Let $g(x)$ be the density of some random variable $x$. I am looking at the following expected value: $\int_a^b f(x, a) g(x) dx$.
I am curious what happens if I upwards shift the lower bound:
$$ \frac{d}{da}\int_a^b f(x, a) g(x) dx $$
I know that in general, this should have two effects: the effect of reducing the probability space (since $a$ appears in the boundary), and the direct effect of reducing $f(x, a)$ throughout the probability space. Per Leibniz:
$$ \frac{d}{da}\int_a^b f(x, a) g(x) dx = -f(a, a) g(a) + \int_a^b f_a(x, a) g(x) dx$$
Now, let's look at the specific case where $f(x, a) = x - a$. In this case, the first term becomes 0. It appears that the marginal effect of shrinking the probability space is zero if the function yields 0 when evaluated at the boundary. But then there is no effect of reducing the probability space? All the effect is coming from the change in $f_a(x, a)$?
Intuitively, I find it difficult to wrap my head around this. Could someone help me understand this result?
Consider the simpler case where $g$ is uniform with support containing $[a,b]$. Then your integral (up to a constant factor) is just
$$\int_a^b f(x,a)dx$$
Let $F$ be such that $F_1(x,a)=f(x,a)$. Then we have
$$\int_a^b f(x,a)dx=F(b,a)-F(a,a)=-F(a,a)+F(b,a) \tag{1}$$
The derivative with respect to $a$ is
$$\frac{\partial }{\partial a}\left(\int_a^b f(x,a)dx\right)=-F_1(a,a)+[F_2(b,a)-F_2(a,a)]=-f(a,a)+[F_2(b,a)-F_2(a,a)] \tag{2}$$
If you assume $f(a,a)=F_1(a,a)=0$, then that amounts to assuming that a marginal change in $x$ from $x=a$ has no effect on $F(x,a)$.
If it helps, consider the case where $a$ does not enter the integrand:
$$\int_a^bf(x)dx=F(b)-F(a)$$
Then $$\frac{\partial }{\partial a}\left(\int_a^b f(x)dx\right)=-F'(a)=-f(a)$$
If $f(a)=0$, then a marginal change in $a$ has no effect on the value of the integral $F(b)-F(a)$. This is simply because when $f(a)=0$, a marginal change in $x$ from $x=a$ has no effect on the value of $F(x)$ (as $a$ is a stationary point of $F$).