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 $n$:
$$
hat{theta}_n = bar{X} + begin{cases}
0 & text{with probability } 1−1/n,\
n & text{with probability } 1/n.
end{cases}
$$
- Is $hat{theta}_n$ consistent? Prove or disprove.
- Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.
Any possible hints??
asymptotics sampling parameter-estimation estimation-theory sampling-theory
asked Nov 28 at 10:28
Newt
237
add a comment |
up vote
0
down vote
favorite
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 $n$:
$$
hat{theta}_n = bar{X} + begin{cases}
0 & text{with probability } 1−1/n,\
n & text{with probability } 1/n.
end{cases}
$$
- Is $hat{theta}_n$ consistent? Prove or disprove.
- Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.
Any possible hints??
asymptotics sampling parameter-estimation estimation-theory sampling-theory
asked Nov 28 at 10:28
Newt
237
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
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 $n$:
$$
hat{theta}_n = bar{X} + begin{cases}
0 & text{with probability } 1−1/n,\
n & text{with probability } 1/n.
end{cases}
$$
- Is $hat{theta}_n$ consistent? Prove or disprove.
- Is $hat{theta}_n$ asymptotically unbiased? Prove or disprove.
Any possible hints??
asymptotics sampling parameter-estimation estimation-theory sampling-theory
asked Nov 28 at 10:28
Newt
237
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 $n$:
$$
hat{theta}_n = bar{X} + begin{cases}
0 & text{with probability } 1−1/n,\
n & text{with probability } 1/n.
end{cases}
$$
- Is $hat{theta}_n$ consistent? Prove or disprove.
- 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
asked Nov 28 at 10:28
Newt
237
asked Nov 28 at 10:28
Newt
237
asked Nov 28 at 10:28
Newt
237
asked Nov 28 at 10:28
Newt
237
asked Nov 28 at 10:28
Newt
237
237
add a comment |
add a comment |
1 Answer
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up vote
2
down vote
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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 $ntoinfty$.
answered Nov 28 at 10:59
Alex Hodges
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
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
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 $ntoinfty$.
answered Nov 28 at 10:59
Alex Hodges
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
add a comment |
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 $ntoinfty$.
answered Nov 28 at 10:59
Alex Hodges
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
add a comment |
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 $ntoinfty$.
answered Nov 28 at 10:59
Alex Hodges
6113
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 $ntoinfty$.
answered Nov 28 at 10:59
Alex Hodges
6113
answered Nov 28 at 10:59
Alex Hodges
6113
answered Nov 28 at 10:59
Alex Hodges
6113
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
add a comment |
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
add a comment |
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