Can I derive a well-known distribution from another divergence?












0












$begingroup$


A large class of distributions can be derived from



$max_{p(x)} H(p)$



s.t.



$E_x{x}=mu$



$E_x{x^n}=c_n$



where $H(p)$ denotes the Shannon (differential) entropy and are called maximum entropy. E.g., a Gaussian in canonical form. Do you know any well-known distribution that is derived from another entropic measure? E.g., Renyi Entropy, etc?










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$endgroup$

















    0












    $begingroup$


    A large class of distributions can be derived from



    $max_{p(x)} H(p)$



    s.t.



    $E_x{x}=mu$



    $E_x{x^n}=c_n$



    where $H(p)$ denotes the Shannon (differential) entropy and are called maximum entropy. E.g., a Gaussian in canonical form. Do you know any well-known distribution that is derived from another entropic measure? E.g., Renyi Entropy, etc?










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      A large class of distributions can be derived from



      $max_{p(x)} H(p)$



      s.t.



      $E_x{x}=mu$



      $E_x{x^n}=c_n$



      where $H(p)$ denotes the Shannon (differential) entropy and are called maximum entropy. E.g., a Gaussian in canonical form. Do you know any well-known distribution that is derived from another entropic measure? E.g., Renyi Entropy, etc?










      share|cite|improve this question









      $endgroup$




      A large class of distributions can be derived from



      $max_{p(x)} H(p)$



      s.t.



      $E_x{x}=mu$



      $E_x{x^n}=c_n$



      where $H(p)$ denotes the Shannon (differential) entropy and are called maximum entropy. E.g., a Gaussian in canonical form. Do you know any well-known distribution that is derived from another entropic measure? E.g., Renyi Entropy, etc?







      probability-distributions entropy






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Jan 10 at 21:09









      J. ReinhardJ. Reinhard

      184




      184






















          1 Answer
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          $begingroup$

          There's a proof here that the Shannon entropy $H$ is maximal when $ln p$ is linear in the functions of $X$ (be they powers thereof or otherwise) whose means are specified as constraints. But you already knew that. If you adapt that proof, you can similarly show other results of the kind you seek. I'll talk about one outcome from doing this, but if you want to consider other alternatives to $H$ I'll leave you to try the calculations yourself. (The example I'll discuss is actually a bit hard if you're new to functional calculus, but I'll only present the result.)



          For example, you can show the Rényi entropy $H_alpha$ with $alphane 1$ is maximised when $frac{p^{1-alpha}-1}{1-alpha}$ is linear in the constrained functions. This bizarre-looking quantity, which by L'Hôpital's rule becomes $ln p$ in the $alphato 1$ limit, is called the Tsallis-$alpha$ logarithm of $p$ (although with that terminology the parameter is usually called $q$).



          The idea is that this function differentiates to $p^{-alpha}$, and equals $0$ when $p=1$. (Yes, I realise $p=1$ has no special meaning for probability densities, but the Tsallis-$q$ logarithm is often worth computing for quantities other than probabilities or densities thereof; in fact, you'd be amazed how many approximate but imperfect power laws in the applied sciences can be expressed in terms of Tsallis statistics, though I don't think they usually are.)



          For example, what maximises $H_alpha$ for known mean and variance? Well, $p$ would have to be the Tsallis $alpha$-exponential of a quadratic, and this is (up to a linear transformation of $X$) just a Student's-$t$ distribution, with $nu$ satisfying, if my mental arithmetic is right, $alpha=frac{nu+3}{nu+1}$. Unsurprisingly, we recover the usual Gaussian in the $alphato 1,,nutoinfty$ limit.






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            1 Answer
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            1 Answer
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            active

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            0












            $begingroup$

            There's a proof here that the Shannon entropy $H$ is maximal when $ln p$ is linear in the functions of $X$ (be they powers thereof or otherwise) whose means are specified as constraints. But you already knew that. If you adapt that proof, you can similarly show other results of the kind you seek. I'll talk about one outcome from doing this, but if you want to consider other alternatives to $H$ I'll leave you to try the calculations yourself. (The example I'll discuss is actually a bit hard if you're new to functional calculus, but I'll only present the result.)



            For example, you can show the Rényi entropy $H_alpha$ with $alphane 1$ is maximised when $frac{p^{1-alpha}-1}{1-alpha}$ is linear in the constrained functions. This bizarre-looking quantity, which by L'Hôpital's rule becomes $ln p$ in the $alphato 1$ limit, is called the Tsallis-$alpha$ logarithm of $p$ (although with that terminology the parameter is usually called $q$).



            The idea is that this function differentiates to $p^{-alpha}$, and equals $0$ when $p=1$. (Yes, I realise $p=1$ has no special meaning for probability densities, but the Tsallis-$q$ logarithm is often worth computing for quantities other than probabilities or densities thereof; in fact, you'd be amazed how many approximate but imperfect power laws in the applied sciences can be expressed in terms of Tsallis statistics, though I don't think they usually are.)



            For example, what maximises $H_alpha$ for known mean and variance? Well, $p$ would have to be the Tsallis $alpha$-exponential of a quadratic, and this is (up to a linear transformation of $X$) just a Student's-$t$ distribution, with $nu$ satisfying, if my mental arithmetic is right, $alpha=frac{nu+3}{nu+1}$. Unsurprisingly, we recover the usual Gaussian in the $alphato 1,,nutoinfty$ limit.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              There's a proof here that the Shannon entropy $H$ is maximal when $ln p$ is linear in the functions of $X$ (be they powers thereof or otherwise) whose means are specified as constraints. But you already knew that. If you adapt that proof, you can similarly show other results of the kind you seek. I'll talk about one outcome from doing this, but if you want to consider other alternatives to $H$ I'll leave you to try the calculations yourself. (The example I'll discuss is actually a bit hard if you're new to functional calculus, but I'll only present the result.)



              For example, you can show the Rényi entropy $H_alpha$ with $alphane 1$ is maximised when $frac{p^{1-alpha}-1}{1-alpha}$ is linear in the constrained functions. This bizarre-looking quantity, which by L'Hôpital's rule becomes $ln p$ in the $alphato 1$ limit, is called the Tsallis-$alpha$ logarithm of $p$ (although with that terminology the parameter is usually called $q$).



              The idea is that this function differentiates to $p^{-alpha}$, and equals $0$ when $p=1$. (Yes, I realise $p=1$ has no special meaning for probability densities, but the Tsallis-$q$ logarithm is often worth computing for quantities other than probabilities or densities thereof; in fact, you'd be amazed how many approximate but imperfect power laws in the applied sciences can be expressed in terms of Tsallis statistics, though I don't think they usually are.)



              For example, what maximises $H_alpha$ for known mean and variance? Well, $p$ would have to be the Tsallis $alpha$-exponential of a quadratic, and this is (up to a linear transformation of $X$) just a Student's-$t$ distribution, with $nu$ satisfying, if my mental arithmetic is right, $alpha=frac{nu+3}{nu+1}$. Unsurprisingly, we recover the usual Gaussian in the $alphato 1,,nutoinfty$ limit.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                There's a proof here that the Shannon entropy $H$ is maximal when $ln p$ is linear in the functions of $X$ (be they powers thereof or otherwise) whose means are specified as constraints. But you already knew that. If you adapt that proof, you can similarly show other results of the kind you seek. I'll talk about one outcome from doing this, but if you want to consider other alternatives to $H$ I'll leave you to try the calculations yourself. (The example I'll discuss is actually a bit hard if you're new to functional calculus, but I'll only present the result.)



                For example, you can show the Rényi entropy $H_alpha$ with $alphane 1$ is maximised when $frac{p^{1-alpha}-1}{1-alpha}$ is linear in the constrained functions. This bizarre-looking quantity, which by L'Hôpital's rule becomes $ln p$ in the $alphato 1$ limit, is called the Tsallis-$alpha$ logarithm of $p$ (although with that terminology the parameter is usually called $q$).



                The idea is that this function differentiates to $p^{-alpha}$, and equals $0$ when $p=1$. (Yes, I realise $p=1$ has no special meaning for probability densities, but the Tsallis-$q$ logarithm is often worth computing for quantities other than probabilities or densities thereof; in fact, you'd be amazed how many approximate but imperfect power laws in the applied sciences can be expressed in terms of Tsallis statistics, though I don't think they usually are.)



                For example, what maximises $H_alpha$ for known mean and variance? Well, $p$ would have to be the Tsallis $alpha$-exponential of a quadratic, and this is (up to a linear transformation of $X$) just a Student's-$t$ distribution, with $nu$ satisfying, if my mental arithmetic is right, $alpha=frac{nu+3}{nu+1}$. Unsurprisingly, we recover the usual Gaussian in the $alphato 1,,nutoinfty$ limit.






                share|cite|improve this answer









                $endgroup$



                There's a proof here that the Shannon entropy $H$ is maximal when $ln p$ is linear in the functions of $X$ (be they powers thereof or otherwise) whose means are specified as constraints. But you already knew that. If you adapt that proof, you can similarly show other results of the kind you seek. I'll talk about one outcome from doing this, but if you want to consider other alternatives to $H$ I'll leave you to try the calculations yourself. (The example I'll discuss is actually a bit hard if you're new to functional calculus, but I'll only present the result.)



                For example, you can show the Rényi entropy $H_alpha$ with $alphane 1$ is maximised when $frac{p^{1-alpha}-1}{1-alpha}$ is linear in the constrained functions. This bizarre-looking quantity, which by L'Hôpital's rule becomes $ln p$ in the $alphato 1$ limit, is called the Tsallis-$alpha$ logarithm of $p$ (although with that terminology the parameter is usually called $q$).



                The idea is that this function differentiates to $p^{-alpha}$, and equals $0$ when $p=1$. (Yes, I realise $p=1$ has no special meaning for probability densities, but the Tsallis-$q$ logarithm is often worth computing for quantities other than probabilities or densities thereof; in fact, you'd be amazed how many approximate but imperfect power laws in the applied sciences can be expressed in terms of Tsallis statistics, though I don't think they usually are.)



                For example, what maximises $H_alpha$ for known mean and variance? Well, $p$ would have to be the Tsallis $alpha$-exponential of a quadratic, and this is (up to a linear transformation of $X$) just a Student's-$t$ distribution, with $nu$ satisfying, if my mental arithmetic is right, $alpha=frac{nu+3}{nu+1}$. Unsurprisingly, we recover the usual Gaussian in the $alphato 1,,nutoinfty$ limit.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 10 at 22:42









                J.G.J.G.

                25.7k22539




                25.7k22539






























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