The half-normal distribution is a special case of the folded normal distribution. If $X$ follows an ordinary normal distribution $N(0, \sigma^{2})$, then $Y = |X|$ follows a half-normal distribution.

The probability density function associated with a half-normal distribution $Y$ is is given by: $$f_{Y} (y, \sigma) = \frac{\sqrt{2}}{\sigma \sqrt{\pi}} e^{- \frac{y^{2}}{2 \sigma^{2}}} \quad .$$

Let $H(0, \sigma^{2})$ denote the half-normal distribution. Let $Y_{1} \sim H(0, \sigma_{1}^{2}) $ and $Y_{2} \sim H(0, \sigma_{2}^{2})$. I'd like to figure out what the probability density function is that is associated with the sum of these two random variables: $Z = Y_{1} + Y_{2}$. In other words, I want to compute the convolution $h(z) = f_{Z} (z) = (f_{Y_{1}} * f_{Y_{2}}) (z)$.

Here's what I tried so far:

$\begin{equation} \begin{split} h(z) & = (f_{Y_{1}} * f_{Y_{2}}) (z) \\ &= \int_{- \infty}^{\infty} f(t) g(z-t) dt \\ & = \int_{- \infty}^{\infty} \Big{(} \frac{1}{\sigma_{1}} \sqrt{\frac{2}{\pi}} e^{-\frac{t^{2}}{2 \sigma_{1}^{2}}} \Big{)} \Big{(} \frac{1}{\sigma_{2}} \sqrt{ \frac{2}{\pi} } e^{- \frac{(z-t)^{2}}{2 \sigma_{2}^{2}}} \Big{)} dt \\ &= \int_{- \infty}^{\infty} \frac{1}{\sigma_{1} \sigma_{2}} \cdot \frac{2}{\pi} e^{- \frac{t^{2}}{2 \sigma_{1}^{2}}} e^{- \frac{z^{2}-2t + t^{2}}{2 \sigma_{2}^{2}}} dt \\ &= \int_{- \infty}^{\infty} \frac{1}{\sigma_{1} \sigma_{2}} \cdot \frac{2}{\pi} e^{-\frac{1}{2} \big{(}\frac{\sigma_{1}^{2}z^{2} + (\sigma_{1}^{2} + \sigma_{2}^{2})t^{2} - 2 \sigma \sigma_{1}^{2} t}{\sigma_{1}^{2} \sigma_{2}^{2}} \big{)}} dt \\ &= \frac{1}{\sigma_{1} \sigma_{2}} \cdot \frac{2}{\pi} e^{-\frac{1}{2} \big{(} \frac{\sigma_{1}^{2} z^{2}}{\sigma_{1}^{2} \sigma_{2}^{2}} \big{)}} \int_{\ -\infty}^{\infty} e^{- \frac{1}{2} \big{(} \frac{(\sigma_{1}^{2}+\sigma_{2}^{2})t^{2} - 2 \sigma_{1}^{2}t}{\sigma_{1}^{2} \sigma_{2}^{2}} \big{)}} dt \qquad (1) \\ &= \frac{1}{\sigma_{1} \sigma_{2}} \cdot \frac{2}{\pi} e^{-\frac{1}{2} \big{(} \frac{\sigma_{1}^{2} z^{2}}{\sigma_{1}^{2} \sigma_{2}^{2}} \big{)}} \Bigg{[} \frac{\sqrt{\frac{\pi}{2}} \sigma_{1} \sigma_{2} e^{\frac{\sigma_{1}^{2}}{2\sigma_{2}^{2}(\sigma_{1}^{2}+\sigma_{2}^{2})}}erf\big{(} \frac{\sigma_{1}^{2}(t-1) + \sigma_{2}^{2}t}{\sqrt{2}\sigma_{1} \sigma_{2}} \big{)}} \qquad {\sqrt{\sigma_{1}^{2} + \sigma_{2}^{2}}} \Bigg{]}^{t = \infty}_{t=-\infty} \qquad (2) \\ &= \frac{2 \sqrt{\frac{2}{\pi}}}{\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}} e^{- \frac{1}{2} \big{(} \frac{z^{2}}{\sigma_{2}^{2}} - \frac{\sigma_{1}^{2}}{\sigma_{2}^{2} ( \sigma_{1}^{2} + \sigma_{2}^{2})} \big{)}} \\ &= \frac{2 \sqrt{\frac{2}{\pi}}}{\sqrt{\sigma_{1}^{2}+\sigma_{2}^{2}}} e^{- \frac{1}{2} \big{(} \frac{z^{2}(\sigma_{1}^{2} + \sigma_{2}^{2})}{\sigma_{2}^{2}(\sigma_{1}^{2} + \sigma_{2}^{2})} + \frac{\sigma_{1}^{2}}{\sigma_{2}^{2}(\sigma_{1}^{2}+\sigma_{2}^{2})} \big{)}} \qquad (3) \end{split} \end{equation} .$

In $(2)$, $erf(\cdot)$ is the so-called error function. I have to admit I did not calculate the integral in $(1)$ (which yielded equality $(2)$) myself; I did it with Wolframalpha (see the calculation here, with $\sigma_{1}$ and $\sigma_{2}$ changed to $a$ and $b$ respectively for convenience).

After having arrived at equality $(3)$, I don't know how to proceed anymore. If $\frac{\sigma_{1}^{2}}{\sigma_{2}^{2}(\sigma_{1}^{2} + \sigma_{2}^{2})} $ in the exponent would have been multiplied by $z^{2}$, then it seems I would get close to showing that $Z \sim H(0, \sigma_{1}^{2} + \sigma_{2}^{2})$, but the whole calculation does not support the validity of this extra multiplication (and there's still the extra factor $2$ to be dealt with).

I have a number of questions:

  1. Did I make a mistake somewhere in my calcuations? Or should I perhaps have done something a bit different in one or more steps?
  2. Did Wolframalpha perhaps make a mistake in calculating equation $(1)$ ?

  3. Is there perhaps an (other) way to calculate the convolution of two half-normal distributions? Perhaps by means of Laplace/Fourier transforms and the convolution theorem? Can you show it over here directly or can you point me towards an external reference?


I'm late to the party, but intrigued.

The convolution of two half-normal densities is not a half-normal density. Intuitively, this is because there isn't enough "mass" (in fact, there is no mass) on the left-hand side of zero to contribute to the convolution. In particular, the mass of the convolution around zero "slumps". An explicit computation yields a term with error functions that accounts for this "slump" about zero (plot it and see!). $$ (f_1 * f_2)(y) = 2\sqrt{\frac{\sigma^2 + \tau^2}{2\pi}} \exp\Big(\frac{-y^2}{2(\sigma^2 + \tau^2)}\Big) \Big( \mathrm{erf}\big(\frac{y\tau}{\sqrt{2\sigma^4 + 2\sigma^2\tau^2}}\big) + \mathrm{erf}\big(\frac{y\sigma}{2\sigma^2\tau^2 + 2 \tau^4}\big)\Big) $$ Indeed, from the note following (23) of On the folded normal distribution:

The folded normal distribution is not a stable distribution. That is, the distribution of the sum of its random variables do not form a folded normal distribution. We can see this from the characteristic (or the moment) generating function Equation (22) or Equation (23).

Two notes:

  • I used $\sigma$ and $\tau$ in place of $\sigma_1$ and $\sigma_2$ for notational brevity;
  • and used $y$ as the convolution variable in place of $z$).

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