Robust estimation of the mean











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Problem:




Suppose we want to estimate the mean µ of a random variable $X$ from a
sample $X_1 , dots , X_N$ drawn independently from the distribution
of $X$. We want an $varepsilon$-accurate estimate, i.e. one that
falls in the interval $(mu − varepsilon, mu + varepsilon)$.



Show that a sample of size $N = O( log (delta^{−1} ), sigma^2 / varepsilon^2 )$ is sufficient to compute an $varepsilon$-accurate
estimate with probability at least $1 −delta$.



Hint: Use the median of $O(log(delta^{−1}))$ weak estimates.




It is easy to use Chebyshev's inequality to find a weak estimate of $N = O( sigma^2 / delta varepsilon^2 )$.



However, I do not how to find inequality about their median. The wikipedia of median (https://en.wikipedia.org/wiki/Median#The_sample_median) says sample median asymptotically normal but this does not give a bound for specific $N$. Any suggestion is welcome.










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    up vote
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    Problem:




    Suppose we want to estimate the mean µ of a random variable $X$ from a
    sample $X_1 , dots , X_N$ drawn independently from the distribution
    of $X$. We want an $varepsilon$-accurate estimate, i.e. one that
    falls in the interval $(mu − varepsilon, mu + varepsilon)$.



    Show that a sample of size $N = O( log (delta^{−1} ), sigma^2 / varepsilon^2 )$ is sufficient to compute an $varepsilon$-accurate
    estimate with probability at least $1 −delta$.



    Hint: Use the median of $O(log(delta^{−1}))$ weak estimates.




    It is easy to use Chebyshev's inequality to find a weak estimate of $N = O( sigma^2 / delta varepsilon^2 )$.



    However, I do not how to find inequality about their median. The wikipedia of median (https://en.wikipedia.org/wiki/Median#The_sample_median) says sample median asymptotically normal but this does not give a bound for specific $N$. Any suggestion is welcome.










    share|cite|improve this question
























      up vote
      2
      down vote

      favorite









      up vote
      2
      down vote

      favorite











      Problem:




      Suppose we want to estimate the mean µ of a random variable $X$ from a
      sample $X_1 , dots , X_N$ drawn independently from the distribution
      of $X$. We want an $varepsilon$-accurate estimate, i.e. one that
      falls in the interval $(mu − varepsilon, mu + varepsilon)$.



      Show that a sample of size $N = O( log (delta^{−1} ), sigma^2 / varepsilon^2 )$ is sufficient to compute an $varepsilon$-accurate
      estimate with probability at least $1 −delta$.



      Hint: Use the median of $O(log(delta^{−1}))$ weak estimates.




      It is easy to use Chebyshev's inequality to find a weak estimate of $N = O( sigma^2 / delta varepsilon^2 )$.



      However, I do not how to find inequality about their median. The wikipedia of median (https://en.wikipedia.org/wiki/Median#The_sample_median) says sample median asymptotically normal but this does not give a bound for specific $N$. Any suggestion is welcome.










      share|cite|improve this question













      Problem:




      Suppose we want to estimate the mean µ of a random variable $X$ from a
      sample $X_1 , dots , X_N$ drawn independently from the distribution
      of $X$. We want an $varepsilon$-accurate estimate, i.e. one that
      falls in the interval $(mu − varepsilon, mu + varepsilon)$.



      Show that a sample of size $N = O( log (delta^{−1} ), sigma^2 / varepsilon^2 )$ is sufficient to compute an $varepsilon$-accurate
      estimate with probability at least $1 −delta$.



      Hint: Use the median of $O(log(delta^{−1}))$ weak estimates.




      It is easy to use Chebyshev's inequality to find a weak estimate of $N = O( sigma^2 / delta varepsilon^2 )$.



      However, I do not how to find inequality about their median. The wikipedia of median (https://en.wikipedia.org/wiki/Median#The_sample_median) says sample median asymptotically normal but this does not give a bound for specific $N$. Any suggestion is welcome.







      probability median






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      asked Nov 15 at 17:02









      Rikeijin

      878




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