Convolution of a gaussian and the derivative of the inverse error function

Question

Problem Definition

I need to calculate the convolution between a Gaussian function $g(x)$ and $d(x)$, the derivative of the inverse error function,

$$d(x) * g(x) = \int_{-\infty}^{\infty} d(\tau) g(x-\tau) \text{d}\tau = \int_{-\infty}^{\infty} d(x-\tau) g(\tau) \text{d}\tau $$

with

$$g(x) = \frac{1}{\sqrt{2 \pi}} \text{e}^{-\frac{1}{2} x^2},$$

$$ d(x) = \left\{ \begin{array}{ll} 0 & x < a \\ \frac{\text{d}}{\text{d}x} erf^{-1}(x) & -a\leq x\leq a \\ 0 & x > a \\ \end{array} \right. $$

with the constant $a<1$.

Since $d(x)$ is only non-zero on the interval $x \in [-a, a] $ I assume I can take these as limits for the convolution integral instead of $[-\infty, \infty]$ (please correct me here).

Attempted solution:

Since $d(x)$ is defined as the derivative of a function, I thought about using integration by parts:

$$\int_{-a}^{a} \frac{\text{d}}{\text{d}\tau} erf^{-1}(\tau) g(x-\tau) \text{d}\tau = \Big[ erf^{-1}(\tau) g(x-\tau) \Big]_{-a}^{a} - \int_{-a}^{a} erf^{-1}(\tau) \frac{\text{d}}{\text{d}\tau} g(x-\tau) \text{d}\tau $$

Then, using the convolution-derivative theorem (see below), we see that the last term is equal to the negative original term,

$$ \frac{\text{d}}{\text{d}x} erf^{-1}(x) * g(x) = erf^{-1}(x) * \frac{\text{d}}{\text{d}x} g(x)$$

Moreover, $erf^{-1}(-x) = - er^{-1}(x)$ (property of an odd function).

Rearranging:

$$ 2 \frac{\text{d}}{\text{d}\tau} erf^{-1}(x) * g(x) = 2 erf^{-1}(a)\Big[g(x-a) + g(x+a) \Big],$$

so, (surprisingly!) the result is a sum of two Gaussian functions,

$$d(x)*g(x) = erf^{-1}(a)\Big[g(x-a) + g(x+a) \Big]$$

I believe this can not be correct, as the numerical convolution doesn't look like a sum of Gaussians at all (see attached plots). There has to be a mistake in my math! Can anyone please point it out to me? Also, if you have suggestions on how to attack the problem, let me know!

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Convolution-derivative theorem:

For any two functions $f_1(x)$ and $f_2(x)$, with ' denoting the derivative $\frac{\text{d}}{\text{d}x} $:

$$ ( f_1(x) * f_2(x) )' = f_1(x) * f_2'(x) = f_1'(x) * f_2(x) $$

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If it helps, I found that the derivative of the inverse funciton is defined as follows:

$$\frac{\text{d}}{\text{d}u} erf^{-1}(u) = \frac{1}{\frac{\text{d}}{\text{d}u} erf\Big(erf^{-1}(u)\Big)}, $$

with

$$\frac{\text{d}}{\text{d}u} erf(u) = \frac{2}{\sqrt{\pi}} \text{e}^{-x^2}. $$

This would give

$$d(x) = \frac{\sqrt{\pi}}{2} \text{e}^{\Big(\text{erf}^{-1}(u) \Big)^2}, $$

and the following integral for the convolution,

$$ d(x) * g(x) = \frac{\sqrt{\pi}}{2} \frac{1}{\sqrt{2 \pi}} \int_{-\infty}^{\infty} \text{e}^{\Big(\text{erf}^{-1}(\tau) \Big)^2} \text{e}^{-\frac{1}{2} (x-\tau)^2} \text{d}\tau. $$

It looks difficult to attach directly, which is why I'm guessing playing with the derivatives might be the way to go.

Answer 1

I think the convolution-derivative theorem assumes that your functions vanish at the boundary, so in that sense where you are mistaken is in $$\int_{-a}^{a} \frac{\text{d}}{\text{d}\tau} erf^{-1}(\tau) g(x-\tau) \text{d}\tau = \Big[ erf^{-1}(\tau) g(x-\tau) \Big]_{-a}^{a} - \int_{-a}^{a} erf^{-1}(\tau) \frac{\text{d}}{\text{d}\tau} g(x-\tau) \text{d}\tau$$ which - after changing the $\tau$-derivative against the negative $x$-derivative - just becomes your convolution-derivative theorem when you neglect the boundary term. So I think you can not apply this theorem here in that form or you have to assume $\pm \infty$ as boundary in which case it vanishes. I think you have a sign error also, since you add these terms up to 2 times, but they actually cancel.

Also note that the integral as you wrote it above actually diverges for $a=1$, so you have to either keep track of the boundary term or use the derivative with respect to $x$ on the Gaussian.

In any case for the integration of the original problem I doubt you will find closed form. But you can get some approximation for $a\rightarrow 1$ by first substituting $t={\rm erf}^{-1}(\tau)$ and then integrating by parts \begin{align} &\quad \, \, \int_{-a}^{a} {\rm erf}^{-1}(\tau) \, \frac{{\rm e}^{-\frac{(x-\tau)^2}{2}}}{\sqrt{2\pi}} \, {\rm d}\tau \tag{1} \\ &= \int_{-{\rm erf}^{-1}(a)}^{{\rm erf}^{-1}(a)} \frac{{\rm e}^{-\frac{(x-{\rm erf}(t))^2}{2}}}{\sqrt{2\pi}} \, \frac{2t}{\sqrt{\pi}} \, {\rm e}^{-t^2} \, {\rm d}t \\ &= -\frac{{\rm e}^{-\frac{(x-{\rm erf}(t))^2}{2}}}{\sqrt{2\pi}} \, \frac{ {\rm e}^{-t^2}}{\sqrt{\pi}} \Bigg|_{-{\rm erf}^{-1}(a)}^{{\rm erf}^{-1}(a)} - \frac{2}{{\pi}} \frac{{\rm d}}{{\rm d}x} \int_{-{\rm erf}^{-1}(a)}^{{\rm erf}^{-1}(a)} \frac{{\rm e}^{-\frac{(x-{\rm erf}(t))^2}{2}}}{\sqrt{2\pi}} \, {\rm e}^{-2t^2} \, {\rm d}t \, . \end{align}

Approximating ${\rm erf}(t) \approx {\alpha t}$ with $\alpha=\frac{2}{\sqrt{\pi}}$ the latter integral becomes \begin{align} &\quad \,\, \frac{2}{\pi} \int_{-{\rm erf}^{-1}(a)}^{{\rm erf}^{-1}(a)} \frac{{\rm e}^{-\frac{(x-{\rm erf}(t))^2}{2}}}{\sqrt{2\pi}} \, {\rm e}^{-2t^2} \, {\rm d}t \\ &\approx \frac{2}{\pi} \int_{-{\rm erf}^{-1}(a)}^{{\rm erf}^{-1}(a)} \frac{{\rm e}^{-\frac{\left(x-{\alpha t}\right)^2}{2}}}{\sqrt{2\pi}} \, {\rm e}^{-2t^2} \, {\rm d}t \\ &=\frac{{\rm e}^{-\frac{2 x^2}{4+\alpha^2}}}{\pi\sqrt{4+\alpha^2}} \left\{ {\rm erf}\left( \frac{x \alpha + T(4+\alpha^2)}{\sqrt{2(4+\alpha^2)}} \right) - {\rm erf}\left( \frac{x \alpha - T(4+\alpha^2)}{\sqrt{2(4+\alpha^2)}} \right) \right\} \end{align} where $T={\rm erf}^{-1}(a)$. For $a=1$ this yields $$ \sim \frac{2}{\pi} \frac{{\rm e}^{-\frac{2 x^2}{4+\alpha^2}}}{\sqrt{4+\alpha^2}} \tag{2} \, . $$ The result fits qualitatively, but underestimates a little bit for $\alpha=\frac{2}{\sqrt{\pi}}$. In fact by choosing $\alpha=\frac{1.685}{\sqrt{\pi}}$ you get an almost perfect match. I doubt you get much closer with any other approximation. In the below figure you don't see the difference between $-\frac{{\rm d}}{{\rm d}x} (2)$ and the exact result of (1).

For the derivative of the inverse error function as in your question you finally need to derive (1) with respect to $x$ another time as in the convolution-derivative theorem.