Derivative of conditional expectation on an interval

Question

Let the random variable $X$ have a bounded and strictly positive density $f\left( x\right)$ for all $x\in % \left[ \underline{x},\bar{x}\right] $. Consider the conditional expectation $% \mathbb{E}\left( X|X\in \left[ a,b\right] \right) $ for some $a$ and $b$ satisfying $\underline{x}<a<b<\bar{x}$ (think of $a$ and $b$ as two close numbers). Then is it true that \begin{equation} \frac{\partial }{\partial b}\mathbb{E}\left( X|X\in \left[ a,b\right] \right) =0.5+O\left( b-a\right) \end{equation} where $O\left( \cdot \right) $ stands for the Big O notation? I believe that the answer is yes, but is there a standard reference to this result? If I'm wrong, are there simple conditions under which this is true?

Answer 1

Since $$ \mathbb{E}[X \mid X \in [a, b]] = \frac {1} {F(b) - F(a)}\int_a^b xf(x)dx $$

Hence

$$ \frac {\partial} {\partial b} \mathbb{E}[X \mid X \in [a, b]] = \frac {\displaystyle f(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\}} {[F(b) - F(a)]^2} \triangleq \frac {g(b)} {h(b)} $$

where $g, h$ are defined as the numerator and denominator of the above derivative respectively. We compute the derivatives of $g, h$:

$$ \begin{align} g'(b) &= f'(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\} + f(b)\left\{F(b) - F(a) + bf(b) - bf(b)\right\} \\ &= f'(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\} + f(b)\left[F(b) - F(a)\right] \end{align}$$

$$ \begin{align} g''(b) =&~ f''(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\} + f'(b)\left[F(b) - F(a)\right] \\ & + f'(b)[F(b) - F(a)] + f(b)^2 \\ =&~ f''(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\} + 2f'(b)\left[F(b) - F(a)\right] + f(b)^2 \end{align}$$

$$ \begin{align} g^{(3)}(b) =&~ f^{(3)}(b)\left\{b[F(b) - F(a)] - \int_a^b xf(x)dx \right\} + f''(b)\left[F(b) - F(a)\right] \\ & + 2f''(b)[F(b) - F(a)] 2f'(b)^2 + 2f(b)f'(b) \\ \end{align}$$

$$ h'(b) = 2[F(b) - F(a)]f(b)$$

$$ h''(b) = 2\{f(b)^2 + [F(b) - F(a)]f'(b)\}$$

$$ h^{(3)}(b) = 2\{2f(b)f'(b) + f(b)f'(b) + [F(b) - F(a)]f''(b)\} = 2\{3f(b)f'(b) + [F(b) - F(a)]f''(b)\} $$

As

$$g(a) = h(a) = 0, g'(a) = h'(a) = 0, g''(a) = f(a)^2, h''(a) = 2f(a)^2$$ $$g^{(3)}(a) = 2f(a)f'(a), h^{(3)}(a) = 6f(a)f'(a)$$

by L'Hôpital's rule,

$$ \lim_{b\to a} \frac {g(b)} {h(b)} = \lim_{b\to a} \frac {g'(b)} {h'(b)} = \lim_{b\to a} \frac {g''(b)} {h''(b)} = \frac {g''(a)} {h''(a)} = \frac {f(a)^2} {2f(a)^2} = \frac {1} {2} $$

Similarly, $$ \begin{align} &~\lim_{b\to a} \frac {\displaystyle \frac {g(b)} {h(b)} - \frac {1} {2}} {b - a} \\ =&~ \lim_{b\to a} \frac {g'(b)h(b) - g(b)h'(b)} {h(b)^2} \\ =&~ \lim_{b\to a} \frac {g''(b)h(b) - g(b)h''(b)} {2h(b)h'(b)} \\ =&~ \lim_{b\to a} \frac {g^{(3)}(b)h(b) + g''(b)h'(b) - g'(b)h''(b) - g(b)h^{(3)}(b)} {2[h(b)h''(b) + h'(b)^2]} \\ =&~ \lim_{b\to a} \frac {g^{(4)}(b)h(b) + 2g^{(3)}(b)h'(b) - 2g'(b)h^{(3)}(b) - g(b)h^{(4)}(b)} {2[h(b)h^{(3)}(b) + 3h'(b)h''(b)]} \\ =&~ \lim_{b\to a} \frac {g^{(5)}(b)h(b) + 3g^{(4)}(b)h'(b) + 2g^{(3)}(b)h''(b) - 2g''(b)h^{(3)}(b) - 3g'(b)h^{(4)}(b) - g(b)h^{(5)}(b)} {2[h(b)h^{(4)}(b) + 4h'(b)h^{(3)}(b) + 3h''(b)^2]} \\ =&~ \frac {g^{(3)}(a)h''(a) - g''(a)h^{(3)}(a)} {3h''(a)^2} \\ =&~ \frac {2f(a)f'(a)2f(a)^2 - f(a)^26f(a)f'(a)} {12f(a)^4} \\ =&~ - \frac {f'(a)} {6f(a)} \end{align}$$

The above calculation base on the assumption that $f$ is strictly positive in the support, i.e. $f(a) > 0$. And thus we have

$$ \frac {\partial} {\partial b} \mathbb{E}[X \mid X \in [a, b]] \approx \frac {1} {2} - \frac {f'(a)} {6f(a)} (b - a) + O((b-a)^2)$$

as $b \to a$.