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Let $X$ and $Y$ be two continuous real random variables with common support $(0,x_{\max}]$ and with PDF $f_X(x)$ and $f_Y(y)$. Assume that $\Pr [Y\geq\beta \mid X<\beta] \leq k$ and that $\Pr [Y<\beta \mid X\geq\beta] \leq k$ for any $\beta$ in the support of $X$ and $Y$, where $0 < k < 1$ is a constant. Consider function $Z(X) = \log(1+X)$.

What can we say on mean and variance of $Z(Y)$ based on the moments of $Z(X)$? Exact expressions, bounds?

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  • $\begingroup$ Could you please describe a bit about the background of this problem? $\endgroup$
    – Chee
    Commented Mar 13, 2016 at 4:36
  • $\begingroup$ Consider $Z(X)$ as a function of random variable $X>0$. Imagine that we are estimating random variable $X$ by another random variable, called $Y$, such as the false alarm and misdetection rates become less than $k$ w.r.t. any reference point $\beta$. We are interested to investigate distribution of $Z(Y)$ w.r.t. parameters of the distribution of $Z(X)$. This abstract problem shows itself in many applications such as wireless communications. In this problem, I asked for a simple example of $Z(X)=\log⁡(1+X)$ and characterizing average of $Z(Y)$ based on distribution of $Z(X)$. $\endgroup$
    – Jeff
    Commented Mar 13, 2016 at 9:07

2 Answers 2

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You'd have better luck bounding the median of $Z(Y)$, or other quantiles and L-moments.

Let $Q_X$ and $Q_Y$ be the quantile functions for $X$ and $Y$. Then $Q_Y(p)$ is between $Q_X((p-k)/(1-k))$ and $Q_X(p/(1-k))$, and similarly for the $Z$'s.

As an example, say $k$ is 10%. Then the median of log($Y$) is between the logs of the 44th and 56th percentiles of $X$.

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Besides some trivial lower bound because of the range constraint, there is not much you can say here. I can let $Y$ be $X$ with probability $> k$ for each $X$ value and anything else otherwise. There is a lot of freedom no matter how close $k$ is to $1$.

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  • $\begingroup$ not clear to me how we can let $Y=X$ with probability higher $k$ for each $X$. $\endgroup$
    – Jeff
    Commented Mar 14, 2016 at 22:59
  • $\begingroup$ Define Y = X 1{Z < k} + W 1{Z>=k}, where Z is a uniform [0,1], X, W, Z mutually independent. By manipulating W you can get arbitrarily large moments for Y regardless of moments of X. $\endgroup$
    – John Jiang
    Commented Mar 15, 2016 at 2:02
  • $\begingroup$ $Y=X \, \mathbb{1}_{Z<k} + W \, 1_{Z \geq k}$ will not necessarily ensure both $\Pr[Y\geq \beta \mid X<\beta] \leq k$ and $\Pr[Y< \beta \mid X \geq \beta] \leq k$. $\endgroup$
    – Jeff
    Commented Mar 16, 2016 at 20:45
  • $\begingroup$ My Y satisfies P(Y > X|X=x) < k for all little x. Now integrate over $x < \beta$. Same holds for the other inequality. What's your counterexample? $\endgroup$
    – John Jiang
    Commented Mar 16, 2016 at 23:32
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    $\begingroup$ Can you first start by assuming $(X,Y)$ is bivariance Gaussian? Then you can look at a bivariate Lancaster family; then your problem relates to extremal points in the corresponding family. Without a joint distribution on $(X,Y)$, how can one get exact expression? Think about it this way: you are tying to expand a function in terms of the moments of another random variable. $\endgroup$
    – Chee
    Commented Mar 25, 2016 at 23:10

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