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I'm reading a text - Roger Koenker (2005) Quantile Regression [page 8] - that goes like this:

Consider the function $$R(\xi) = \sum_{i=1}^n \rho_\tau(y_i-\xi)$$ where $$\rho_\tau(y_i-\xi) =(y_i-\xi)(\tau - I[y_i-\xi<0])$$ If $\hat \xi$ solves $\min_\xi R(\xi)$ then the objective function must be increasing as one moves away from $\hat \xi$. This requires that the left and right derivatives are both non-negative at $\hat \xi$. Thus

$$R'(\xi+) := \lim_{h \rightarrow 0} \frac{R(\xi + h) - R(\xi)}{h} = \sum_i( I[y_i-\xi<0]- \tau ) $$ with $R'(\hat \xi+) \geq 0$ and

$$R'(\xi-) := \lim_{h \rightarrow 0} \frac{R(\xi - h) - R(\xi)}{h} = \sum_i (\tau - I[y_i-\xi<0] ) $$

with $R'(\hat \xi- )\geq 0$.

My question is: Shouldn't $R'(\xi+) = \sum_i( I[y_i-\xi\leq 0]- \tau )$ with a weak inequality in the indicator function rather than the strict?

Because for $y_i - \xi = 0$ the limit

$$ \lim_{h \rightarrow 0^+} \frac{(y_i-\xi-h)(\tau - I[y_i-\xi-h<0]) - (y_i-\xi)(\tau - I[y_i-\xi<0])}{h} = \\[8pt] \lim_{h \rightarrow 0^+} \frac{-h(\tau - I[-h<0]) }{h} = -(\tau - \lim_{h \rightarrow 0^+} I[-h < 0]) = 1-\tau$$ so the i'th summand in $R'(\xi+)$ should be $I[y_i-\xi\leq 0] - \tau$.

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