If $X \sim \mathcal{N}(\mu, \sigma^2)$, what is the derivative of the probability that $X \geq 0$ w.r.t. $\mu$ and $\sigma$

Question

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

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

Are the above equations correct?

Answer 1

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

you'll get \begin{align*} \int_0^{\infty}\frac{\partial}{\partial \mu}(f^X(x,\mu,\sigma)) dx &= \int_0^{\infty}\frac{1}{\sqrt{2\pi\sigma^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{2\pi\sigma^2}} t \exp(-\frac{t^2}{2}) dt. \\ &= \frac{\sigma^2}{\sqrt{2\pi\sigma^2}} \int_{-\frac{\mu}{\sigma^2}}^{\infty} t \exp(-\frac{t^2}{2}) dt\\ &= \frac{- \sigma^2}{\sqrt{2\pi\sigma^2}} \big[t \exp (-t^2/2)\big]_{-\mu/\sigma^2}^\infty\\ &= \frac{- \sigma^2}{\sqrt{2\pi\sigma^2}} \Big( 0 - \big( \frac{- \mu}{\sigma^2}\exp{\frac{-\mu^2}{2 \sigma^4}}\big) \Big) \\ &= \frac{- \mu}{\sqrt{2\pi\sigma^2}}\exp{\frac{-\mu^2}{2 \sigma^4}}. \end{align*}

Answer 2

Since $Z:=\frac{X-\mu}{\sigma}\sim\mathcal{N}(0,\,1)$ has pdf $\phi(z):=\frac{1}{\sqrt{2\pi}}\exp-\frac{z^2}{2}$, $P(X\ge 0)=1-\int_{-\infty}^{-\mu/\sigma}\phi(z)dz$. 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.