Comparing two MGFs












0












$begingroup$


Suppose I have two MGFs given by



$MGF_1 = e^{-2 (p-1) p x^2}$



and



$MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.



Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?



(How similar are the probability distributions in this case?)










share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Suppose I have two MGFs given by



    $MGF_1 = e^{-2 (p-1) p x^2}$



    and



    $MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.



    Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?



    (How similar are the probability distributions in this case?)










    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Suppose I have two MGFs given by



      $MGF_1 = e^{-2 (p-1) p x^2}$



      and



      $MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.



      Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?



      (How similar are the probability distributions in this case?)










      share|cite|improve this question









      $endgroup$




      Suppose I have two MGFs given by



      $MGF_1 = e^{-2 (p-1) p x^2}$



      and



      $MGF_2 = e^{8 (p-1)^2 p^2 x^2}$.



      Is there an accepted way to compare the corresponding probability distributions to obtain a metric for similarity?



      (How similar are the probability distributions in this case?)







      probability-theory






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 30 '18 at 0:03









      user120911user120911

      231110




      231110






















          1 Answer
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          $begingroup$

          Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).



          Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Are you sure they both are Gaussian?
            $endgroup$
            – user120911
            Dec 30 '18 at 1:10










          • $begingroup$
            If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
            $endgroup$
            – Lee David Chung Lin
            Dec 30 '18 at 8:16










          • $begingroup$
            @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
            $endgroup$
            – user120911
            Dec 30 '18 at 8:32











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          1 Answer
          1






          active

          oldest

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          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1












          $begingroup$

          Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).



          Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Are you sure they both are Gaussian?
            $endgroup$
            – user120911
            Dec 30 '18 at 1:10










          • $begingroup$
            If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
            $endgroup$
            – Lee David Chung Lin
            Dec 30 '18 at 8:16










          • $begingroup$
            @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
            $endgroup$
            – user120911
            Dec 30 '18 at 8:32
















          1












          $begingroup$

          Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).



          Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Are you sure they both are Gaussian?
            $endgroup$
            – user120911
            Dec 30 '18 at 1:10










          • $begingroup$
            If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
            $endgroup$
            – Lee David Chung Lin
            Dec 30 '18 at 8:16










          • $begingroup$
            @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
            $endgroup$
            – user120911
            Dec 30 '18 at 8:32














          1












          1








          1





          $begingroup$

          Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).



          Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.






          share|cite|improve this answer









          $endgroup$



          Both of your MGF's indicate the underlying distributions are Gaussian, with mean $0$, and variance in terms of $p$ (lazy to check MGF table).



          Once you do this, you'd have $P_1:N(0,sigma_1^2)$ and $P_2:N(0,sigma_2^2)$. Now, you can compare them, one metric would be to measure the total variation (TV divergence between them). Another, would be Kullback-Leibler. In general, you can use any $f$-divergence.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 30 '18 at 0:28









          AaronAaron

          1,912415




          1,912415












          • $begingroup$
            Are you sure they both are Gaussian?
            $endgroup$
            – user120911
            Dec 30 '18 at 1:10










          • $begingroup$
            If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
            $endgroup$
            – Lee David Chung Lin
            Dec 30 '18 at 8:16










          • $begingroup$
            @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
            $endgroup$
            – user120911
            Dec 30 '18 at 8:32


















          • $begingroup$
            Are you sure they both are Gaussian?
            $endgroup$
            – user120911
            Dec 30 '18 at 1:10










          • $begingroup$
            If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
            $endgroup$
            – Lee David Chung Lin
            Dec 30 '18 at 8:16










          • $begingroup$
            @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
            $endgroup$
            – user120911
            Dec 30 '18 at 8:32
















          $begingroup$
          Are you sure they both are Gaussian?
          $endgroup$
          – user120911
          Dec 30 '18 at 1:10




          $begingroup$
          Are you sure they both are Gaussian?
          $endgroup$
          – user120911
          Dec 30 '18 at 1:10












          $begingroup$
          If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
          $endgroup$
          – Lee David Chung Lin
          Dec 30 '18 at 8:16




          $begingroup$
          If you don't like the table in wikipedia then how about Eq.(17) of Wolfram MathWorld? You can find this in most textbooks, or just calculate it yourself (it's pretty easy).
          $endgroup$
          – Lee David Chung Lin
          Dec 30 '18 at 8:16












          $begingroup$
          @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
          $endgroup$
          – user120911
          Dec 30 '18 at 8:32




          $begingroup$
          @LeeDavidChungLin The two MGFs are equal when p = 0.5 and p = 1. If p sufficiently captured variance, and we are dealing with two Gaussians, then I would guess that other values of p would make these MGFs equal too. Can you explain that?
          $endgroup$
          – user120911
          Dec 30 '18 at 8:32


















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