \, dy \, dz = f_{Y, Z}(xu, x(1-u)) \left|\det \frac{\partial(y, z)}{\partial(x, u)}\right| \, dx \, du = |x|f_Y(xu) f_Z(x(1-u)) \, dx \, du,$$ where in the last step we have used the independence of $Y$ and $Z$ to factor the joint density. If we suppose $Y$ and $Z$ are nonnegative random variables, then we have $$f_{Y, Z}(y, z) \, dy \, dz = x f_Y(xu) f_Z(x(1-u)) \, dx \, du$$ where $(y, z) \in \mathbb{R}_{>0}^2$ and $(x, u) \in \mathbb{R} \times [0,1]$.

Thus, we need $f_Y$ and $f_Z$ to be such that $f_Y(xu) f_Z(x(1-u))$ can be written as $g(x) h(u)$. Passing to logarithms, this becomes $$\log f_Y(xu) + \log f_Z(x(1-u)) = \log g(x) + \log h(u).$$

  • If we impose that $\log f_Y(y) = -\lambda y + c$ and $\log f_Z(z) = -\mu z + c'$, we recover the constraint $\lambda=\mu$ and obtain $Y, Z \overset{i.i.d.}{\sim} \text{Exponential}(\lambda)$.
  • With the more general form $\log f_Y(y) = -\lambda y + (\alpha-1) \log y + c$ and $\log f_Z(z) = - \mu z + (\beta - 1) \log z + c'$, we again have the constraint $\lambda=\mu$ and recover the more general case of $Y \sim \text{Gamma}(\alpha, \lambda)$ and $Z \sim \text{Gamma}(\beta, \lambda)$.
angryavian
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  • You're very close to finishing the solution. (i) You can drop the $x$ factor in "$xf_Y(xu)f_Z(x(1-u))$ can be written as $g(x)h(u)$" by replacing $g$ with $g/x$. (ii) If you take logarithms, it's obvious how to achieve this. – J.G. Jan 17 '23 at 18:34
  • @J.G. Thanks for the pointers! Sorry, it is still not entirely obvious to me. If I impose certain guesses on the form of the log density, then I can recover certain examples, but it is not clear to me that I haven't missed other possibilities. – angryavian Jan 18 '23 at 02:19
  • Thanks angryavian for your ideas, it is clear to me now how one gets to the Gamma example. However, I am also curious, how one gets to the general case, if it is possible to somehow characterise the densities for which that logarithmic equality holds. The Gamma case comes quite naturally as an example, but @J.G. suggests that it can be adapted into somethnig nicer that describes all cases, can it? – Albert Paradek Jan 18 '23 at 10:17