Jtdylktuy

I have a variable, $X$, which is normally distributed

$$X sim mathcal{N}(mu, sigma^2)$$

I also have an event, $A$, which measures the probability that $X$ is greater than or equal to 0.
$$P[A] = P[X geq 0] = lim_{b to +infty} int_{x=0}^{x=b} f(x),dx quad text{f(x) is PDF of X}$$

I'd like to know what the derivative of $P[A]$ is w.r.t. $mu$ and $sigma^2$.

$$frac{partial P[A]}{partial mu} $$
$$frac{partial P[A]}{partial sigma^2}$$

I think this is really simple because I'm looking for the derivative of an integral, but would like to make sure I'm doing it right:

$$frac{partial P[A]}{partial mu} = frac{partial}{partial mu} lim_{b->+infty} f(b) - f(0) = frac{partial}{partial mu} f(0) $$
$$frac{partial P[A]}{partial sigma^2} = frac{partial}{partial sigma^2} lim_{b->+infty} f(b) - f(0) = frac{partial}{partial sigma^2} f(0)$$

where

$$f(x) = frac{1}{sqrt{2pisigma^2}}e^{-frac{(x-mu)^2}{2sigma^2}}$$

Are the above equations correct?

No you can't to do that.You're mixing up the fundamental theorem of calculus among other things. You could use Leibnitz integral rule ! Remember by definition of partial derivative :

$$ frac{partial }{partial mu} int_0^infty f^X(x,mu,sigma) dx = lim_{h to 0} int_0^infty frac{f^X(x,mu+h,sigma) - f^X(x,mu,sigma)}{h} dx$$

Under certain conditions you can pass the limit under the integral sign and you'll get

$$ frac{partial }{partial mu} int_0^infty f^X(x,mu,sigma) dx = int_0^infty frac{partial}{partial mu}(f^X(x,mu,sigma))dx.$$

Since $$ frac{partial}{partial mu}(f^X(x,mu,sigma)) = frac{1}{sqrt{2pisigma^2}}frac{-1}{2sigma^2}(2mu -2x)e^{-frac{(x-mu)^2}{2 sigma^2}} =
frac{1}{sqrt{2pisigma^2}}frac{x - mu}{sigma^2}e^{-frac{(x-mu)^2}{2 sigma^2}}$$

you'll get
begin{align*}
int_0^{infty}frac{partial}{partial mu}(f^X(x,mu,sigma)) dx &=
int_0^{infty}frac{1}{sqrt{2pisigma^2}}frac{x - mu}{sigma^2}e^{-frac{(x-mu)^2}{2 sigma^2}} dx. \
end{align*}

Change of variable $ t = frac{x- mu}{sigma^2} iff x = sigma^2 t + mu $ and "$dx = sigma^2 dt$ ", $x = 0 rightarrow t = - frac{mu}{sigma^2}$

begin{align*}
int_0^{infty}frac{partial}{partial mu}(f^X(x,mu,sigma)) dx &=
int_{-frac{mu}{sigma^2}}^{infty} frac{sigma^2}{sqrt{2pisigma^2}} t exp(-frac{t^2}{2}) dt. \
&= frac{sigma^2}{sqrt{2pisigma^2}} int_{-frac{mu}{sigma^2}}^{infty} t exp(-frac{t^2}{2}) dt\
&= frac{- sigma^2}{sqrt{2pisigma^2}} big[t exp (-t^2/2)big]_{-mu/sigma^2}^infty\
&= frac{- sigma^2}{sqrt{2pisigma^2}} Big( 0 - big( frac{- mu}{sigma^2}exp{frac{-mu^2}{2 sigma^4}}big) Big) \
&= frac{- mu}{sqrt{2pisigma^2}}exp{frac{-mu^2}{2 sigma^4}}.
end{align*}

Since $Z:=frac{X-mu}{sigma}simmathcal{N}(0,,1)$ has pdf $phi(z):=frac{1}{sqrt{2pi}}exp-frac{z^2}{2}$, $P(Xge 0)=1-int_{-infty}^{-mu/sigma}phi(z)dz$. Differentiation with respect to $mu$ gives $frac{1}{sigma}phi(-frac{mu}{sigma})$; differentiation with respect to $sigma$ gives $-frac{mu}{sigma^2}phi(-frac{mu}{sigma})$. In both cases, since $phi$ is even we can delete the $-$ sign in its argument.