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When $x$ and $y$ are two iid random variables, I want show if $$E(x\mid x>y)<1$$ can be determined:

[1] without knowledge of the distribution of the random variables; and

[2] if $x$ and $y$ have standard normal distribution

Moreover, I also want to show whether $$E(x\mid x>c)<E(1\mid x>c)$$ where $c$ is some constant.

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up vote 1 down vote accepted

There is no chance that any universal bound on $E(x\mid x\gt y)$ can hold: try $x$ and $y$ with support included in $[a,b]$, then $E(x\mid x\gt y)$ belongs to $[a,b]$ as well. If $a\geqslant1$, then $E(x\mid x\gt y)\lt1$ is wrong. Thus:

[1] does not hold in general.

If $x$ and $y$ are i.i.d. standard normal, then $s=x+y$ and $t=x-y$ are i.i.d. centered normal hence $E(x\mid x\gt y)=\frac12E(s+t\mid t\gt0)=\frac12(E(s)+E(t\mid t\gt0))$. Since $s$ is centered and $t$ has variance $2$, $E(x\mid x\gt y)=\frac12E(t\mid t\gt0)=\frac12\sqrt2E(x\mid x\gt0)$. Since $E(x;x\gt0)=1/\sqrt{2\pi}$ and $P(x\gt0)=\frac12$, one gets $E(x\mid x\gt y)=1/\sqrt\pi$. Thus:

[2] holds, namely, $E(x\mid x\gt y)\lt1$ when $x$ and $y$ are i.i.d. standard normal, but maybe not for the reasons you imagined.

The third inequality you are interested in is equivalent to $E(x;x\gt c)\lt P(x\gt c)$. Once again, there is no chance this can hold in general since, for example, $E(x;x\gt c)\geqslant c P(x\gt c)$. Thus:

[3] does not hold in general.

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