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Given two Gaussian mixture models with \begin{equation} \begin{aligned} f(x) &=\sum_{k=1}^{K} \pi_{k} \mathcal{N}\left(x \mid \mu_{k}, \sigma_{k}\right), \\ g(x) &=\sum_{i=1}^{N} \lambda_{i} \mathcal{N}\left(x \mid \mu_{i}, \sigma_{i}\right), \end{aligned} \end{equation} I want to compute their $L_1$ distance, \begin{equation} \delta_{1}(f,g) = \int_{x} |f(x) - g(x)|. \end{equation} However, the $L_1$ distance cannot be computed directly, and I use the squared $L_2$ distance instead, i.e., \begin{equation} \delta_{2}(f,g) = \int_{x} (f(x) - g(x))^2. \end{equation} The closed-form expression of $L_2$ distance is available (c.f. Distance between two Gaussian mixtures to evaluate cluster solutions), but I want to know: does there exist any relationship between $L_1$ distance and $L_2$ distance in this case?

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  • $\begingroup$ What kind of relationship are you hoping for? an upper and lower bound should be attainable, but the constants would depend (plausibly, I guess) on the "diameter" $\max_{i,k} |\mu_k-\mu_i|$ as well as on the ratio $\frac{\min_j \sigma_j}{\max_j\sigma_j}$ (hopefully not on $K,N$). $\endgroup$ Commented May 27, 2021 at 8:52
  • $\begingroup$ @leomonsaingeon, thanks for the comment. Actually, I want to find an inequality relation between $L_1$ and $L_2$ distance, and how could I attain these upper and lower bounds? $\endgroup$
    – Ze-Nan Li
    Commented May 27, 2021 at 10:03
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    $\begingroup$ Inside a big ball you can use Young/Cauchy-Schwarz to bound $L^2\lesssim L^1$. And outside of a big ball your mixtures behave essentially as the single Gaussian with the smallest $\sigma$, so you can compare $L^2$ and $L^1$ explicitly. Then you should take care of the details, optimizing the radius of the ball, taking into account the shifts $\mu_i=\mu_j$ etc, but I guess this should work $\endgroup$ Commented May 27, 2021 at 11:02
  • $\begingroup$ @leomonsaingeon Thanks for the hint. $\endgroup$
    – Ze-Nan Li
    Commented May 27, 2021 at 14:31

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