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Let $X\geq0$, $\eta\geq0$ and $X,\eta$ independent. We measure $X$ with a one-sided error: $\widetilde{X} = X - \eta$.

Is $E[X|\widetilde{X}=\widetilde{x}]$ increasing in $\widetilde{x}$?

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  • $\begingroup$ I was reading $\eta$ as a parameter (in accordance with statistical convention, and for which the answer is trivial), but then it makes no sense to declare it as independent of $X$ (since it's a fixed quantity). ... so just to confirm, $\eta$ is another random variate? [If so, why not stick to the convention of uppercase Roman letters?] $\endgroup$ – Glen_b Sep 6 '18 at 7:19
  • $\begingroup$ Yes, $\eta$ is a random variable. $\eta$ looks like an n as in noise and is typically chosen in the context of measurement error, e.g. $\eta$ on Wikipedia. $\endgroup$ – robust Sep 6 '18 at 22:59
  • $\begingroup$ Ah, thanks, in the context of an error term it makes sense. Which variable are you taking expectation over in $E[X|\widetilde{X}=\widetilde{x}]$? I guess this is $E_X[X|\widetilde{X}=\widetilde{x}]$ but did you intend $E_\eta[X|\widetilde{X}=\widetilde{x}]$ (or perhaps something else) instead? $\endgroup$ – Glen_b Sep 6 '18 at 23:26
  • $\begingroup$ In this case it doesn't matter what variable the expectation is taken over, because conditioning on $\widetilde{X}$ implies that $X$ and $\eta$ determine each other. $\endgroup$ – robust Sep 7 '18 at 4:55
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Assuming the expectation is over $X$:

$E_X[X|\widetilde{X}=\widetilde{x}]=E_X[X|X-\eta=\widetilde{x}]=E_X[X|X=\widetilde{x}+\eta]=\widetilde{x}+\eta$,

which is not increasing in $\widetilde{x}$ because of the $\eta$.

However, $E_\eta[E_X(X|\widetilde{x})]=\widetilde{x}+E(\eta)$ would (naturally) be increasing in $\widetilde{x}$.

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  • $\begingroup$ Can you clarify what you mean by "because of the $\eta$"? Also, why is $E_\eta[E_X(X|\widetilde{x})] = \widetilde{x}$? $\endgroup$ – robust Sep 7 '18 at 4:47
  • $\begingroup$ The second thing is an error on my part (now corrected): I took something you'd refer to as a noise term to be mean $0$ but it's not in your case. With the respect to the first thing, because $\eta$ is a random variable; you could increase $\tilde{x}$ but $\tilde{x}+\eta$ might not increase with it. $\endgroup$ – Glen_b Sep 7 '18 at 8:44
  • $\begingroup$ I don't follow $E_\eta[E_X(X|\widetilde{x})]=\widetilde{x}+E(\eta)$: Taking $E_\eta$ on the outside expectation implies that the conditioning variable $\widetilde{X}$ in the inside expectation also varies, and hence the right-hand side cannot contain a simple $\widetilde{x}$. On the other hand, $E_{\eta|\widetilde{X}=\widetilde{x}}[E_X(X|\widetilde{x})]=\widetilde{x} + E(\eta|\widetilde{X}=\widetilde{x})$, which by itself doesn't provide a definite answer to the question. $\endgroup$ – robust Sep 7 '18 at 17:30
  • $\begingroup$ "Taking Eη on the outside expectation implies that the conditioning variable X˜ in the inside expectation also varies," ... -- I don't think so $\endgroup$ – Glen_b Sep 8 '18 at 2:13
  • $\begingroup$ Take the setup from my answer, and let $\widetilde{x}=4$. In that setup $E[\eta]=1$, and hence $\widetilde{x} + E[\eta]=5$, which is larger than any value in the support of $X$, $\eta$ and $\widetilde{X}$. $\endgroup$ – robust Sep 8 '18 at 3:19
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No, not in general.

Let $P(X=3)=P(X=4)=P(\eta=0)=P(\eta=2)=0.5$.

Then,

  • $E[X|\widetilde{X}=1] = 3$
  • $E[X|\widetilde{X}=2] = 4$
  • $E[X|\widetilde{X}=3] = 3$
  • $E[X|\widetilde{X}=4] = 4$

which is neither increasing nor decreasing in $\widetilde{x}$.

If we modify the original question, and require $X$ and $\eta$ to have continuous or common support, the counterexample remains valid if sufficent probability mass is put on above values.

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