standard deviation - linear coding
$begingroup$
I have the following question on an exercise sheet:
Let $x = 4y$. One student claims $sigma_x = 4sigma_y$. Another claims that since $ x = y + y + y + y$, then $sigma_x = 2sigma_y$. Who is correct and why is the other person wrong?
I'm think the first student is correct because just intuitively, if you multiply your data points by 4, then the spread becomes 4 times larger (I also did linear coding before and was told this).
However, I am not sure how the student even gets from $ x = y + y + y + y$ to $sigma_x = 2sigma_y$. Can anyone help me see what the mistake is or even how they have made this leap?
(also a proof of the first student's claim or a link to a proof would be appreciated!)
proof-verification standard-deviation
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
$endgroup$
add a comment |
$begingroup$
I have the following question on an exercise sheet:
Let $x = 4y$. One student claims $sigma_x = 4sigma_y$. Another claims that since $ x = y + y + y + y$, then $sigma_x = 2sigma_y$. Who is correct and why is the other person wrong?
I'm think the first student is correct because just intuitively, if you multiply your data points by 4, then the spread becomes 4 times larger (I also did linear coding before and was told this).
However, I am not sure how the student even gets from $ x = y + y + y + y$ to $sigma_x = 2sigma_y$. Can anyone help me see what the mistake is or even how they have made this leap?
(also a proof of the first student's claim or a link to a proof would be appreciated!)
proof-verification standard-deviation
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
$endgroup$
add a comment |
$begingroup$
I have the following question on an exercise sheet:
Let $x = 4y$. One student claims $sigma_x = 4sigma_y$. Another claims that since $ x = y + y + y + y$, then $sigma_x = 2sigma_y$. Who is correct and why is the other person wrong?
I'm think the first student is correct because just intuitively, if you multiply your data points by 4, then the spread becomes 4 times larger (I also did linear coding before and was told this).
However, I am not sure how the student even gets from $ x = y + y + y + y$ to $sigma_x = 2sigma_y$. Can anyone help me see what the mistake is or even how they have made this leap?
(also a proof of the first student's claim or a link to a proof would be appreciated!)
proof-verification standard-deviation
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
$endgroup$
I have the following question on an exercise sheet:
Let $x = 4y$. One student claims $sigma_x = 4sigma_y$. Another claims that since $ x = y + y + y + y$, then $sigma_x = 2sigma_y$. Who is correct and why is the other person wrong?
I'm think the first student is correct because just intuitively, if you multiply your data points by 4, then the spread becomes 4 times larger (I also did linear coding before and was told this).
However, I am not sure how the student even gets from $ x = y + y + y + y$ to $sigma_x = 2sigma_y$. Can anyone help me see what the mistake is or even how they have made this leap?
(also a proof of the first student's claim or a link to a proof would be appreciated!)
proof-verification standard-deviation
proof-verification standard-deviation
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
asked Jan 7 at 12:10
PhysicsMathsLovePhysicsMathsLove
1,307515
1,307515
add a comment |
add a comment |
1 Answer
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$begingroup$
The wrong argument would be that if you take the sum of independent random variables, then their variance ($sigma^2$) adds up. So the new variance should be $4sigma^2$, and then the deviation is the square root, which is $2sigma$.
But this is a different setup, when independent random variables are added up, nothing to do with multiplication by $4$.
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
$endgroup$
add a comment |
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1 Answer
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1 Answer
1
active
oldest
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active
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$begingroup$
The wrong argument would be that if you take the sum of independent random variables, then their variance ($sigma^2$) adds up. So the new variance should be $4sigma^2$, and then the deviation is the square root, which is $2sigma$.
But this is a different setup, when independent random variables are added up, nothing to do with multiplication by $4$.
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
$endgroup$
add a comment |
$begingroup$
The wrong argument would be that if you take the sum of independent random variables, then their variance ($sigma^2$) adds up. So the new variance should be $4sigma^2$, and then the deviation is the square root, which is $2sigma$.
But this is a different setup, when independent random variables are added up, nothing to do with multiplication by $4$.
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
$endgroup$
add a comment |
$begingroup$
The wrong argument would be that if you take the sum of independent random variables, then their variance ($sigma^2$) adds up. So the new variance should be $4sigma^2$, and then the deviation is the square root, which is $2sigma$.
But this is a different setup, when independent random variables are added up, nothing to do with multiplication by $4$.
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
$endgroup$
The wrong argument would be that if you take the sum of independent random variables, then their variance ($sigma^2$) adds up. So the new variance should be $4sigma^2$, and then the deviation is the square root, which is $2sigma$.
But this is a different setup, when independent random variables are added up, nothing to do with multiplication by $4$.
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
answered Jan 7 at 12:26
A. PongráczA. Pongrácz
6,0821929
6,0821929
add a comment |
add a comment |
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