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Lehmann, in Theory of Point Estimation p.212, defines scale median as the solution to: $${E(X)I(X\le c)} = {E(X)I(X\ge c)}$$

given $X$ is a positive random variable, and ${E(X)}< \infty$.

Now based on this: $${E(X)I(X\le c)} = \int_0^c x\,f(x)\,dx \le c $$ and thus conversely, $${E(X)I(X\ge c)} = \int_c^\infty x\,f(x)\,dx \ge c $$

Therefore, the equality will never hold unless $X$ is a discrete point at $c$ w.p. 1.

So what's wrong with my reasoning?

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Again, the definition of the scale median is not what Lehman writes and should be modified. – Did Feb 28 '13 at 7:30
Thanks! I saw how you defined scale median. Much better and easier to understand than how Lehmann did. Thanks again! – RVC Feb 28 '13 at 7:52
You are welcome. As I already mentioned, what you write is NOT what Lehmann and Casella did. – Did Feb 28 '13 at 13:43

"Conversely" is wrong and misses the fact that one uses two inequalities to get an upper bound of $\mathbb E(X;X\leqslant c)$, namely:

(1) If $x\leqslant c$, then $xf(x)\leqslant cf(x)$.

(2) $\displaystyle\int_{-\infty}^cf(x)\mathrm dx\leqslant1$.

This approach fails with $\mathbb E(X;X\geqslant c)$ because the inequality in (1) is reversed while the inequality in (2) is not, to wit:

(1') If $x\geqslant c$, then $xf(x)\geqslant cf(x)$.

(2') $\displaystyle\int_c^{+\infty}f(x)\mathrm dx\leqslant1$.

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Got it! As a quick example, if $f(x)=0$ for say, $c \le x \le +\infty$, the integral would be 0, and not within the range. Thanks! – RVC Feb 28 '13 at 7:49

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