Limit in distribution of increasing variance normal random variable












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Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.



Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.










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    $begingroup$


    Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.



    Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.










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      $begingroup$


      Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.



      Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.










      share|cite|improve this question









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      Suppose you have a sequence of normal random variables $X_n sim N(0, n)$. Is there any random variable $X$ such that $X_n to X$ in distribution.



      Here, I use the following definition: $X_n to X$ in distribution if $P(X_n leq x) to P(X leq x) equiv F_X(x)$ for every $x$ for which $F_x$ is continuous.







      functional-analysis stochastic-calculus






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      asked Jan 21 at 18:53









      BrisãoBrisão

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          $X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
          $$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
          Now show that no CDF can equal $1/2$ wherever it is continuous.






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            $begingroup$

            $X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
            $$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
            Now show that no CDF can equal $1/2$ wherever it is continuous.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              $X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
              $$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
              Now show that no CDF can equal $1/2$ wherever it is continuous.






              share|cite|improve this answer









              $endgroup$
















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                $begingroup$

                $X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
                $$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
                Now show that no CDF can equal $1/2$ wherever it is continuous.






                share|cite|improve this answer









                $endgroup$



                $X_n overset{d}{=} sqrt{n} Z$ where $Z sim N(0, 1)$ so
                $$P(X_n le x) = P(Z le x / sqrt{n}) to frac{1}{2}.$$
                Now show that no CDF can equal $1/2$ wherever it is continuous.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Jan 21 at 19:02









                angryavianangryavian

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