How can this adjustment of sum be correct?












2












$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36
















2












$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36














2












2








2





$begingroup$


In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory







share|cite|improve this question











$endgroup$




In our Probability and statistics materials, I run into this equation:equation



$X_i$ is one result (I hope)



$n$ is a count of the results



$overline{X_n}$ is sample mean of the results



It is connected with point estimations, we are trying to estimate variance, and this adjustment should help. But I don't understand how we can do it.



I expect it has something to do with this adjustment:




$(a-b)^2=a^2-2ab+b^2$




But I just don't see how (especially the minus sign confuses me). I understand that the $(overline{X_n})^2$ was removed from the sum. But where is the $2ab$ part?



So, How can the adjustment be accurate?






algebra-precalculus statistics estimation-theory




algebra-precalculus statistics estimation-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 20:39







TGar

















asked Jan 12 at 20:27









TGarTGar

1558




1558












  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36


















  • $begingroup$
    If you expand the sum you will see how.
    $endgroup$
    – John Douma
    Jan 12 at 20:36
















$begingroup$
If you expand the sum you will see how.
$endgroup$
– John Douma
Jan 12 at 20:36




$begingroup$
If you expand the sum you will see how.
$endgroup$
– John Douma
Jan 12 at 20:36










1 Answer
1






active

oldest

votes


















4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45











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1 Answer
1






active

oldest

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1 Answer
1






active

oldest

votes









active

oldest

votes






active

oldest

votes









4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45
















4












$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$













  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45














4












4








4





$begingroup$

It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!






share|cite|improve this answer









$endgroup$



It's actually pretty simple but the brackets in your picture might be a bit confusing. For simplicity reasons I'll only look at the sum.



We know that $$sum^{n}_{i=1}(X_i - bar{X_{n}})^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2$$



Which in turn reduces to $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}sum^{n}_{i=1}X_i+sum^{n}_{i=1}bar{X_n}^2=(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2$$



Because ofcourse we know that $frac{1}{n}sum^{n}_{i=1}X_i=bar{X_n}$ so $sum^{n}_{i=1}X_i=nbar{X_n}$. Here is now the thing we want because $$(sum^{n}_{i=1}X_i^2)-2bar{X_n}nbar{X_{n}}+nbar{X_n}^2=(sum^{n}_{i=1}X_i^2)-nbar{X_n}^2$$



Put the constant of $frac{1}{1-n}$ that we left out for simplicity, back in front, and you're done!







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 12 at 20:38









S. CrimS. Crim

404212




404212












  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45


















  • $begingroup$
    How can you put $2bar{Xn}$ outside the sum in the first part?
    $endgroup$
    – TGar
    Jan 12 at 20:42










  • $begingroup$
    Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
    $endgroup$
    – S. Crim
    Jan 12 at 20:43












  • $begingroup$
    Sure, of course, you're right! I understand. Thanks!
    $endgroup$
    – TGar
    Jan 12 at 20:45
















$begingroup$
How can you put $2bar{Xn}$ outside the sum in the first part?
$endgroup$
– TGar
Jan 12 at 20:42




$begingroup$
How can you put $2bar{Xn}$ outside the sum in the first part?
$endgroup$
– TGar
Jan 12 at 20:42












$begingroup$
Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
$endgroup$
– S. Crim
Jan 12 at 20:43






$begingroup$
Because there is no subscript $i$ anymore, so you are not summing over anything. Instead, try looking at the $bar{X_n}$ as just a constant $c$ (ofcourse it isn't actually constant, but for simplicity). Do you agree that you could then take this constant $c$ out of the sum?
$endgroup$
– S. Crim
Jan 12 at 20:43














$begingroup$
Sure, of course, you're right! I understand. Thanks!
$endgroup$
– TGar
Jan 12 at 20:45




$begingroup$
Sure, of course, you're right! I understand. Thanks!
$endgroup$
– TGar
Jan 12 at 20:45


















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