5
$\begingroup$

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?

$\endgroup$
2
  • $\begingroup$ Could you please describe a bit about the background of this problem? $\endgroup$
    – Chee
    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
    Mar 13, 2016 at 9:07

2 Answers 2

1
$\begingroup$

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 $X$.

$\endgroup$
-1
$\begingroup$

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 $1$.

$\endgroup$
6
  • $\begingroup$ not clear to me how we can let $Y=X$ with probability higher $k$ for each $X$. $\endgroup$
    – Jeff
    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
    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
    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
    Mar 16, 2016 at 23:32
  • 1
    $\begingroup$ 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
    Mar 25, 2016 at 23:10

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.