Convergence in distribution off by one correction?












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Suppose that $mu > 0$ is given. I know that $mathsf{Ber}(n, mu/n)to mathsf{Poi}(mu)$ in distribution. Is it also true that $mathsf{Ber}(n-1, mu/n) to mathsf{Poi}(mu)$? If possible, I'd like to show this using the simpler characterization of distributional convergence (ie., not the version which says something about integration of bounded continuous functions). One idea I had was to show that the total variation distance between $mathsf{Ber}(n, mu/n)$ and $mathsf{Ber}(n-1, mu/n)$ goes to zero, which would be sufficient by the triangle inequality, but I'm having trouble getting the details to work out.






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    $begingroup$


    Suppose that $mu > 0$ is given. I know that $mathsf{Ber}(n, mu/n)to mathsf{Poi}(mu)$ in distribution. Is it also true that $mathsf{Ber}(n-1, mu/n) to mathsf{Poi}(mu)$? If possible, I'd like to show this using the simpler characterization of distributional convergence (ie., not the version which says something about integration of bounded continuous functions). One idea I had was to show that the total variation distance between $mathsf{Ber}(n, mu/n)$ and $mathsf{Ber}(n-1, mu/n)$ goes to zero, which would be sufficient by the triangle inequality, but I'm having trouble getting the details to work out.






    probability-theory







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      0












      0








      0





      $begingroup$


      Suppose that $mu > 0$ is given. I know that $mathsf{Ber}(n, mu/n)to mathsf{Poi}(mu)$ in distribution. Is it also true that $mathsf{Ber}(n-1, mu/n) to mathsf{Poi}(mu)$? If possible, I'd like to show this using the simpler characterization of distributional convergence (ie., not the version which says something about integration of bounded continuous functions). One idea I had was to show that the total variation distance between $mathsf{Ber}(n, mu/n)$ and $mathsf{Ber}(n-1, mu/n)$ goes to zero, which would be sufficient by the triangle inequality, but I'm having trouble getting the details to work out.






      probability-theory







      share|cite|improve this question









      $endgroup$




      Suppose that $mu > 0$ is given. I know that $mathsf{Ber}(n, mu/n)to mathsf{Poi}(mu)$ in distribution. Is it also true that $mathsf{Ber}(n-1, mu/n) to mathsf{Poi}(mu)$? If possible, I'd like to show this using the simpler characterization of distributional convergence (ie., not the version which says something about integration of bounded continuous functions). One idea I had was to show that the total variation distance between $mathsf{Ber}(n, mu/n)$ and $mathsf{Ber}(n-1, mu/n)$ goes to zero, which would be sufficient by the triangle inequality, but I'm having trouble getting the details to work out.






      probability-theory




      probability-theory






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      asked Dec 23 '18 at 12:22









      Drew BradyDrew Brady

      731315




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          2 Answers
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          2












          $begingroup$

          Poisson Theorem says:




          $forall n geq 1, X_{n,k}$ for $k=1...n,$ are independent random variables identically distributed as a $B(p_n)$ and $np_n longrightarrow lambda > 0$ for $n rightarrow infty$ and $p_n rightarrow 0$. If so, then:




          $$ S_{n,n} = sum_{k=1}^{n}X_{n,k} overset L longrightarrow X = P(lambda) Longrightarrow B(n,p_n) simeq P(lambda)$$



          If we change $n$ for $n-1$, the hypothesis $(n-1)p_{n-1}$ is $np_{n-1} - p_{n-1}$ and for $n rightarrow infty$ and $p_n rightarrow 0$ (so $p_{n-1} rightarrow 0)$, is $lambda - 0 = lambda$. So here we have the importance of limit to $infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, that's pretty much what I needed, thanks!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 13:22



















          2












          $begingroup$

          Let $X_nsim B(n-1,mu/n)$ and $Y_nsim B(n,mu/n)$; you want to show the TV distance between the laws of $X_n$ and $Y_n$ goes like $mu/nto0$. I think this is a good idea.



          Here's one way to see why this is so.
          Construct a coupling $Y_n=X_n+Z_n$, where $Z_nsim B(1,mu/n)$ is independent of $X_n$. Then the TV distance is bounded by $P(X_nne Y_n) =P(Z_n=1)=mu/n$. For given any set $A$, when conditioning on $Z_n$ you get (after a tiny bit of algebra) $$|P(X_nin A) - P(Y_nin A)|le frac mu n .$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 19:32














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          2 Answers
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          active

          oldest

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          2 Answers
          2






          active

          oldest

          votes









          active

          oldest

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          active

          oldest

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          2












          $begingroup$

          Poisson Theorem says:




          $forall n geq 1, X_{n,k}$ for $k=1...n,$ are independent random variables identically distributed as a $B(p_n)$ and $np_n longrightarrow lambda > 0$ for $n rightarrow infty$ and $p_n rightarrow 0$. If so, then:




          $$ S_{n,n} = sum_{k=1}^{n}X_{n,k} overset L longrightarrow X = P(lambda) Longrightarrow B(n,p_n) simeq P(lambda)$$



          If we change $n$ for $n-1$, the hypothesis $(n-1)p_{n-1}$ is $np_{n-1} - p_{n-1}$ and for $n rightarrow infty$ and $p_n rightarrow 0$ (so $p_{n-1} rightarrow 0)$, is $lambda - 0 = lambda$. So here we have the importance of limit to $infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, that's pretty much what I needed, thanks!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 13:22
















          2












          $begingroup$

          Poisson Theorem says:




          $forall n geq 1, X_{n,k}$ for $k=1...n,$ are independent random variables identically distributed as a $B(p_n)$ and $np_n longrightarrow lambda > 0$ for $n rightarrow infty$ and $p_n rightarrow 0$. If so, then:




          $$ S_{n,n} = sum_{k=1}^{n}X_{n,k} overset L longrightarrow X = P(lambda) Longrightarrow B(n,p_n) simeq P(lambda)$$



          If we change $n$ for $n-1$, the hypothesis $(n-1)p_{n-1}$ is $np_{n-1} - p_{n-1}$ and for $n rightarrow infty$ and $p_n rightarrow 0$ (so $p_{n-1} rightarrow 0)$, is $lambda - 0 = lambda$. So here we have the importance of limit to $infty$.






          share|cite|improve this answer











          $endgroup$













          • $begingroup$
            Yeah, that's pretty much what I needed, thanks!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 13:22














          2












          2








          2





          $begingroup$

          Poisson Theorem says:




          $forall n geq 1, X_{n,k}$ for $k=1...n,$ are independent random variables identically distributed as a $B(p_n)$ and $np_n longrightarrow lambda > 0$ for $n rightarrow infty$ and $p_n rightarrow 0$. If so, then:




          $$ S_{n,n} = sum_{k=1}^{n}X_{n,k} overset L longrightarrow X = P(lambda) Longrightarrow B(n,p_n) simeq P(lambda)$$



          If we change $n$ for $n-1$, the hypothesis $(n-1)p_{n-1}$ is $np_{n-1} - p_{n-1}$ and for $n rightarrow infty$ and $p_n rightarrow 0$ (so $p_{n-1} rightarrow 0)$, is $lambda - 0 = lambda$. So here we have the importance of limit to $infty$.






          share|cite|improve this answer











          $endgroup$



          Poisson Theorem says:




          $forall n geq 1, X_{n,k}$ for $k=1...n,$ are independent random variables identically distributed as a $B(p_n)$ and $np_n longrightarrow lambda > 0$ for $n rightarrow infty$ and $p_n rightarrow 0$. If so, then:




          $$ S_{n,n} = sum_{k=1}^{n}X_{n,k} overset L longrightarrow X = P(lambda) Longrightarrow B(n,p_n) simeq P(lambda)$$



          If we change $n$ for $n-1$, the hypothesis $(n-1)p_{n-1}$ is $np_{n-1} - p_{n-1}$ and for $n rightarrow infty$ and $p_n rightarrow 0$ (so $p_{n-1} rightarrow 0)$, is $lambda - 0 = lambda$. So here we have the importance of limit to $infty$.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Dec 23 '18 at 15:49

























          answered Dec 23 '18 at 13:00









          Miguel AnguitaMiguel Anguita

          485




          485












          • $begingroup$
            Yeah, that's pretty much what I needed, thanks!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 13:22


















          • $begingroup$
            Yeah, that's pretty much what I needed, thanks!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 13:22
















          $begingroup$
          Yeah, that's pretty much what I needed, thanks!
          $endgroup$
          – Drew Brady
          Dec 23 '18 at 13:22




          $begingroup$
          Yeah, that's pretty much what I needed, thanks!
          $endgroup$
          – Drew Brady
          Dec 23 '18 at 13:22











          2












          $begingroup$

          Let $X_nsim B(n-1,mu/n)$ and $Y_nsim B(n,mu/n)$; you want to show the TV distance between the laws of $X_n$ and $Y_n$ goes like $mu/nto0$. I think this is a good idea.



          Here's one way to see why this is so.
          Construct a coupling $Y_n=X_n+Z_n$, where $Z_nsim B(1,mu/n)$ is independent of $X_n$. Then the TV distance is bounded by $P(X_nne Y_n) =P(Z_n=1)=mu/n$. For given any set $A$, when conditioning on $Z_n$ you get (after a tiny bit of algebra) $$|P(X_nin A) - P(Y_nin A)|le frac mu n .$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 19:32


















          2












          $begingroup$

          Let $X_nsim B(n-1,mu/n)$ and $Y_nsim B(n,mu/n)$; you want to show the TV distance between the laws of $X_n$ and $Y_n$ goes like $mu/nto0$. I think this is a good idea.



          Here's one way to see why this is so.
          Construct a coupling $Y_n=X_n+Z_n$, where $Z_nsim B(1,mu/n)$ is independent of $X_n$. Then the TV distance is bounded by $P(X_nne Y_n) =P(Z_n=1)=mu/n$. For given any set $A$, when conditioning on $Z_n$ you get (after a tiny bit of algebra) $$|P(X_nin A) - P(Y_nin A)|le frac mu n .$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 19:32
















          2












          2








          2





          $begingroup$

          Let $X_nsim B(n-1,mu/n)$ and $Y_nsim B(n,mu/n)$; you want to show the TV distance between the laws of $X_n$ and $Y_n$ goes like $mu/nto0$. I think this is a good idea.



          Here's one way to see why this is so.
          Construct a coupling $Y_n=X_n+Z_n$, where $Z_nsim B(1,mu/n)$ is independent of $X_n$. Then the TV distance is bounded by $P(X_nne Y_n) =P(Z_n=1)=mu/n$. For given any set $A$, when conditioning on $Z_n$ you get (after a tiny bit of algebra) $$|P(X_nin A) - P(Y_nin A)|le frac mu n .$$






          share|cite|improve this answer









          $endgroup$



          Let $X_nsim B(n-1,mu/n)$ and $Y_nsim B(n,mu/n)$; you want to show the TV distance between the laws of $X_n$ and $Y_n$ goes like $mu/nto0$. I think this is a good idea.



          Here's one way to see why this is so.
          Construct a coupling $Y_n=X_n+Z_n$, where $Z_nsim B(1,mu/n)$ is independent of $X_n$. Then the TV distance is bounded by $P(X_nne Y_n) =P(Z_n=1)=mu/n$. For given any set $A$, when conditioning on $Z_n$ you get (after a tiny bit of algebra) $$|P(X_nin A) - P(Y_nin A)|le frac mu n .$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Dec 23 '18 at 13:25









          kimchi loverkimchi lover

          11.9k31229




          11.9k31229












          • $begingroup$
            Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 19:32




















          • $begingroup$
            Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
            $endgroup$
            – Drew Brady
            Dec 23 '18 at 19:32


















          $begingroup$
          Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
          $endgroup$
          – Drew Brady
          Dec 23 '18 at 19:32






          $begingroup$
          Yep, that makes sense. I figured there would be a correction term (i.e., something like your $Z_n$ to add to $X_n$ to make it equal to $Y_n$ in distribution; thanks for giving me a direction to go here!
          $endgroup$
          – Drew Brady
          Dec 23 '18 at 19:32




















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