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Let $U$ denote a random variable with PDF $$f_U(u)=ce^{-u\sqrt{u}}\,\mathbf 1_{u>0}$$ Does there exist a random variable $W$ independent of $U$ such that $U+W$ is distributed like $2U$?

This question is related to something I asked here.

In other words, can we find a random variable $W$ such that \begin{align} 2 U' =U+W \end{align} where $W$ and $U$ are independent and $U'$ has the same distribution as $U$.

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Boby
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  • It is the same question as is $2U_1 - U_2$ independent of $U_2$ ?. I think it suffices to take $U_1$ and $U_2$ independent to have a counter example. –  Dec 19 '16 at 14:49
  • @Gribouillis thanks for you comment. Perhaps you can elaborate more by putting your thoughts as an answer. – Boby Dec 19 '16 at 15:11
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    Now that you received an answer to your other question, did you try to apply it here? – Did Dec 19 '16 at 17:11
  • @Gribouillis Of course assuming that $U_1$ should be independent of $U_2$ is putting such a severe constraint on the problem that it becomes without solution. But this is not the question asked at all... – Did Dec 19 '16 at 17:12
  • @Did then the question should be Is it possible to find a variable $W$ independent from $U_2$ such that $(W+U_2)/2$ follows the same law ? As the question is formulated, the answer is no in general (that is to say in general, given 2 variables following this law, there is no W). Of course, in general is not defined :) –  Dec 19 '16 at 17:24
  • @Gribouillis The question is clearly: Let $U$ with PDF $f_U$, does there exist $W$ independent of $U$ such that $U+W$ is distributed like $2U$? – Did Dec 19 '16 at 17:26
  • @Did Adding an independent variable to obtain a certain law is indeed very different from adding an independent variable to obtain a certain variable, this point could be clarified by the OP. –  Dec 19 '16 at 17:32
  • @Did would this work take $f(x)=\frac{1}{\sqrt{x}}$ this a convex function. Then by the answer here $E[ f(2 U_1)] \ge E[ f(U_2)]$ this implis that $\frac{1}{\sqrt{2}} \ge 1$. Which leads to a contradiction? – Boby Dec 19 '16 at 18:21
  • Not directly since you would need to appeal to $f(x)$ for $x>0$ and for $x<0$ and $f$ is not convex on any open interval containing $0$. – Did Dec 19 '16 at 21:56
  • @Did is there a better choice of $f(x)$. Because I don't think that $f(x)=|x|^p, p\ge1$ will work. – Boby Dec 19 '16 at 21:59
  • You should first center your variables, meaning that you would have $$E\big[f\big(2(U_1 - E[U_1])\big)\big]\ge E\big[f(U_2 - E[U_2])\big].$$ Try $f(x) = (x-c)^+= \max(x-c,0)$ for different $c$. – zhoraster Dec 20 '16 at 06:25
  • @zhoraster I tried the function that you suggested by I do no get a contradiction. – Boby Dec 20 '16 at 18:38
  • Well, if the inequality is true for each $c$, then there is the convex dominance, so you should try other approaches. – zhoraster Dec 20 '16 at 18:47
  • @zhoraster but is in it also possible that such a $W$ exists? – Boby Dec 20 '16 at 18:49
  • I doubt it does. By the way, it seems that for a centered $U$, $U\preceq_{cv} 2U$ always. So my suggestion was doomed to fail. Sorry for that. – zhoraster Dec 20 '16 at 18:59
  • @zhoraster No problem thanks for the ideas. Do you have an intuition for why such a $W$ might not exist. – Boby Dec 20 '16 at 19:41
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    Yes, I do have an intuition. It is, however, just an intuition, so hard to explain. It may as well be wrong. Speaking of ideas, there is another approach based on Laplace transforms. It is spiritually related to characteristic functions. I'll update my answer elsewhere. By the way, you didn't explain your motivation yet. – zhoraster Dec 20 '16 at 19:51
  • Boby: Why the interest for the PDF on $u>0$ proportional to $\exp(-u\sqrt{u})$? – Did Dec 21 '16 at 18:04
  • @Did sorry, I didn't understand what you meant? – Boby Dec 21 '16 at 18:23
  • Really? Again: Why are you interested in this specific PDF? – Did Dec 21 '16 at 18:27
  • @Did There is no particular reason. I am trying to generate a few examples to see how to approach this question: http://math.stackexchange.com/questions/1877093/existence-of-additive-transformation-of-random-variables The exponential seemed like a good choice. – Boby Dec 21 '16 at 18:34
  • @Did To motivation this example, I add some more details and examples to the question I asked before: http://math.stackexchange.com/questions/1877093/existence-of-additive-transformation-of-random-variables – Boby Dec 23 '16 at 14:45
  • I doubt many people would call this motivation, this is more a near duplicate question... – Did Dec 23 '16 at 17:50
  • @Did I was wondering if you could take a look at this question that I recently asked I give any suggestion on how to show it. The question is here: http://math.stackexchange.com/questions/2155030/e-sqrt1-t-v-sqrtt-w-sqrt1-t-vz-1-sqrtt-wz-2-e-sqrt1 – Boby Feb 23 '17 at 01:46
  • @Boby What is your source for it? – Did Feb 23 '17 at 05:48
  • @Did In communication theory, this comes up in the context of something called maximum ratio combining. The basic ideas that it is enough to look at a linear combination of outputs of two antennas instead of each output individually. However, I think this only true when signals are assumed to be Gaussian. I wanted to see that this only optimal if random variables a Gaussian. Sorry, if this is too brief, I will be happy to fill in more details if you want. It would be great if you could point me in some direction of how to approach this problem. – Boby Feb 23 '17 at 19:57
  • @zhoraster you mentioned that you had an idea of using Laplace transform. Do you still think it will work? – Boby Feb 24 '17 at 22:50

1 Answers1

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In general the answer is such a $W$ doesn't exist.

To see this let assume $U_1$ and $U_2$ are independent and $W$ exists. Then $$U_2 = 2U_1 - W$$ but $2U_1 - W$ is independent of $U_2$ by assumption so $U_2$ is independent of itself hence $U_2$ is constant what is a contradiction.

So such a $W$ cannot exist.

Gono
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  • It seems that your argument applies to any $U_1$ and $U_2$ and does not depend on the properties of the pdf. But then we can take $U_1$ and $U_2$ to be standard normal and we can have $2U_1=W+U_2$ by taking $W \sim \mathcal{N}(0, 1)$. – Boby Dec 19 '16 at 15:58
  • The argument applies to any independent $U_1$ and $U_2$. If $W$ and $U_2$ follow ${\cal N}(0,1)$ and are independent, then $U_1=(W+ U_2)/2$ follows the same law but it is not independent from $U_2$. –  Dec 19 '16 at 17:12
  • Gono: Please see comments to main. – Did Dec 19 '16 at 17:27
  • Note: The question has been edited since this answer was written; the question once contained $U_1$ and $U_2$ like this answer. – Jeppe Stig Nielsen Jan 17 '17 at 17:29