Showing that an estimator is consistent











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Let $X_1,X_2,ldots,X_n$ be a random sample from $mathcal{N}(theta,1)$. Consider the following (randomized) estimator of $theta$ given a sample of size $n$:
$$
hat{theta}_n = bar{X} + begin{cases}
0 & text{with probability } 1−1/n,\
n & text{with probability } 1/n.
end{cases}
$$




  1. Is $hat{theta}_n$ consistent? Prove or disprove.

  2. Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.


Any possible hints??






asymptotics sampling parameter-estimation estimation-theory sampling-theory







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    Let $X_1,X_2,ldots,X_n$ be a random sample from $mathcal{N}(theta,1)$. Consider the following (randomized) estimator of $theta$ given a sample of size $n$:
    $$
    hat{theta}_n = bar{X} + begin{cases}
    0 & text{with probability } 1−1/n,\
    n & text{with probability } 1/n.
    end{cases}
    $$




    1. Is $hat{theta}_n$ consistent? Prove or disprove.

    2. Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.


    Any possible hints??






    asymptotics sampling parameter-estimation estimation-theory sampling-theory







    share|cite|improve this question
























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      0
      down vote

      favorite
      1









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      Let $X_1,X_2,ldots,X_n$ be a random sample from $mathcal{N}(theta,1)$. Consider the following (randomized) estimator of $theta$ given a sample of size $n$:
      $$
      hat{theta}_n = bar{X} + begin{cases}
      0 & text{with probability } 1−1/n,\
      n & text{with probability } 1/n.
      end{cases}
      $$




      1. Is $hat{theta}_n$ consistent? Prove or disprove.

      2. Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.


      Any possible hints??






      asymptotics sampling parameter-estimation estimation-theory sampling-theory







      share|cite|improve this question













      Let $X_1,X_2,ldots,X_n$ be a random sample from $mathcal{N}(theta,1)$. Consider the following (randomized) estimator of $theta$ given a sample of size $n$:
      $$
      hat{theta}_n = bar{X} + begin{cases}
      0 & text{with probability } 1−1/n,\
      n & text{with probability } 1/n.
      end{cases}
      $$




      1. Is $hat{theta}_n$ consistent? Prove or disprove.

      2. Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.


      Any possible hints??






      asymptotics sampling parameter-estimation estimation-theory sampling-theory




      asymptotics sampling parameter-estimation estimation-theory sampling-theory






      share|cite|improve this question













      share|cite|improve this question











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      asked Nov 28 at 10:28









      Newt

      237




      237






















          1 Answer
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          I can't comment (yet), so I'll add this as an answer.



          I will assume that $bar{X}_n =frac{1}{n}sum_{i=1}^n X_i$.



          1) In this setting, consistency means that $hat{theta}_nto theta$ in probability. For a first hint, try looking at the weak law of large numbers: https://en.wikipedia.org/wiki/Law_of_large_numbers and note that (it is easy to prove that) if $(Z_n)$ and $(Y_n)$ are sequences of random variables which converge in probability to $Z$ and $Y$ respectively, then $(Z_n + Y_n)$ converges in probability to $Z+Y$. In your setting it should be easy to show (directly from the definition) that the random variable:



          $$W_n = begin{cases}
          0 & text{with probability } 1 -1/n\
          n & text{with probability } 1/n
          end{cases}$$
          converges to 0 in probability. Together, these should allow to answer the question.



          2) Asymptotic unbiasedness requires that $mathbb{E}(hat{theta}_n) - theta to 0$ as $ntoinfty$. Here, compute $mathbb{E}(hat{theta}_n) - theta$ and see what you can conclude about it's limit as $ntoinfty$.






          share|cite|improve this answer





















          • Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
            – Newt
            Nov 28 at 11:06






          • 1




            Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
            – Alex Hodges
            Nov 28 at 11:09












          • Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
            – Newt
            Nov 28 at 11:13






          • 1




            In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
            – Alex Hodges
            Nov 28 at 11:17












          • And why does Wn converges to 0 in probability??
            – Newt
            Nov 28 at 11:17











          Your Answer





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          up vote
          2
          down vote



          accepted










          I can't comment (yet), so I'll add this as an answer.



          I will assume that $bar{X}_n =frac{1}{n}sum_{i=1}^n X_i$.



          1) In this setting, consistency means that $hat{theta}_nto theta$ in probability. For a first hint, try looking at the weak law of large numbers: https://en.wikipedia.org/wiki/Law_of_large_numbers and note that (it is easy to prove that) if $(Z_n)$ and $(Y_n)$ are sequences of random variables which converge in probability to $Z$ and $Y$ respectively, then $(Z_n + Y_n)$ converges in probability to $Z+Y$. In your setting it should be easy to show (directly from the definition) that the random variable:



          $$W_n = begin{cases}
          0 & text{with probability } 1 -1/n\
          n & text{with probability } 1/n
          end{cases}$$
          converges to 0 in probability. Together, these should allow to answer the question.



          2) Asymptotic unbiasedness requires that $mathbb{E}(hat{theta}_n) - theta to 0$ as $ntoinfty$. Here, compute $mathbb{E}(hat{theta}_n) - theta$ and see what you can conclude about it's limit as $ntoinfty$.






          share|cite|improve this answer





















          • Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
            – Newt
            Nov 28 at 11:06






          • 1




            Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
            – Alex Hodges
            Nov 28 at 11:09












          • Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
            – Newt
            Nov 28 at 11:13






          • 1




            In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
            – Alex Hodges
            Nov 28 at 11:17












          • And why does Wn converges to 0 in probability??
            – Newt
            Nov 28 at 11:17















          up vote
          2
          down vote



          accepted










          I can't comment (yet), so I'll add this as an answer.



          I will assume that $bar{X}_n =frac{1}{n}sum_{i=1}^n X_i$.



          1) In this setting, consistency means that $hat{theta}_nto theta$ in probability. For a first hint, try looking at the weak law of large numbers: https://en.wikipedia.org/wiki/Law_of_large_numbers and note that (it is easy to prove that) if $(Z_n)$ and $(Y_n)$ are sequences of random variables which converge in probability to $Z$ and $Y$ respectively, then $(Z_n + Y_n)$ converges in probability to $Z+Y$. In your setting it should be easy to show (directly from the definition) that the random variable:



          $$W_n = begin{cases}
          0 & text{with probability } 1 -1/n\
          n & text{with probability } 1/n
          end{cases}$$
          converges to 0 in probability. Together, these should allow to answer the question.



          2) Asymptotic unbiasedness requires that $mathbb{E}(hat{theta}_n) - theta to 0$ as $ntoinfty$. Here, compute $mathbb{E}(hat{theta}_n) - theta$ and see what you can conclude about it's limit as $ntoinfty$.






          share|cite|improve this answer





















          • Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
            – Newt
            Nov 28 at 11:06






          • 1




            Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
            – Alex Hodges
            Nov 28 at 11:09












          • Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
            – Newt
            Nov 28 at 11:13






          • 1




            In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
            – Alex Hodges
            Nov 28 at 11:17












          • And why does Wn converges to 0 in probability??
            – Newt
            Nov 28 at 11:17













          up vote
          2
          down vote



          accepted







          up vote
          2
          down vote



          accepted






          I can't comment (yet), so I'll add this as an answer.



          I will assume that $bar{X}_n =frac{1}{n}sum_{i=1}^n X_i$.



          1) In this setting, consistency means that $hat{theta}_nto theta$ in probability. For a first hint, try looking at the weak law of large numbers: https://en.wikipedia.org/wiki/Law_of_large_numbers and note that (it is easy to prove that) if $(Z_n)$ and $(Y_n)$ are sequences of random variables which converge in probability to $Z$ and $Y$ respectively, then $(Z_n + Y_n)$ converges in probability to $Z+Y$. In your setting it should be easy to show (directly from the definition) that the random variable:



          $$W_n = begin{cases}
          0 & text{with probability } 1 -1/n\
          n & text{with probability } 1/n
          end{cases}$$
          converges to 0 in probability. Together, these should allow to answer the question.



          2) Asymptotic unbiasedness requires that $mathbb{E}(hat{theta}_n) - theta to 0$ as $ntoinfty$. Here, compute $mathbb{E}(hat{theta}_n) - theta$ and see what you can conclude about it's limit as $ntoinfty$.






          share|cite|improve this answer












          I can't comment (yet), so I'll add this as an answer.



          I will assume that $bar{X}_n =frac{1}{n}sum_{i=1}^n X_i$.



          1) In this setting, consistency means that $hat{theta}_nto theta$ in probability. For a first hint, try looking at the weak law of large numbers: https://en.wikipedia.org/wiki/Law_of_large_numbers and note that (it is easy to prove that) if $(Z_n)$ and $(Y_n)$ are sequences of random variables which converge in probability to $Z$ and $Y$ respectively, then $(Z_n + Y_n)$ converges in probability to $Z+Y$. In your setting it should be easy to show (directly from the definition) that the random variable:



          $$W_n = begin{cases}
          0 & text{with probability } 1 -1/n\
          n & text{with probability } 1/n
          end{cases}$$
          converges to 0 in probability. Together, these should allow to answer the question.



          2) Asymptotic unbiasedness requires that $mathbb{E}(hat{theta}_n) - theta to 0$ as $ntoinfty$. Here, compute $mathbb{E}(hat{theta}_n) - theta$ and see what you can conclude about it's limit as $ntoinfty$.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 28 at 10:59









          Alex Hodges

          6113




          6113












          • Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
            – Newt
            Nov 28 at 11:06






          • 1




            Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
            – Alex Hodges
            Nov 28 at 11:09












          • Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
            – Newt
            Nov 28 at 11:13






          • 1




            In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
            – Alex Hodges
            Nov 28 at 11:17












          • And why does Wn converges to 0 in probability??
            – Newt
            Nov 28 at 11:17


















          • Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
            – Newt
            Nov 28 at 11:06






          • 1




            Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
            – Alex Hodges
            Nov 28 at 11:09












          • Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
            – Newt
            Nov 28 at 11:13






          • 1




            In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
            – Alex Hodges
            Nov 28 at 11:17












          • And why does Wn converges to 0 in probability??
            – Newt
            Nov 28 at 11:17
















          Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
          – Newt
          Nov 28 at 11:06




          Actually, Im having trouble getting E($hat{theta}_n$) and Var($hat{theta}_n$)... how can i calculate them??
          – Newt
          Nov 28 at 11:06




          1




          1




          Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
          – Alex Hodges
          Nov 28 at 11:09






          Since expectations are linear, $mathbb{E}(hat{theta}_n) = frac{1}{n}sum_{i=1}^nmathbb{E}(X_i) + mathbb{E}(W_n)$. Why do you need to compute the variance?
          – Alex Hodges
          Nov 28 at 11:09














          Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
          – Newt
          Nov 28 at 11:13




          Oh, now I understood your method.. But in order to prove that Wn converges in probability to some value, don't we need its variance? I mean for chebyshev inequality
          – Newt
          Nov 28 at 11:13




          1




          1




          In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
          – Alex Hodges
          Nov 28 at 11:17






          In this case it's probably easier to show it directly from the definition of convergence in probability. That is, for any $epsilon>0$ we have $P(|W_n|>epsilon)to 0$ as $ntoinfty$
          – Alex Hodges
          Nov 28 at 11:17














          And why does Wn converges to 0 in probability??
          – Newt
          Nov 28 at 11:17




          And why does Wn converges to 0 in probability??
          – Newt
          Nov 28 at 11:17


















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