Is it possible to prove this if the function is not invertible?












0














Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



I have successfully shown:




  1. $T_n to T_infty$ in probability


  2. $G(T_n) to g(T_infty)$ in probability


  3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


  4. Derived the expression of $sigma (t)$.



However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



Does anyone see a way forward?



Edit:



Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






probability convergence self-learning







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    Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



    Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



    My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



    I have successfully shown:




    1. $T_n to T_infty$ in probability


    2. $G(T_n) to g(T_infty)$ in probability


    3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


    4. Derived the expression of $sigma (t)$.



    However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



    Does anyone see a way forward?



    Edit:



    Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






    probability convergence self-learning







    share|cite|improve this question



























      0












      0








      0







      Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



      Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



      My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



      I have successfully shown:




      1. $T_n to T_infty$ in probability


      2. $G(T_n) to g(T_infty)$ in probability


      3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


      4. Derived the expression of $sigma (t)$.



      However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



      Does anyone see a way forward?



      Edit:



      Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






      probability convergence self-learning







      share|cite|improve this question















      Let $f(t,X) = log (1+t^2 X)/t$ and let $G_n(t) = n^{-1} sum_{i=1}^n f(t,X_i)$



      Define $T_n$ as the point that maximizes $G_n(t)$ and let $T_infty$ be defined as the value that maximises the non-random function $g(t) = EG_n (t)$.



      My goal is to find the asymptotic distribution of $sqrt{n}(T_n - T_infty)$.



      I have successfully shown:




      1. $T_n to T_infty$ in probability


      2. $G(T_n) to g(T_infty)$ in probability


      3. $sqrt{n}(G(T_n) - G_n(T_infty)) to N(0,sigma(t))$


      4. Derived the expression of $sigma (t)$.



      However, the last step, getting from the aymptotic distribution of $(3)$ to the desired result is hindered by the fact $f(t,X)$ and $G_n(t)$ are not invertible, and so the $delta$-method cannot be used.



      Does anyone see a way forward?



      Edit:



      Looking at this further I think the trick is to use a Taylor expansion, so I''m going down that route for now!






      probability convergence self-learning




      probability convergence self-learning






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      edited Nov 27 at 7:44









      daw

      24k1544




      24k1544










      asked Nov 27 at 7:19









      Xiaomi

      1,016115




      1,016115






















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          I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
          $$
          G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
          $$          ($f_1$ is partial derivative with respect to $t$.)
          What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
          $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
          $$           almost surely. Moreover, we know from CLT the limiting distribution of $S$:
          $$sqrt{n}S_n to N(0,sigma^2),$$
          where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
          $$          From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
          $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
          $$          (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
          $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
          $$






          share|cite|improve this answer





















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            1 Answer
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            1 Answer
            1






            active

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            active

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            active

            oldest

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            1














            I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
            $$
            G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
            $$            ($f_1$ is partial derivative with respect to $t$.)
            What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
            $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
            $$             almost surely. Moreover, we know from CLT the limiting distribution of $S$:
            $$sqrt{n}S_n to N(0,sigma^2),$$
            where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
            $$            From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
            $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
            $$            (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
            $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
            $$






            share|cite|improve this answer


























              1














              I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
              $$
              G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
              $$              ($f_1$ is partial derivative with respect to $t$.)
              What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
              $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
              $$               almost surely. Moreover, we know from CLT the limiting distribution of $S$:
              $$sqrt{n}S_n to N(0,sigma^2),$$
              where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
              $$              From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
              $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
              $$              (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
              $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
              $$






              share|cite|improve this answer
























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                I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
                $$
                G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
                $$                ($f_1$ is partial derivative with respect to $t$.)
                What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
                $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
                $$                 almost surely. Moreover, we know from CLT the limiting distribution of $S$:
                $$sqrt{n}S_n to N(0,sigma^2),$$
                where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
                $$                From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
                $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
                $$                (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
                $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
                $$






                share|cite|improve this answer












                I suggest to mimic derivation of Maximum Likelihood Estimator: The First-Order Condition is given by
                $$
                G_n'(T_n) =frac{1}{n}sum_{ileq n}f_1(T_n,X_i) = 0 cdots(*).
                $$                ($f_1$ is partial derivative with respect to $t$.)
                What we know is that, roughly speaking, under sufficient regularity of $f$, it holds that
                $$S_n := frac{1}{n}sum_{ileq n}f_1(T_infty,X_i)to E[f_1(T_infty,X_i)] =g'(T_infty)=0,
                $$                 almost surely. Moreover, we know from CLT the limiting distribution of $S$:
                $$sqrt{n}S_n to N(0,sigma^2),$$
                where $$sigma^2 = var(f_1(T_infty,X_1)) = E[f_1(T_infty,X_1)^2].
                $$                From $(*)$, we know $$sqrt{n}S_n = sqrt{n}(S_n - G_n'(T_n)) = frac{1}{sqrt{n}}sum_{ileq n}(f_1(T_infty,X_i)-f_1(T_n,X_i))=frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i)cdotsqrt{n}(T_infty-T_n),$$ where in the last equation, we used mean value theorem($0<s<1$). Under sufficient condition, we have
                $$frac{1}{n}sum_{ileq n}f_{11}(sT_infty+(1-s)T_n,X_i) to E[f_{11}(T_infty,X_1)]:= tau neq 0
                $$                (being non-zero is assumed.) Finally, Slutsky's theorem gives us the result:
                $$sqrt{n}(T_n-T_infty) to N(0, frac{sigma^2}{tau^2}).
                $$







                share|cite|improve this answer












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                answered Nov 27 at 8:02









                Song

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                4,200316






























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