What does $dF(y)$ mean?

Question

What does $dF(y)$ mean?

Sorry for the silly question. To put it into context. I am trying to figure out how to do solve this problem but I'm not sure how and I think my understanding of notation might be stopping me. Also, if anyone knows the step to solve this I would also appreciate that.

$$ 0 = ( 1 - \tau ) \int _ { - \infty } ^ { \hat { x } } d F ( x ) - \tau \int _ { \hat { x } } ^ { \infty } d F ( x ) = F ( \hat { x } ) - \tau $$

Answer 1

Kavi Rama Murthy is correct. So when the integrand is $1$ as in the OP we simiplify $$ \int _ { - \infty } ^ { \hat { x } } d F ( x ) = F(\hat(x))-F(-\infty) = F(\hat{x}) \\ \int _ { \hat { x } } ^ { \infty } d F ( x ) = F(+\infty) - F(\hat{x}) = 1-F(\hat{x}) $$ so that $$ ( 1 - \tau ) \int _ { - \infty } ^ { \hat { x } } d F ( x ) - \tau \int _ { \hat { x } } ^ { \infty } d F ( x ) = (1-\tau)F(\hat{x}) - \tau\;(1-F(\hat{x})) =F(\hat{x}) - \tau $$

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Note: If $F$ is not continuous at the point $\hat{x}$, then one of those integrals will infolve a one-sided limit of $F$ at $\hat{x}$, and this answer will be wrong.

Answer 2

Since this involves Probability Theory I assume that $F$ is a distribution funciton. There is a unique probability measure $\mu$ on the Borel sets of $\mathbb R$ such that $\mu (-\infty,x]=F(x)$ for all $x$ and the notation $\int h(x)dF(x)$ stands for $\int hd\mu$.

Answer 3

What is $F$ in your exercise? If $F$ is a cumulative probability distribution, then an expression of the form $\int_a^b g(x)dF(x)$ is a Riemann-Stieltjes integral (https://en.wikipedia.org/wiki/Riemann%E2%80%93Stieltjes_integral).

For instance, $\int_a^{b} dF(x) = F(b) - F(a)$.

In the particular case that $F$ is absolutely continuous and its density probability distribution is $f$, then $\int_a^b g(x)dF(x) = \int_a^b g(x)f(x)dx$.