# Definition of Scale median

Lehmann, in Theory of Point Estimation p.212 (and also on p.169), 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 ${E(X)}< \infty$.

However, the expression does not make sense as it ${E(X)I(X\le c)}$ and ${E(X)I(X\ge c)}$, are two random variables whose support intersect only at c. Hence a solution can not exist in general. Am I missing something here? Should it be defined as the solution to $${E(XI(X\le c))} = {E(XI(X\ge c))},$$ instead?

• See this question answered here: math.stackexchange.com/questions/316644/… Oct 21, 2014 at 19:49
• @TrynnaDoStat : I did see that, but are the definitions of ${E(X)I(X\le c)}$ and ${E(X)I(X\ge c)}$ in the question not wrong? The OP there seems to use these quantities as numbers, while they are random variables. Oct 21, 2014 at 19:54

3.7 Let $X$ be a positive random variable. (a) If $EX < ∞$, then the value of $c$ that minimizes $E|X/c − 1|$ is a solution to $EXI (X ≤ c) = EXI (X ≥ c)$, which is known as a scale median.
Note carefully that there are no parentheses separating $X$ and $I (X ≤ c)$ or enclosing them. This is confusing notation alright, and uncharacteristic of the specific book where in other cases of the use of the $E$ operator, parentheses, brackets and curly brackets are used (some authors do write $EX$ instead of $E(X)$, but not here).
Taking clues from Did's answer and corrective comments in the math.SE linked thread, as well as here, it is indeed $E\big[XI (X ≤ c)\big]$, since Did uses $E(X;X\leq c)$ which is equal to $E\big[XI (X ≤ c)\big]$, being the expected value restricted in the set $\{X\leq c\}$. The right-hand side integrals are also equal to $E\big[XI (X ≤ c)\big]$ and its counterpart, indicating just a typo from the part of the OP there.