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For me it sounds intuitive, but I am having a hard time to come up with a proper proof for the following statement (if it can be demonstrated).

Let $y$ and $c$ be two strictly positive random, stationary variables, not necessarily having the same pdf. How can I show that $\text{Var}(y \times c)$ increases by increasing the covariance $\text{Cov}(y , c)$ between both random variables $c$ and $y$ *? $\text{Var}[.]$ and $\text{Cov}[.]$ are the stochastic variance and covariance operators, as commonly referred to in the literature (e.g. Papoulis).

Intuition: if $y$ and $c$ share a positive covariance structure, larger (or smaller) values of $y$ will mainly correspond to larger (or smaller) values of $c$, and vice versa. Thus, by increasing the covariance between both variables the range for which the random variable formed by the multiplication $y \times c$ is defined will likely increase, due to the "strong" reinforcement of maxima of $c$ by large values of $y$, and the "weak" reinforcement of minima of $c$ by small values of $y$.

(*) = Assume this covariance is increased externally, by a mechanism that adds a linear dependence between $c$ and $y$, forcing, for example, $y = a\times c + b$, where $a$ and $b$ are constant weights. The degree of dependence can be "increased" by adjusting the weights $a$ and $b$. Remember, the random variables $c$ and $y$ are strictly positive.

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  • $\begingroup$ Please explain what you mean by the variance and covariance of a series. Several possibilities are natural. Are these series of numbers or series of data values? Are the variances and covariances underlying properties of a stochastic process or estimates from data? Are you assuming stationarity? $\endgroup$ – whuber Nov 17 '17 at 22:22
  • $\begingroup$ After your latest edit, the question appears to have been substantially changed and no longer is clear. Precisely how do you propose to "increase the covariance" of $c$ and $y$? By what mechanism will you do that? It would be well to take the idea behind Escachator's answer to heart: it shows that multiplying $c$ by $y$ does not "reinforce" the values of $c$ in any sense when $y$ is small; it does just the opposite. $\endgroup$ – whuber Nov 17 '17 at 23:57
  • $\begingroup$ Hello Whuber, I don't think it is relevant to the question the mechanism by which we increase the covariance between $c$ and $y$. I can specify further that it is up to us the degree of, say, linear dependence assumed by these two variables. Moreover, given we define the multiplied variable $yc$ beforehand, there is no need to bother with Escachator's point that $y$ (or $c$) can lie in $[0,1]$. Given the problem as is now, we are no longer concerned with the individual variance of $c$ (or $y$), but solely with the variance of the multiplied variable $yc$. $\endgroup$ – antamoeba Nov 18 '17 at 0:34
  • $\begingroup$ It's not only relevant, it's essential, because the answer depends on it. $\endgroup$ – whuber Nov 18 '17 at 0:35
  • $\begingroup$ C'mon Whuber, I get your point of writing the problem precisely. But, in many textbooks, we don't have to specify the dependence mechanism, for example, to show that var(x+y) = var(x) + var(y) + 2cov(x,y) increases with cov(x,y) given the assumption cov(x,y)>0. I didn't want to define the problem super detailed (which I did here before, and no one cared to read the whole problem).The point is to show how c and y relate assuming there's a mechanism which, somehow, creates a dependence between them that can be seen as a positive covariance. Anyway, OK, I will specify how this dependence is added $\endgroup$ – antamoeba Nov 18 '17 at 13:15
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I don't think so because I can find a counterexample: if the two random variables are always between 0 and 1 when you multiply them the result will be smaller than the two original random variables. Hence the variance of the result will be smaller (or equal).

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  • $\begingroup$ I think your argument might be false because you did not center the RVs. Can you write a more concrete example? $\endgroup$ – Yair Daon Nov 17 '17 at 23:20
  • $\begingroup$ What is a stationary variable? The following seems to be a counter example: Let X be N(0,1) and let Y be X/100. Clearly COV(X,Y)>0 but VAR(XY)<VAR(X). $\endgroup$ – Zahava Kor Nov 17 '17 at 23:34
  • $\begingroup$ Hello Escachator, thanks for your reply. I rewrote the original post, so as to account for this problem. I think now the original post, as is, captures the problem well. The two random variables are positive, can be between 0 and 1. The idea that bugs me is that the "reinforcement" of large and small values of $c$ by $y$ should be higher in case of a positive covariance. If $y$ lies in $[0,1]$ it still will weight "weakly" small values of $c$ and "strongly" large of values $c$. $\endgroup$ – antamoeba Nov 17 '17 at 23:37
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What is a stationary variable? The following seems to be a counter example: Let X be N(0,1) and let Y be X/100. Clearly COV(X,Y)>0 but VAR(XY) is smaller than VAR(X).

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  • $\begingroup$ Yes, that's what I wrote. $\endgroup$ – Escachator Nov 17 '17 at 23:38
  • $\begingroup$ Hello Zahava, thanks for your reply. I edited the original post. Roughly speaking, A stationary random variable (RV) is a RV with a invariant pdf (so its moments are constant). The OP, as is now, concerns the problem of increasing the dependence (considered as the covariance) between two RVs X and Y and checking a possible increase in variance in the resulting variable XY. The two RVs are already being multiplied, so the effect of between-zero-and-one variables is already taken into account. Thanks again for you reply. $\endgroup$ – antamoeba Nov 17 '17 at 23:40
  • $\begingroup$ No normal variate is "strictly positive" as required by the question. $\endgroup$ – whuber Nov 17 '17 at 23:58
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    $\begingroup$ Every RV is stationary! $\endgroup$ – Zahava Kor Nov 18 '17 at 0:00
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    $\begingroup$ Sorry - I did not notice the positive RVs. I can change the N(0,1) to Uniform(0,1) and the inequality still holds. Unfortunately, the question now changed. $\endgroup$ – Zahava Kor Nov 18 '17 at 0:06

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