KL divergence between two Asymmetric Laplace distributions?

Question

Consider two asymmetric Laplace distribution $L_1(\mu_1,\sigma_1,\tau_1)$ and $L_2(\mu_2,\sigma_2,\tau_2)$ where

\begin{equation} L(x;\mu,\sigma,\tau) =\frac{\tau(1-\tau)}{\sigma} \begin{cases} \text{exp}(- \frac{x-\mu}{\sigma}(\tau - 1)) & \text{if $x < \mu$}\\ \text{exp}(-\frac{x-\mu}{\sigma}\tau) & \text{if $x \ge \mu$} \end{cases}\\ =\frac{\tau(1-\tau)}{\sigma}\text{exp}(-\rho_{\tau}(\frac{x-\mu}{\sigma})) \end{equation}

where $\rho_{\tau}(\frac{x-\mu}{\sigma}) = \frac{x-\mu}{\sigma}(\tau - \mathbb{1}(x<\mu) )$

I'm wondering if there is an closed form solution to the KL-divergence between two laplaces ?

\begin{equation} \mathbb{E}_{L_1}[\text{log}(L_1) - \text{log}(L_2)]=\text{log}(\frac{\tau_1}{\tau_2}) + \text{log}(\frac{1-\tau_1}{1-\tau_2}) + \text{log}(\frac{\sigma_1}{\sigma_2}) + \mathbb{E}_{L_1}[T] \end{equation}

where

\begin{equation} T = \rho_{\tau_2}(\frac{x-\mu_2}{\sigma_2}) -\rho_{\tau_1}(\frac{x-\mu_1}{\sigma_1})\\ = \frac{x-\mu_2}{\sigma_2}(\tau_2 - \mathbb{1}(x<\mu_2)) - \frac{x-\mu_1}{\sigma_1}(\tau_1 - \mathbb{1}(x<\mu_1))\\ =\frac{\tau_2}{\sigma_2}x - \frac{\tau_2\mu_2}{\sigma_2} - \frac{\mathbb{1}(x<\mu_2)}{\sigma_2}x + \mathbb{1}(x<\mu_2)\frac{\mu_2}{\sigma_2} - \frac{\tau_1}{\sigma_1}x + \frac{\tau_1\mu_1}{\sigma_1} + \frac{\mathbb{1}(x<\mu_1)}{\sigma_1}x -\mathbb{1}(x<\mu_1)\frac{\mu_1}{\sigma_1} \end{equation}

So now; by linearity of expectation, $\mathbb{E}_{L_1}[T]$ leads to compute

$$\frac{\tau_2}{\sigma_2}\mathbb{E}_{L_1}[x] - \frac{\tau_2\mu_2}{\sigma_2} - \frac{1}{\sigma_2}\mathbb{E}_{L_1}[x\mathbb{1}(x<\mu_2)] + \mathbb{E}_{L_1}[\mathbb{1}(x<\mu_2)]\frac{\mu_2}{\sigma_2} - \frac{\tau_1}{\sigma_1}\mathbb{E}_{L_1}[x] + \frac{\tau_1\mu_1}{\sigma_1} + \frac{1}{\sigma_1}\mathbb{E}_{L_1}[x\mathbb{1}(x<\mu_1)] -\mathbb{E}_{L_1}[\mathbb{1}(x<\mu_1)]\frac{\mu_1}{\sigma_1}$$

where $ \mathbb{E}_{L_1}[\mathbb{1}(x<\mu_1)] = \mathbb{P}(x<\mu_1),\mathbb{E}_{L_1}[\mathbb{1}(x<\mu_2)] = \mathbb{P}(x<\mu_2)$; the CDF of $L_1$ Laplace distribution given by https://en.wikipedia.org/wiki/Asymmetric_Laplace_distribution when $\lambda = \frac{\sqrt{\tau(1-\tau)}}{\sigma}$ and $\kappa = \sqrt{\frac{\tau}{1-\tau}}$

Now; it remains to compute $\mathbb{E}_{L_1}[x\mathbb{1}(x<\mu_1)]$ and $\mathbb{E}_{L_1}[x\mathbb{1}(x<\mu_2)]$

Answer 1

Expectations are derived by standard exponential integrations \begin{align*}\mathbb E_{L_1}[\mathbb 1(X<\mu_1)]&=\int_{-\infty}^{\mu_1} \frac{\tau_1(1-\tau_1)}{\sigma_1}\text{exp}\left(- \frac{x-\mu_1}{\sigma_1}(\tau_1 - 1)\right)\text{d}x\\ &=\tau_1\int_{-\infty}^{\mu_1}\frac{(1-\tau_1)}{\sigma_1}\text{exp}\left(\frac{1-\tau_1}{\sigma_1}(x-\mu_1)\right)\text{d}x \\ &=\tau_1\int_0^\infty \frac{(1-\tau_1)}{\sigma_1}\text{exp}\left(-\frac{1-\tau_1}{\sigma_1}y\right)\text{d}y \\ &=\tau_1 \end{align*} \begin{align*}\mathbb E_{L_1}[X\mathbb 1(X<\mu_1)]&=\int_{-\infty}^{\mu_1}x \frac{\tau_1(1-\tau_1)}{\sigma_1}\text{exp}\left(- \frac{x-\mu_1}{\sigma_1}(\tau_1 - 1)\right)\text{d}x\\ &=\tau_1\int_{-\infty}^{\mu_1}x\frac{(1-\tau_1)}{\sigma_1}\text{exp}\left(\frac{x-\mu_1}{\sigma_1}(1-\tau_1)\right)\text{d}x \\ &=\tau_1\int_{-\infty}^{\mu_1}(x-\mu_1+\mu_1)\frac{(1-\tau_1)}{\sigma_1}\text{exp}\left(\frac{x-\mu_1}{\sigma_1}(1-\tau_1)\right)\text{d}x \\ &=\mu_1\mathbb E_{L_1}[\mathbb 1(X<\mu_1)]+\tau_1\int_0^{\infty}y\frac{(1-\tau_1)}{\sigma_1}\text{exp}\left(-\frac{1-\tau_1}{\sigma_1}y\right)\text{d}\\ &=\mu_1\tau_1+\tau_1\frac{\sigma_1}{1-\tau_1} \end{align*} \begin{align*}\mathbb E_{L_1}[\mathbb 1(X<\mu_2)]&=\mathbb E_{L_1}[\mathbb 1(X<\mu_1)]+\mathbb E_{L_1}[\mathbb 1(\mu_1<X<\mu_2)]\qquad\qquad(\mu_1<\mu_2)\\ &=\tau_1+\int_{\mu_1}^{\mu_2} \frac{\tau_1(1-\tau_1)}{\sigma_1}\text{exp}\left(- \frac{\tau_1}{\sigma_1} (x-\mu_1)\right)\text{d}x\\ &=\tau_1+(1-\tau_1)\int_0^{\mu_2-\mu_1} \frac{\tau_1}{\sigma_1}\text{exp}\left(-\frac{\tau_1}{\sigma_1}y\right)\text{d}y \\ &=\tau_1+(1-\tau_1)\left\{1-\exp(-\tau_1(\mu_2-\mu_1)/\sigma_1) \right\} \end{align*} \begin{align*}\mathbb E_{L_1}[X\mathbb 1(X<\mu_2)]&=\mathbb E_{L_1}[X\mathbb 1(X<\mu_1)]+\mathbb E_{L_1}[X\mathbb 1(\mu_1<X<\mu_2)]\qquad\qquad(\mu_1<\mu_2)\\ &=\tau_1\left\{\mu_1+\frac{\sigma_1}{1-\tau_1}\right\}+\int_{\mu_1}^{\mu_2} x\frac{\tau_1(1-\tau_1)}{\sigma_1}\text{exp}\left(- \frac{\tau_1}{\sigma_1} (x-\mu_1)\right)\text{d}x\\ &=\tau_1\left\{\mu_1+\frac{\sigma_1}{1-\tau_1}\right\}+(1-\tau_1)\int_0^{\mu_2-\mu_1} (y+\mu_1)\frac{\tau_1}{\sigma_1}\text{exp}\left(-\frac{\tau_1}{\sigma_1}y\right)\text{d}y \\ &=\tau_1\left\{\mu_1+\frac{\sigma_1}{1-\tau_1}\right\}+(1-\tau_1)\mu_1\left\{1-\exp(-\tau_1(\mu_2-\mu_1)/\sigma_1) \right\}\\ &\quad+(1-\tau_1)\big\{-(\mu_2-\mu_1)\exp(-\tau_1(\mu_2-\mu_1)/\sigma_1)\\ &\qquad\left.+\frac{\sigma_1}{\tau_1}[1-\exp(-\tau_1(\mu_2-\mu_1)/\sigma_1)]\right\} \end{align*} Alternatively, most of these computations stem from the (basic) observation that an asymmetric Laplace distribution $\mathcal L^a(\mu,\tau,\sigma)$ is a mixture of two drifted Exponential distributions$$\tau \left\{\mu-\mathcal E([1-\tau]/\sigma)\right\}+(1-\tau)\left\{\mu+\mathcal E(\tau/\sigma)\right\}$$