# Hellinger distance for two shifted log-normal distributions

If I am not mistaken, Hellinger distance between P and Q is generally given by:

$$H^2(P, Q) = \frac12 \int \left( \sqrt{dP} - \sqrt{dQ} \right)^2 .$$

If P and Q, however, are two differently shifted log-normal distributions of the following form $${\frac {1}{(x-\gamma)\sigma {\sqrt {2\pi \,}}}}\exp \left(-{\frac {[\ln (x-\gamma)-\mu ]^{2}}{2\sigma ^{2}}}\right) ,$$

how would the Hellinger distance then be formed?

in terms of: $$\gamma1,\gamma2, \mu1, \mu2 .. etc$$

• It's the same formula. Are you asking for an expression in terms of the parameters of the two distributions? – Dougal Aug 8 '18 at 13:41

Note that \begin{align} H^2(P, Q) &= \frac12 \int (\sqrt{dP} - \sqrt{dQ})^2 \\&= \frac12 \int dP + \frac12 \int dQ - \int \sqrt{dP} \sqrt{dQ} \\&= 1 - \int \sqrt{dP} \sqrt{dQ} ,\end{align} and that the density function is 0 if $x \le \gamma$. Thus your question asks to compute \begin{align} 1 - H^2(P, Q) &= \int_{\max(\gamma_1,\gamma_2)}^\infty \sqrt{\frac{1}{(x - \gamma_1) \sigma_1 \sqrt{2 \pi}} \exp\left( - \frac{\left(\ln(x - \gamma_1) - \mu_1\right)^2}{2 \sigma_1^2} \right)} \\&\qquad\qquad \sqrt{\frac{1}{(x - \gamma_2) \sigma_2 \sqrt{2 \pi}} \exp\left( - \frac{\left(\ln(x - \gamma_2) - \mu_2\right)^2}{2 \sigma_2^2} \right)} dx \\&= \sqrt{\frac{1}{2 \pi \sigma_1 \sigma_2}} {\huge\int}_{\max(\gamma_1,\gamma_2)}^\infty \frac{\exp\left( - \frac{\left(\ln(x - \gamma_1) - \mu_1\right)^2}{4 \sigma_1^2} - \frac{\left(\ln(x - \gamma_2) - \mu_2\right)^2}{4 \sigma_2^2} \right)}{\sqrt{(x - \gamma_1)(x - \gamma_2)}} dx .\end{align} Assume (WLOG) that $\gamma_1 \ge \gamma_2$, and do a change of variables to $y = \ln(x - \gamma_1)$, $dy = \frac{1}{x - \gamma_1} dx$. Let $\Delta = \gamma_1 - \gamma_2$, so we have $x - \gamma_2 = \exp(y) + \Delta .$ Then we get $1 - H^2(P, Q)$ as \begin{align} \sqrt{\frac{1}{2 \pi \sigma_1 \sigma_2}} \int_{-\infty}^\infty \exp\left( - \frac{\left(y - \mu_1\right)^2}{4 \sigma_1^2} - \frac{\left(\ln(\exp(y) + \Delta) - \mu_2\right)^2}{4 \sigma_2^2} \right) \sqrt{\frac{\exp(y)}{\exp(y) + \Delta}} dy .\end{align}
If $\gamma_1 = \gamma_2$ so $\Delta = 0$, this works out to $$H^2(P, Q) = 1 - \sqrt{\frac{2 \sigma_1 \sigma_2}{\sigma_1^2 + \sigma_2^2}} \exp\left( - \frac{(\mu_1 - \mu_2)^2}{4 (\sigma_1^2 + \sigma_2^2)} \right) .$$ For $\Delta \ne 0$, though, neither I nor Mathematica made any immediate headway.