# Recurrence differential equations arising from the Normal PDF

Let $$a,b,c:\mathbb{R}\rightarrow\mathbb{R}$$ be differentiable functions, with $$a(t)\rightarrow -\infty$$ and $$c(t)\rightarrow -\infty$$ as $$t\rightarrow -\infty$$, and with $$a(t)\rightarrow\infty$$ and $$c(t)\rightarrow\infty$$ as $$t\rightarrow\infty$$.

Let $$\phi:\mathbb{R}\rightarrow\mathbb{R}^+$$ be the density of a standard normal random variable, meaning that for all $$z\in\mathbb{R}$$: $$\phi(z)=\frac{1}{\sqrt{2\pi}} \exp{\left(-\frac{z^2}{2}\right)}.$$

For $$j\in\mathbb{N}$$, let $$z_j:\mathbb{R}\rightarrow\mathbb{R}$$ with: $$z_j(t)=\int_{-\infty}^t{\exp{\left(a(\tau)-a(t)\right)}\,\left(b(\tau)+c(t)-c(\tau)\right)^j\,\phi\left(b(\tau)+c(t)-c(\tau)\right)\,d\tau},$$ for all $$t\in\mathbb{R}$$, where you may assume the integral exists and is finite for all $$j\in\mathbb{N}$$ and $$t\in\mathbb{R}$$.

Finally, let $$z_{-1}:\mathbb{R}\rightarrow\mathbb{R}$$.

Then from differentiating $$z_j(t)$$ (in $$t$$), we have that for $$j\in\mathbb{N}$$ and $$t\in\mathbb{R}$$: $$z_j'(t)=\left(b(t)\right)^j\,\phi\left(b(t)\right)-a'(t)z_j(t)+c'(t)\left(jz_{j-1}(t)-z_{j+1}(t)\right),$$ where, as usual, dashes denote first derivatives.

I want to derive a differential equation for $$z_0(t)$$ that does not depend on $$z_1(t),z_2(t),\dots$$. Is there a way to solve the recurrence equation (holding $$t$$ fixed) here to leave a differential equation? Or is there some other direct way to get a differential equation for $$z_0(t)$$?

The solution should be somehow related to the recurrence relation for $$z_j$$ ($$j\in\mathbb{N}$$): $$0=\beta^j\varphi-\alpha z_j+\gamma\left(jz_{j-1}-z_{j+1}\right).$$ When $$\phi=0$$ or $$\beta=0$$, and $$\alpha=-\gamma$$, this gives: $$z_{j+1}=z_j+jz_{j-1},$$ which is the sequence here and here (the number of involutions on a finite set) for appropriate initial conditions.