# Can I apply Fundamental Theorem of Calculus?

What are the assumptions that I need on the function $P( X | Y = y )$ (where $P$ is the conditional probability density function of X knowing an observation y of Y - so is bounded by one and always positive) in order to be able to write the following?

$$\frac{\partial }{\partial \tau} \left( \int_{-\infty}^{\tau} P( X = x | Y = y ) \right) = P( X = \tau | Y = y ).$$

The partial derivative $$\frac{\partial}{\partial x}\int_{-\infty}^{x}f_{X|Y}(x^\prime|y) \mathrm{d}x^\prime=P(X=x|Y=y)=f_{X|Y}(X=x|Y=y)$$ always if the conditional distribution $F_{X|Y}(x|y)=\int_{-\infty}^{x}f_{X|Y}(x^\prime|y) \mathrm{d}x^\prime$ is differentiable at every $X=x, \forall y \in \Omega_y$; which simply means that the Conditional CDF has to be well defined to be differentiable everywhere. At those places where the derivatives do not exist, the distribution $f_{X|Y}$ cannot be defined as we want to and hence does not exist.
• I understand, so there is no need to consider the fact that the integral is from $-\infty$ and so there is a limit in there? Oct 17, 2013 at 7:38
• No it's totally ok because any PDF to be valid should have $\int_{-\infty}^{+\infty}f_X \mathrm{d}x=1$. i.e. The real line is the support for any PDF where in $f_X$ can be zero at some places or can even be left undefined (which will lead to weird cases like the Random variable cannot take that value at all). Oct 17, 2013 at 19:43
• It boils down to the Differentiability of the CDF $F_X(x)$. If it is continuous and differentiable everywhere $f_X(x)$ is well defined. If $F_X(x)$ is continuous but not differentiable at a few places, then at those places the PDF is undefined i.e. $X$ cannot take those values at all. If $F_X(x)$ is not continuous at a place then it means that there is a probability mass for $f_X(x)$ at that place and it is defined using the well known $\delta$ function; so it's still a well defined point in terms of the PDF. Oct 17, 2013 at 19:47