limiting distribution and continuous transformation











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Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.



I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?



edit: I guess the delta method suits well into this case, I will proceed using the delta method.






statistics probability-distributions convergence







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    up vote
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    favorite












    Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.



    I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?



    edit: I guess the delta method suits well into this case, I will proceed using the delta method.






    statistics probability-distributions convergence







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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.



      I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?



      edit: I guess the delta method suits well into this case, I will proceed using the delta method.






      statistics probability-distributions convergence







      share|cite|improve this question















      Assume that $sqrt{n}(b-beta)$ converges in distribution to a normal variable with zero mean and some variance S (here $b$ is a random variable and $beta$ is a constant, both are scalar valued). I want to find the limiting distribution of $sqrt{n}(e^{b}-e^beta)$.



      I know the conversation property of convergence for continuous transformations, thus if I can rewrite $sqrt{n}(e^{b}-e^beta)$ in terms of $sqrt{n}(b-beta)$ then the limiting distribution is easy to find. But I couldn't write it in terms of $sqrt{n}(b-beta)$. Does anybody have any idea about how to proceed?



      edit: I guess the delta method suits well into this case, I will proceed using the delta method.






      statistics probability-distributions convergence




      statistics probability-distributions convergence






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      edited 2 days ago

























      asked 2 days ago









      Osman Bulut

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      556






















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          Yes, delta method applies here.



          $$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$



          then we have



          $$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$



          if $g'(theta)$ exists and non-zero.



          Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.



          Hence



          $$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$






          share|cite|improve this answer





















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            up vote
            1
            down vote



            accepted










            Yes, delta method applies here.



            $$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$



            then we have



            $$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$



            if $g'(theta)$ exists and non-zero.



            Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.



            Hence



            $$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$






            share|cite|improve this answer

























              up vote
              1
              down vote



              accepted










              Yes, delta method applies here.



              $$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$



              then we have



              $$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$



              if $g'(theta)$ exists and non-zero.



              Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.



              Hence



              $$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$






              share|cite|improve this answer























                up vote
                1
                down vote



                accepted







                up vote
                1
                down vote



                accepted






                Yes, delta method applies here.



                $$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$



                then we have



                $$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$



                if $g'(theta)$ exists and non-zero.



                Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.



                Hence



                $$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$






                share|cite|improve this answer












                Yes, delta method applies here.



                $$sqrt{n}(X_n - theta) xrightarrow{D} N(0, sigma^2)$$



                then we have



                $$sqrt{n}(g(X_n) - g(theta)) xrightarrow{D} N(0, sigma^2[g'(theta)]^2)$$



                if $g'(theta)$ exists and non-zero.



                Here, $sigma^2 = S, theta = beta$, $g(x)=exp(x)=g'(x)$.



                Hence



                $$sqrt{n}(e^{b_n} - e^beta) xrightarrow{D} N(0, Sexp(2beta)).$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered 2 days ago









                Siong Thye Goh

                93.3k1462114




                93.3k1462114






























                     

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