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Let $\xi, \eta, \eta'$ be non-negative random variables such that:

  • $\eta \stackrel{\mathcal{L}}{=} \eta'$,
  • $\xi + \eta \stackrel{\mathcal{L}}{=} \xi + \eta'$,
  • $\xi$ and $\eta'$ are independent.

Does this imply that $\xi$ and $\eta$ are independent? Can one construct a counter-example? Any sort of reference would be of great help, too.

In terms of Laplace functions, does $\mathrm{E}\, e^{t(\xi+\eta)} = \mathrm{E}\, e^{t\xi} \mathrm{E}\, e^{t\eta}, \forall t<0$ imply $\mathrm{E}\, e^{t_1\xi + t_2\eta} = \mathrm{E}\, e^{t_1 \xi} \mathrm{E}\, e^{t_2 \eta}, \ \forall t_1, t_2 < 0$?

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up vote 8 down vote accepted

No, $\xi$ and $\eta$ need not be independent. For example let $\xi$, $\eta$, and $\eta'$ all be uniformly distributed on $\{1,2,3\}$ with the joint distribution of $\xi$ and $\eta$ is given by the matrix \[ P = \frac{1}{9}\begin{bmatrix} 1 & 2 & 0 \\ 0 & 1 & 2 \\ 2 & 0 & 1 \end{bmatrix}, \] i.e. $\mathbb{P}(\xi = i, \eta = j) = P_{ij}$. You can generate lots of discrete examples like this because your conditions are linear equations on the row sums, column sums, and anti-diagonal sums of $P$. Altogether this is only linearly many conditions on the quadratically many entries of $P$ (as a function of the size of the support of the variables).

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Thanks, that does it! – Anton Mar 17 '14 at 16:39

Let $\xi$ and $\eta'$ be two independent identically Cauchy distributed random variables (Lebesque density $\frac 1\pi \frac{1}{1+x^2}$) and put $\eta=\xi$ (which are trivially not indenpendent). Then $\xi+\eta=2\xi$ and $\xi+\eta'$ have the same distribution which follows by Fourier transformation: The characteristic function of $\xi$ is $\varphi_\xi(t)=e^{-|t|}$ and thus $\varphi_{\xi+\eta}(t)=\varphi_{2\xi}(t)= \varphi_\xi(2t)= e^{-2|t|} =e^{-|t|}e^{-|t|}= \varphi_\xi(t) \varphi_{\eta'}(t)=\varphi_{\xi+\eta'}(t)$.

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True, but non-negativity is missing. – Anton Mar 17 '14 at 15:41

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