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Let $X,Y\sim N(m,1)$ be identically distributed random variables, $m>0$. An observation $(\hat X,\hat Y)$ is obtained from $(X,Y)$ by one of two rules: $(\hat X,\hat Y)=(\lambda X,Y)$ or $(X,\lambda Y)$ for $0<\lambda<1$. One needs to know which rule was applied more likely.

There is a simple answer (an analog of standard scores): the first rule was applied more likely if $\hat X<\hat Y$ and the second rule was applied otherwise. We prove it analytically in the following case. Let $Y=-\sqrt{1-\sigma^2}X+Z$, where $Z\sim N(m(1+\sqrt{1-\sigma^2}),\sigma^2)$ and $X,Z$ are independent. Assume that $\hat X<\hat Y$. Then we calculate $$p((\hat X,\hat Y)|{\rm the\,\, first\,\, rule\,\, is\,\, used})= p(\lambda X<Y)=\Phi\Bigl(\frac{(1-\lambda)m}{\sqrt{\lambda^2+2\lambda\sqrt{1-\sigma^2}+1}}\Bigl)$$ Here $\Phi$ is the standard CDF. Similarly $$p((\hat X,\hat Y)|{\rm the\,\, second\,\, rule\,\, is\,\, used})= p(X<\lambda Y)=\Phi\Bigl(\frac{(\lambda-1)m}{\sqrt{\lambda^2+2\lambda\sqrt{1-\sigma^2}+1}}\Bigl)$$ Clearly the first expression is greater than the second.

This answer is counterintuitive if we know that $X<m$. Indeed, $X=m$ is where $X=Y$ and OLS line $Y=-\sqrt{1-\sigma^2}X+m(1+\sqrt{1-\sigma^2})$ meet. Then for $X<m$ we will have $\hat X<\hat Y$ (if $\lambda$ is small and $X$ is not close to $m$).

My question is how to prove it analytically. One can try to compare $p((\hat X,\hat Y)|{\rm the\,\, first\,\, rule\,\, is\,\, used\,\, and\,\,} \hat X<m)$ and $p((\hat X,\hat Y)|{\rm the\,\, second\,\, rule\,\, is\,\, used\,\, and\,\,} \hat X<m)$ but the calculation there seems to be too complicated. Is there a simpler way to show that the criterium $\hat X<\hat Y$ does not work in this case?

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    $\begingroup$ Please tell us (1) whether the value of $\lambda$ is known or not; (2) whether the value of $m$ is known or not; and (3) whether $X$ and $Y$ are also assumed to be independent (or even bivariate Normal). $\endgroup$
    – whuber
    Commented Jul 28, 2018 at 18:24
  • $\begingroup$ @whuber Thanks. $\lambda$ and $m$ are known. $X$ and $Y$ are not independent as one can see from the relation between $X$, $Y$ and $Z$. $\endgroup$
    – 8k14
    Commented Jul 28, 2018 at 20:48
  • $\begingroup$ Please don't ask your readers to figure out the question by studying your attempt at an answer: be explicit. $\endgroup$
    – whuber
    Commented Jul 28, 2018 at 20:50
  • $\begingroup$ @whuber I am sorry $\endgroup$
    – 8k14
    Commented Jul 29, 2018 at 14:07

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