 See for a roundup.

A common activity for me at the moment is differentiating the integral - for example, through the inverse-CDF lookup.

You see, what I would really like is the derivative of the mass-preserving continuous map $$\phi_{\theta, \tau}$$ such that

$\mathsf{z}\sim F(\cdot;\theta) \Rightarrow \phi_{\theta, \tau}(\mathsf{z})\sim F(\cdot;\tau).$ Now suppose I wish to optimise or otherwise perturb $$\theta$$. This gives me a way of continuously parameterising a change in measure with respect to a realisation, and I can differentiate with respect to the parameterisation at

$\left.\frac{\partial}{\partial \tau} \phi_{\theta, \tau}(\mathsf{z})\right|_{\tau=\theta}$ Let us say I need to differentiate through a monte carlo algorithm to alter its parameters while holding the PRNG fixed.

See the reparameterization trick for a way of making this, more or less, feasible for a class of suitably smooth nonparametric neural net problems by transforming it into a slightly different problem. But what if I am not doing some fluffy non-parametric thing but instead a using real and specific parametric distributions. What to do?

How can I get the derivative of such a map? I can look for candidates for the map. Here I can imagine that our observed rv $${\mathsf{x}}\in \mathbb{R}$$ is generated via lookups from its iCDF $$F(\cdot;\theta)$$ with parameter $$\theta$$:

$\mathsf{x} = F^{-1}(\mathsf{u};\theta)$

where $$\mathsf{u}\sim\operatorname{Uniform}(0,1)$$. Each realization corresponds to a choice of $$u_i\sim \mathsf{u}$$ independently.

But maybe I generated my original variable not by the icdf method but by simulating some variable $${\mathsf{z}}\sim F(\cdot; \theta).$$ In which case I may as well have generated those $$\mathsf{u}_i$$ by taking $$\mathsf{u}_i=F(\mathsf{z}_i;\theta)$$ for some $$\mathsf{z} \sim F(\cdot;\theta)$$ and I am conceptually generating my RV by fixing $$z_i\sim\mathsf{z}_i$$ and taking $$\phi := F^{-1}(F(z_i;\theta);\tau).$$ So to find the effect of my perturbation what I actually need is

\begin{aligned} \left.\frac{\partial}{\partial \tau} F^{-1}(F(z;\theta);\tau)\right|_{\tau=\theta}\\ \end{aligned}

Does this do what we want? Kinda. So suppose that the parameters in question are something boring, such as the location parameter of a location-scale distribution, i.e. $$F(\cdot;\theta)=F(\cdot-\theta;0).$$ Then $$F^{-1}(\cdot;\theta)=F^{-1}(\cdot;0)+\theta$$ and thus

\begin{aligned} \left.\frac{\partial}{\partial \tau} F^{-1}(F(z;\theta);\tau)\right|_{\tau=\theta} &=\left.\frac{\partial}{\partial \tau} F^{-1}(F(z-\theta;0);0)+\tau\right|_{\tau=\theta}\\ &=\left.\frac{\partial}{\partial \tau}\left(z-\theta+\tau\right)\right|_{\tau=\theta}\\ &=1\\ \end{aligned}

OK grand that came out simple enough.

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