There is a probability density function that depends on non-deterministic ($v$) and random ($x$) parameters:
$Pr(v)=\int_{G(v)} dP(v)$,
where $G (v)$ is the "goal" region, the probability of getting into it must be determined and that, in turn, depends on $v$ and $x$.
We consider the normal distribution function $P(v)=N(0,v)$:
$Pr(v)=\frac{1}{\sqrt{2\pi v}}\int_{g(v)}^{m(v)} e^{-\frac{(x)^2}{2v}}dx$
So that by replacing $a=\frac{1}{\sqrt{2\pi v}}$ and $f(x,v)=e^{-\frac{(x)^2}{2v}}$ we have:
$Pr(v)=a\int_{g(v)}^{m(v)} f(x,v)dx$
We assume that f,g, and m are smooth functions. To show that $Pr(v)$ is smooth we need to show that the derivatives of all orders exist. I know how to do it for 1 variable case via Fundamental theorem of calculus: $Pr(v)=a\int_{g(v)}^{m(v)} f(x)dx$, so that $\frac{d}{dx} a\int_{g(v)}^{m(v)} f(x)dx=a(-g'(x)f(g(x))+m'(x)f(m(x)))$, where -g'(x) and m'(x) are smooth, f(g(x)) and f(m(x)) are also smooth and this holds for all derivatives of higher order so $Pr(v)$ is also smooth.
MY QUESTION IS: How to show the same thing for two variables case?
I will really appreciate your help!
Using Leibniz integral rule we can rewrite the derivative of the integral as:
$\frac{d}{dv}(a\int_{g(v)}^{m(v)} f(v,x)dx)=a(f(v,m(v))\frac{d}{dv}m(v)-f(v,g(v))\frac{d}{dv}g(v)+\int_{g(v)}^{m(v)} \frac{d}{dv}f(v,x)dx)$
From our assumptions, we can conclude that $f(v,m(v))\frac{d}{dv}m(v)-f(v,g(v))\frac{d}{dv}g(v)$ part is smooth by definition of the smooth functions.
So the only term we care is $\int_{g(v)}^{m(v)} \frac{d}{dv}f(v,x)dx$, which is also should be smooth.