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SD derivation

Expert replies
by yellowho » Fri Feb 18, 2011 2:20 am
The SD of the population of 5 cities in 1975 is 30,000, what is the SD of the population of the 5 cities in 1985?

1) The total changes is 200,000 and the population of every city grew by at least 10,000.

Here you don't know because you don't know how much is added to each. Essentially there can be different numbers added to every element of the set.

2) The total population of each city grew by 10%

I know that if every number in a set is multiply by X then the new SD is just old SD times X.

However, I'm trying to reconcile that concept with the concept that you don't know the SD of the new set if you add different numbers to every element of the set. By multiplying by percentage, isn't that analogous to adding a different number to the set? The only time its not is if the set has all the same value, which case the SD is zero.


I know thats thinking is wrong but can someone elaborate on why that's wrong?
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Source: — Data Sufficiency |

by stormier » Fri Feb 18, 2011 5:56 am
yellowho wrote:The SD of the population of 5 cities in 1975 is 30,000, what is the SD of the population of the 5 cities in 1985?

1) The total changes is 200,000 and the population of every city grew by at least 10,000.

Here you don't know because you don't know how much is added to each. Essentially there can be different numbers added to every element of the set.

2) The total population of each city grew by 10%

I know that if every number in a set is multiply by X then the new SD is just old SD times X.

However, I'm trying to reconcile that concept with the concept that you don't know the SD of the new set if you add different numbers to every element of the set. By multiplying by percentage, isn't that analogous to adding a different number to the set? The only time its not is if the set has all the same value, which case the SD is zero.


I know thats thinking is wrong but can someone elaborate on why that's wrong?

SD = sqrt [SUM (xi-xav)^2]

let's consider a set

a,b,c where d is the mean

then the terms in SD calculation would be

(a-d)^2 + (b-d)^2 + (c-d)^2

when you increase each term by exactly 10 %, the new terms become

1.1a, 1.1b, 1.1c and the new mean becomes 1.1d

the terms in SD calculation would be

(1.1a-1.1d)^2 + (1.1b-1.1d)^2 + (1.1c-1.1d)^2

= 1.1^2 [(a-d)^2 + (b-d)^2 + (c-d)^2]

and when you take the sqare root, the new SD becomes 1.1 times old SD.


Although we subtract different numbers from new mean, the percentage change is same in all and thus can be taken out of the equation as a common multiplier.

However, if you add different numbers to the set [which are not a fixed % higher or lower], you cannot take a common multiplier out and thus cannot establish a relation between the old and the new SD.

When you add or subtract different numbers to the terms, each of the terms will change by a different percentage amount.

hope it helps
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by ankur.agrawal » Sat Feb 19, 2011 4:19 am
Hey Gr8 insight Into SD. Thanks.
stormier wrote:
yellowho wrote:The SD of the population of 5 cities in 1975 is 30,000, what is the SD of the population of the 5 cities in 1985?

1) The total changes is 200,000 and the population of every city grew by at least 10,000.

Here you don't know because you don't know how much is added to each. Essentially there can be different numbers added to every element of the set.

2) The total population of each city grew by 10%

I know that if every number in a set is multiply by X then the new SD is just old SD times X.

However, I'm trying to reconcile that concept with the concept that you don't know the SD of the new set if you add different numbers to every element of the set. By multiplying by percentage, isn't that analogous to adding a different number to the set? The only time its not is if the set has all the same value, which case the SD is zero.


I know thats thinking is wrong but can someone elaborate on why that's wrong?

SD = sqrt [SUM (xi-xav)^2]

let's consider a set

a,b,c where d is the mean

then the terms in SD calculation would be

(a-d)^2 + (b-d)^2 + (c-d)^2

when you increase each term by exactly 10 %, the new terms become

1.1a, 1.1b, 1.1c and the new mean becomes 1.1d

the terms in SD calculation would be

(1.1a-1.1d)^2 + (1.1b-1.1d)^2 + (1.1c-1.1d)^2

= 1.1^2 [(a-d)^2 + (b-d)^2 + (c-d)^2]

and when you take the sqare root, the new SD becomes 1.1 times old SD.


Although we subtract different numbers from new mean, the percentage change is same in all and thus can be taken out of the equation as a common multiplier.

However, if you add different numbers to the set [which are not a fixed % higher or lower], you cannot take a common multiplier out and thus cannot establish a relation between the old and the new SD.

When you add or subtract different numbers to the terms, each of the terms will change by a different percentage amount.

hope it helps
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by ankur.agrawal » Sat Feb 19, 2011 4:36 am
Key Takeaway from question:

1) If we multiply each element of a set by a common factor/ (or increase/decrease each element by fixed %) / ( increase/decrease each element by the same amount then):

New Mean or New SD = Old Mean or Old SD * common factor.

New Mean or New SD = Old Mean or old SD * (1+% increase/decrease)

New Mean or New SD = Old Mean or Old SD + increase/decrease in amount.

Thanks.
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by Night reader » Sat Feb 19, 2011 5:18 am
Hi Ankur, good note
and further to your observation (just don't label me complicator :) ) in your top MBA class they will be teaching you that old mean vs. new mean and old SD vs. new SD have great impacts on business risk - coefficient of variation (CV) which is SD/Mean will change this way 1) multiplication/division of values by common factor --> result zero (no change) on CV (risk); 2) addition of const. +ve "a" to the set values (increase by a) --> decrease on CV (risk); 3) addition of const. -ve "a" to the set values (decrease by a) --> increase on CV (risk)

Cheers.
ankur.agrawal wrote:Key Takeaway from question:

1) If we multiply each element of a set by a common factor/ (or increase/decrease each element by fixed %) / ( increase/decrease each element by the same amount then):

New Mean or New SD = Old Mean or Old SD * common factor.

New Mean or New SD = Old Mean or old SD * (1+% increase/decrease)

New Mean or New SD = Old Mean or Old SD + increase/decrease in amount.

Thanks.
Last edited by Night reader on Sat Feb 19, 2011 12:23 pm, edited 1 time in total.
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by ankur.agrawal » Sat Feb 19, 2011 11:43 am
Hey Thanks Buddy for CV inputs. I think i read it sumwhere while preparing for CFA level 1 .

One thing: Have u already done ur MBA?
Night reader wrote:Hi Ankur, good note
and further to your observation (just don't label me complicator :) ) in your top MBA class they will be teach you that old mean vs. new mean and old SD vs. new SD have great impacts in business risk - coefficient of variation (CV) which is SD/Mean will change this way 1) multiplication/division of values by common factor --> result zero (no change) on CV/risk; 2) addition of const. +ve "a" to the set values (increase by a) --> decrease on CV/risk; 3) addition of const. -ve "a" to the set values (decrease by a) --> increase on CV/risk

Cheers.
ankur.agrawal wrote:Key Takeaway from question:

1) If we multiply each element of a set by a common factor/ (or increase/decrease each element by fixed %) / ( increase/decrease each element by the same amount then):

New Mean or New SD = Old Mean or Old SD * common factor.

New Mean or New SD = Old Mean or old SD * (1+% increase/decrease)

New Mean or New SD = Old Mean or Old SD + increase/decrease in amount.

Thanks.
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by Night reader » Sat Feb 19, 2011 12:19 pm
please go private - it's off-topic :)
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