Vtyjkyuk

As the title says, I am trying to find the derivative of the inverse cumulative distribution function for the standard normal distribution. I have this figured out for one particular case, but there is an extra layer of complexity that has be stumped.

Let $0 le p le 1$ and let $z = Phi^-1(p)$, where $Phi^-1(p)$ is the inverse cumulative distribution function for the standard normal distribution. Then: $$fracpartial Phi^-1(p)partial p = left(fracpartial Phi(z)partial zright)^-1,$$ where $Phi(z)$ is the cumulative distribution function for the standard normal distribution. This yields: $$= left(frac1sqrt2pi exp(-z^2/2) right)^-1 = fracsqrt2piexp(-z^2/2).$$

I think/hope this is right so far.

But now I have $p_1$ and $p_2$ and I need to find the derivative of $$fracpartial Phi^-1left(fracp_1p_1+p_2right)partial p_1$$ and $$fracpartial Phi^-1left(fracp_1p_1+p_2right)partial p_2.$$ Any help would be appreciated.

This can be done using chain rule, note that

$$
f(f^-1(x)) = x
$$
and therefore
$$
fracddxf(f^-1(x)) = fracddxx = 1 hspace5mm (1)
$$
and also by chain rule
$$
fracddxf(f^-1(x)) = f'(f^-1(x)) cdot (f^-1)'(x) hspace5mm (2)
$$

So by (1) and (2) we have
$$
beginalign
f'(f^-1(x)) cdot (f^-1)'(x) &= 1 \
(f^-1)'(x) &= frac1f'(f^-1(x))
endalign
$$
So in your case, with $f = Phi$ we have $(Phi^-1)'(x) = frac1Phi'(Phi^-1(x))$

where for a standard normal distribution

$$
Phi^-1(x) = sqrt2erf^-1(2x-1)
$$
and $Phi'(x)$ is just the pdf of a standard normal, i.e.
$$
Phi'(x) = frac1sqrt2pie^-fracx^22
$$

$Phi:mathbb R to (0,1)$ and $Phi^-1:(0,1) to mathbb R$ are strictly increasing continuous bijective functions

If $z=Phi^-1(p)$ then $p=Phi(z)$ and $dfracdpdz=Phi^prime(z)=phi(z)$, so $dfracdzdp = dfrac1phi(z)= dfrac1phi(Phi^-1(p))$