distribution of $sum_{i=1}^n (X_i-X_{n+i})^2$ where $X_1,X_2,dots,X_{2n}$ are iid $N(mu,sigma^2)$
$begingroup$
Suppose $X_1,X_2,dots,X_{2n}$ are iid $mathcal N(mu,sigma^2)$ random variables.
How can I find the distribution of $displaystylesum_{i=1}^n (X_i-X_{n+i})^2$?
Should I approach it by taking $Z=dfrac{X-mu}{sigma}$ where $Zsim mathcal N(0,1)$?
Any hint will also help me.
statistics sampling-theory
edited Dec 20 '18 at 15:57
user10354138
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
$endgroup$
add a comment |
$begingroup$
Suppose $X_1,X_2,dots,X_{2n}$ are iid $mathcal N(mu,sigma^2)$ random variables.
How can I find the distribution of $displaystylesum_{i=1}^n (X_i-X_{n+i})^2$?
Should I approach it by taking $Z=dfrac{X-mu}{sigma}$ where $Zsim mathcal N(0,1)$?
Any hint will also help me.
statistics sampling-theory
edited Dec 20 '18 at 15:57
user10354138
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
$endgroup$
add a comment |
$begingroup$
Suppose $X_1,X_2,dots,X_{2n}$ are iid $mathcal N(mu,sigma^2)$ random variables.
How can I find the distribution of $displaystylesum_{i=1}^n (X_i-X_{n+i})^2$?
Should I approach it by taking $Z=dfrac{X-mu}{sigma}$ where $Zsim mathcal N(0,1)$?
Any hint will also help me.
statistics sampling-theory
edited Dec 20 '18 at 15:57
user10354138
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
$endgroup$
Suppose $X_1,X_2,dots,X_{2n}$ are iid $mathcal N(mu,sigma^2)$ random variables.
How can I find the distribution of $displaystylesum_{i=1}^n (X_i-X_{n+i})^2$?
Should I approach it by taking $Z=dfrac{X-mu}{sigma}$ where $Zsim mathcal N(0,1)$?
Any hint will also help me.
statistics sampling-theory
statistics sampling-theory
edited Dec 20 '18 at 15:57
user10354138
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
edited Dec 20 '18 at 15:57
user10354138
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
edited Dec 20 '18 at 15:57
user10354138
7,4322925
edited Dec 20 '18 at 15:57
user10354138
7,4322925
edited Dec 20 '18 at 15:57
user10354138
7,4322925
7,4322925
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
asked Dec 20 '18 at 15:19
Supriyo BanerjeeSupriyo Banerjee
1176
1176
add a comment |
add a comment |
1 Answer
1
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oldest
votes
$begingroup$
Note that $X_i-X_{n+i}$ is an independent sequence. Further note that $$X_i-X_{n+i} sim mathcal{N}(0,2sigma^2),$$
hence $$left(frac{X_i-X_{n+i}}{sqrt2sigma}right)^2simchi^2_{(1)},$$which is also an independent sequence. Therefore $$frac{1}{2sigma^2}sum_{i=1}^n (X_i-X_{n+i})^2simchi^2_{(n)}.$$
Now use this page to see how a scaled chi-dist is related to a Gamma distribution.
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
oldest
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$begingroup$
Note that $X_i-X_{n+i}$ is an independent sequence. Further note that $$X_i-X_{n+i} sim mathcal{N}(0,2sigma^2),$$
hence $$left(frac{X_i-X_{n+i}}{sqrt2sigma}right)^2simchi^2_{(1)},$$which is also an independent sequence. Therefore $$frac{1}{2sigma^2}sum_{i=1}^n (X_i-X_{n+i})^2simchi^2_{(n)}.$$
Now use this page to see how a scaled chi-dist is related to a Gamma distribution.
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
$begingroup$
Note that $X_i-X_{n+i}$ is an independent sequence. Further note that $$X_i-X_{n+i} sim mathcal{N}(0,2sigma^2),$$
hence $$left(frac{X_i-X_{n+i}}{sqrt2sigma}right)^2simchi^2_{(1)},$$which is also an independent sequence. Therefore $$frac{1}{2sigma^2}sum_{i=1}^n (X_i-X_{n+i})^2simchi^2_{(n)}.$$
Now use this page to see how a scaled chi-dist is related to a Gamma distribution.
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
$endgroup$
add a comment |
$begingroup$
Note that $X_i-X_{n+i}$ is an independent sequence. Further note that $$X_i-X_{n+i} sim mathcal{N}(0,2sigma^2),$$
hence $$left(frac{X_i-X_{n+i}}{sqrt2sigma}right)^2simchi^2_{(1)},$$which is also an independent sequence. Therefore $$frac{1}{2sigma^2}sum_{i=1}^n (X_i-X_{n+i})^2simchi^2_{(n)}.$$
Now use this page to see how a scaled chi-dist is related to a Gamma distribution.
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
$endgroup$
Note that $X_i-X_{n+i}$ is an independent sequence. Further note that $$X_i-X_{n+i} sim mathcal{N}(0,2sigma^2),$$
hence $$left(frac{X_i-X_{n+i}}{sqrt2sigma}right)^2simchi^2_{(1)},$$which is also an independent sequence. Therefore $$frac{1}{2sigma^2}sum_{i=1}^n (X_i-X_{n+i})^2simchi^2_{(n)}.$$
Now use this page to see how a scaled chi-dist is related to a Gamma distribution.
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
answered Dec 20 '18 at 15:52
Math-funMath-fun
7,1531527
7,1531527
add a comment |
add a comment |
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