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Suppose that $X$ is a random variable and $\frac{d\mathbb{Q}}{d\mathbb{P}}$ is the Radon-Nikodym derivative. The quantity under consideration is as follows:

\begin{equation} Cov(X, \frac{d\mathbb{Q}}{d\mathbb{P}}) \end{equation}

My question is here: Under which conditions we can say that the above quantity is positive or negative?

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    $\begingroup$ This is too generic. Unless there's something specific to those variable, you're basically asking when the integral of a product of 2 functions, one non-negative is positive. $\endgroup$
    – LazyCat
    Jun 16, 2021 at 12:33
  • $\begingroup$ You can suppose that X is a positive random variable. The reason I am asking this question is here. I want to show when the following inequality holds: \begin{equation} \mathbb{E}^{\mathbb{Q}}[X] < \mathbb{E}^{\mathbb{P}}[X] \end{equation} where $\mathbb{Q}$ and $\mathbb{P}$ are two equivalent measures. We know that there is a relation between $\mathbb{Q}$ and $\mathbb{P}$ via Radon-Nikodym derivative. In other words, using the change of measures technique, we have that $\endgroup$
    – user53249
    Jun 16, 2021 at 12:42
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    $\begingroup$ I guess the question is: is $X$ traded (or attainable)? Any traded asset should yield the risk-free rate under the risk-neutral measure. On the other hand, the risk-free rate is the smallest possible return in an economy (otherwise there would be arbitrage, this is trivial). If the asset is risky, then its return under the physical measure $\mathbb{P}$ should be greater than the risk-free one. Hence, if $X$ is a traded asset, we always have $\mathbb{E}^\mathbb{Q}(X)<\mathbb{E}^\mathbb{P}(X)$. $\endgroup$
    – Daneel Olivaw
    Jun 16, 2021 at 13:08
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    $\begingroup$ If the random variable $X$ represents the loss linked to mortality risks or catastrophe risks (which have nothing to do with the asset price), then is it possible to say something similar? $\endgroup$
    – user53249
    Jun 16, 2021 at 21:50
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    $\begingroup$ @AntoineConze This is sort of an intuitive assumption I think and boils down to the concept of risk-aversion. In the physical world, investors are risk averse, and thus by definition will not purchase an asset with higher (non-zero in this case) volatility than the risk free asset unless $\mathbb{E}^\mathbb{P}[X] > r$. I.e. you will only take on risk if you have some feeling that it is worth it to take on risk. $\endgroup$
    – rubikscube09
    Jun 17, 2021 at 14:21

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