Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.
$begingroup$
Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.
Let $Z_n$ = $X_n$ + $Y_n$
I want to study the law convergence of the sample mean of $Z_n$. That is:
$$ overline{Z_n} = frac{sum X_i + sum Y_i}{n} $$
So, first of all there is an hint: I cannot use the Law of large numbers. Why is that?
Anyway, I am really out of tools to attack this problem. Does someone have a hint?
EDIT: Managed to prove that sample mean of Cauchy is still Cauchy! Still not sure about the solution.
probability weak-convergence law-of-large-numbers
asked Jan 15 at 16:59
qcc101qcc101
629213
$endgroup$
add a comment |
$begingroup$
Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.
Let $Z_n$ = $X_n$ + $Y_n$
I want to study the law convergence of the sample mean of $Z_n$. That is:
$$ overline{Z_n} = frac{sum X_i + sum Y_i}{n} $$
So, first of all there is an hint: I cannot use the Law of large numbers. Why is that?
Anyway, I am really out of tools to attack this problem. Does someone have a hint?
EDIT: Managed to prove that sample mean of Cauchy is still Cauchy! Still not sure about the solution.
probability weak-convergence law-of-large-numbers
asked Jan 15 at 16:59
qcc101qcc101
629213
$endgroup$
add a comment |
$begingroup$
Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.
Let $Z_n$ = $X_n$ + $Y_n$
I want to study the law convergence of the sample mean of $Z_n$. That is:
$$ overline{Z_n} = frac{sum X_i + sum Y_i}{n} $$
So, first of all there is an hint: I cannot use the Law of large numbers. Why is that?
Anyway, I am really out of tools to attack this problem. Does someone have a hint?
EDIT: Managed to prove that sample mean of Cauchy is still Cauchy! Still not sure about the solution.
probability weak-convergence law-of-large-numbers
asked Jan 15 at 16:59
qcc101qcc101
629213
$endgroup$
Let $X_n$ be a uniform distribution on $(-1,1)$. Let$ Y_n$ ~ Cauchy(0,1). Everything independent.
Let $Z_n$ = $X_n$ + $Y_n$
I want to study the law convergence of the sample mean of $Z_n$. That is:
$$ overline{Z_n} = frac{sum X_i + sum Y_i}{n} $$
So, first of all there is an hint: I cannot use the Law of large numbers. Why is that?
Anyway, I am really out of tools to attack this problem. Does someone have a hint?
EDIT: Managed to prove that sample mean of Cauchy is still Cauchy! Still not sure about the solution.
probability weak-convergence law-of-large-numbers
probability weak-convergence law-of-large-numbers
asked Jan 15 at 16:59
qcc101qcc101
629213
asked Jan 15 at 16:59
qcc101qcc101
629213
asked Jan 15 at 16:59
qcc101qcc101
629213
asked Jan 15 at 16:59
qcc101qcc101
629213
asked Jan 15 at 16:59
qcc101qcc101
629213
629213
add a comment |
add a comment |
1 Answer
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Hint: what does $sum X_i / n$ converge to? As you note, the sum $sum Y_i / n$ is equal in distribution to a standard Cauchy.
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
$endgroup$
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Hint: what does $sum X_i / n$ converge to? As you note, the sum $sum Y_i / n$ is equal in distribution to a standard Cauchy.
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
$endgroup$
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
add a comment |
$begingroup$
Hint: what does $sum X_i / n$ converge to? As you note, the sum $sum Y_i / n$ is equal in distribution to a standard Cauchy.
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
$endgroup$
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
add a comment |
$begingroup$
Hint: what does $sum X_i / n$ converge to? As you note, the sum $sum Y_i / n$ is equal in distribution to a standard Cauchy.
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
$endgroup$
Hint: what does $sum X_i / n$ converge to? As you note, the sum $sum Y_i / n$ is equal in distribution to a standard Cauchy.
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
answered Jan 15 at 17:07
Marcus MMarcus M
8,84911047
8,84911047
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
add a comment |
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
I guess it converges to $0$? Not sure how to prove it though. Followup question: can I say something about $Z_n$ belonging to some $L^p?$
$endgroup$
– qcc101
Jan 15 at 17:08
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
@qcc101, yes it converges to $0$. The $X_n$'s are very nice random variables (they're bounded, even), so what theorem can we use to show that their average converges to 0? You can't say anything about the $overline{Z}_n$'s being in a nice $L^p$ space since Cauchy random variables aren't.
$endgroup$
– Marcus M
Jan 15 at 17:10
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
I think I would try to use law of large numbers and then Slutsky to conclude, is that right?
$endgroup$
– qcc101
Jan 15 at 17:15
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
@qcc101, yep that'll do it!
$endgroup$
– Marcus M
Jan 15 at 17:17
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
$begingroup$
Nice, thank you for the follow-ups it was exactly how I wanted to solve it.
$endgroup$
– qcc101
Jan 15 at 17:19
add a comment |
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