Question about limiting distribution for standard normal distribution
$begingroup$
I am studying CLT and meet a question but I have no idea how to solve it,
could you please show me how to prove this question (b)?
Thank you so much!
Wang
statistics central-limit-theorem
edited Dec 2 '18 at 22:22
Henry
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
$endgroup$
add a comment |
$begingroup$
I am studying CLT and meet a question but I have no idea how to solve it,
could you please show me how to prove this question (b)?
Thank you so much!
Wang
statistics central-limit-theorem
edited Dec 2 '18 at 22:22
Henry
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
$endgroup$
add a comment |
$begingroup$
I am studying CLT and meet a question but I have no idea how to solve it,
could you please show me how to prove this question (b)?
Thank you so much!
Wang
statistics central-limit-theorem
edited Dec 2 '18 at 22:22
Henry
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
$endgroup$
I am studying CLT and meet a question but I have no idea how to solve it,
could you please show me how to prove this question (b)?
Thank you so much!
Wang
statistics central-limit-theorem
statistics central-limit-theorem
edited Dec 2 '18 at 22:22
Henry
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
edited Dec 2 '18 at 22:22
Henry
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
edited Dec 2 '18 at 22:22
Henry
99.5k479165
edited Dec 2 '18 at 22:22
Henry
99.5k479165
edited Dec 2 '18 at 22:22
Henry
99.5k479165
99.5k479165
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
asked Dec 2 '18 at 22:16
Wang WangWang Wang
31
31
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Hint for (b):
In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$
Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit
$dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit
Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
$endgroup$
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
add a comment |
$begingroup$
(a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
(b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
$endgroup$
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
add a comment |
Your Answer
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2 Answers
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2 Answers
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$begingroup$
Hint for (b):
In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$
Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit
$dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit
Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
$endgroup$
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
add a comment |
$begingroup$
Hint for (b):
In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$
Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit
$dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit
Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
$endgroup$
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
add a comment |
$begingroup$
Hint for (b):
In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$
Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit
$dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit
Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
$endgroup$
Hint for (b):
In this case you can separate the sum into $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}+dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$
Since the $Z_i$ have independent normal distributions, you should be able to state the distribution of $sumlimits_{i=1}^n Z_i$ and so of $dfrac{sumlimits_{i=1}^n Z_i}{sqrt{n}}$, and thus find the distribution to which this converges in the limit
$dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}}$ is not random, so you can calculate it (noting that the sum is over $i$ not $n$), and so you can find the value to which this converges in the limit
Finally you can combine these to find the limiting distribution of $dfrac{sumlimits_{i=1}^n left(Z_i+frac1nright)}{sqrt{n}}$
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
answered Dec 2 '18 at 22:39
HenryHenry
99.5k479165
99.5k479165
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
add a comment |
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Thank you so much for your comment!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:48
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
Hi, more questiones. I can clearly get the first part of the summation as standard normal distribution. But how can I get the value of the second part of the summation. And how can I combine the standard normal distribution and then get the new distribution? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:41
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
$begingroup$
$sumlimits_{i=1}^n frac1n = ncdot frac1n=1$ so $dfrac{sumlimits_{i=1}^n frac1n}{sqrt{n}} = frac{1}{sqrt{n}}$ and this tends to $0$ as $n$ increases, so you will be adding $0$ to a standard normal
$endgroup$
– Henry
Dec 3 '18 at 9:04
add a comment |
$begingroup$
(a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
(b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
$endgroup$
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
add a comment |
$begingroup$
(a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
(b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
$endgroup$
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
add a comment |
$begingroup$
(a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
(b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
$endgroup$
(a) $P(|1/nsum (X_i-p_i)|>epsilon)leq 1/(n^2epsilon^2)Var(sum X_i)=1/(n^2epsilon^2)sum p_i(1-p_i)leq 1/(n^2epsilon^2)sum 1/4=frac{1}{4nepsilon^2}rightarrow 0.$
(b) $sum (Z_i+1/n)sim N(1,n)Rightarrow 1/sqrt{n}sum (Z_i+1/n)sim N(1/sqrt{n},1)Rightarrow1/sqrt{n}sum (Z_i+1/n)rightarrow_d N(0,1)$
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
answered Dec 2 '18 at 22:42
John_WickJohn_Wick
1,536111
1,536111
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
add a comment |
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
Thanks for your comment! And why 1/(n2ϵ2)∑pi(1−pi)≤1/(n2ϵ2)∑1/4? Where does this 1/4 come from? Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 1:47
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
For $0<x<1$, $x(1-x)leq 1/4.$
$endgroup$
– John_Wick
Dec 3 '18 at 2:08
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
$begingroup$
Ok, I understand now! Thanks!
$endgroup$
– Wang Wang
Dec 3 '18 at 2:13
add a comment |
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