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Suppose we have a random variable $W$ and we want to transform it to a random variable $V$ by using additive transformation, as follows \begin{align} V=U+W \end{align} where $U$ is independent of $W$.

My question is: We are interested in knowing whether such random variable $U$ exists or not and what methods can we used to check this.

Known method: For me, the most obvious way is via characteristic functions.

\begin{align} \phi_V(t)=\phi_U(t) \cdot \phi_W(t) \to \phi_U(t)=\frac{\phi_V(t)}{\phi_W(t)} \end{align} and we have to check if $\phi_U(t)$ is a proper characteristic function.

Motivation: In order to show that there is no such transformation all we need is to check that $\phi_U$ violates one of the properties of characteristic function. For example, by showing that $|\phi_U(t)|>1$ for some $t$.

To show that such a transformation indeed exists one has to check that $\phi_U$ is a proper charcterstic function, which can be done via the following set of theorems (see here).

Unfortunately, these theorems can be difficult to check and I was wondering if there is another method (not through characteristic functions) that can be used to prove or disprove existence of such transformations.

For me, the more important direction is how to disprove the existence, which I guess amounts to showing some necessary condition and then checking if it holds.

Proposed Example

I know that the question I am asking can be fairly difficult to answer. So it might be a good idea to focus on a specific example:

\begin{align} 2V=U+W \end{align} where \begin{align} V &\sim c_ve^{-v^{1.5}}, v\ge0\\ W & \sim c_We^{-w^{1.5-\epsilon_W}}, w\ge0 \end{align} for some $\epsilon_W \in (-\infty, 1.5)$ and \begin{align} c_W= \frac{1.5-\epsilon_W}{\Gamma \left(\frac{1}{1.5-\epsilon_W}\right)} \end{align}

Why did I pick this example: On the one hand, computing anaylytic expression for the ChF's of this example might be impossible and motivates searching for new methods. On the other hand, becuse pdf has an exponential which usually has some nice properties, make me think that there might be hope of answering this question.

To solve the above question we distinguess three case: $\epsilon_W <0$, $\epsilon_W =0$, $1.5\ge \epsilon_W >0$.

Solution for $1.5\ge \epsilon_W >0$. Via a method outlined in one of the anwer we have that for any $n$ \begin{align} 2 \ge \left(\frac{E[V^n]}{E[W^n]} \right)^{1/n}. \end{align} For $1.5\ge \epsilon_W >0$ we can show that the RHS of the above equation can be made as larege as possible by choosing appropriate $n$ and we reach a contradiction.

So, for the case of $1.5\ge \epsilon_W >0$ transformation $2V=U+W$ is impossible.

At this point, however, I do not know how to apporach the case of $\epsilon_W <0$ and the case of $\epsilon_W =0$.

To avoid a really long question. The case of $\epsilon_W =0$ was asked here.

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    $\begingroup$ It would be nice if you provide some context explaining why are you interested in this question and other related things you ask alsewhere. $\endgroup$ – zhoraster Dec 19 '16 at 16:21
  • $\begingroup$ @zhoraster Motivation: I was studying characteristic functions and noticed that while product of characteristic functions is always a characteristic function the ratio of characteristic functions might not be a characteristic function. This question is related to the deconvolution. So, I started wondering how can one check when a function is or is not a characteristic function. I found many theorems that give necessary and sufficient conditions but are generally difficult to check. So, in order to learn more about this I wanted to see some method how to show that a function is not a chf. $\endgroup$ – Boby Dec 20 '16 at 20:04
  • $\begingroup$ 1. It's better to edit the original post adding the motivation. 2. There's a Bochner-Khintchine criterion for a function to be chf. But the positive definiteness is often harder to check than the complete monotonicity. $\endgroup$ – zhoraster Dec 20 '16 at 22:06
  • $\begingroup$ @zhoraster I added some motivation and an example to the question above $\endgroup$ – Boby Dec 23 '16 at 14:44
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In the case if both variables are integrable, there is another simple necessary condition. Let us deduce it.

Assume wlog that $\mathsf E[V]=\mathsf E[W]=0$ (otherwise you can subtract the expectations). Then for any convex function $f$, by Jensen's inequality, $$ \mathsf E[f(V)] = \mathsf E\big[f(U+W)\big]= \mathsf E\big[\mathsf E[f(U+W)\mid W\big]\big]\\ \ge \mathsf E\big[f(\mathsf E[U+W\mid W])\big] = \mathsf E[f(W)]. $$ Therefore, if the variables are centered, $V$ must dominate $W$ in convex stochastic order.


Another approach is to consider Laplace transforms, which are exactly completely monotone functions. Let $r(t)$ be the ratio of Laplace transforms. If there exist some $t_0$ and $n$ such that the $(-1)^nr^{(n)}(t_0)<0$, then there is no such $U$; moreover, this is a criterion.

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  • $\begingroup$ Can you explain what I am doing wrong with the following argument. Take $f(x)=\sqrt{\frac{1}{|x|}}$ which is a convex function. Also take $U$ and $W$ to be standard gaussian. The by your inequality we have that $E[\sqrt{\frac{1}{|V|}} ] \ge E[\sqrt{\frac{1}{|W|}} ]$. This implies that $(1/2)^{1/4} \ge 1$ that no such transformation exists. However, we can transform $V=U+W$. $\endgroup$ – Boby Dec 19 '16 at 18:33
  • $\begingroup$ Never mind. $f(x)=\sqrt{\frac{1}{|x|}}$ is no really convex. $\endgroup$ – Boby Dec 19 '16 at 20:07

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