Solving Variance Problem - Hi All

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SUMMARY

The discussion centers on modifying a dataset V with a mean of 0 and variance var(V)=x to achieve a new variance var(V_hat)=y. The original poster attempted to calculate the necessary error adjustment using the formula delta = sqrt[(y-x)*(n-1) / n], but found it ineffective. A more straightforward solution was provided, suggesting that multiplying each element of V by sqrt(y/x) effectively achieves the desired variance. This method simplifies the process of variance adjustment in datasets.

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Asuralm
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Hi all:
If I have a dataset V of the size n*1. Assume that the mean of the dataset is 0 and var(V)=x is its variance. If I want to modify this dataset so that the variance of the new dataset will be var(V_hat)=y. The errors are spreaded average on each element in the dataset. What I did is first calculate the error which need to be speaded on each elements

v_hat - v = delta = sqrt[(y-x)*(n-1) / n] (0)

because

var(V) = x = sum(v^2)/(n-1) (1)
var(V_hat) = y = sum(v_hat^2)/(n-1) (2)

(2)-(1) and rearrange I got the equation (0);

However, this didn't give me the supposed answer. Could anyone point out what's the error in this or if there is any better methods please?
Thanks a lot!
 
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I am not sure what limits you have on the modification. However, you can multiply each element of V by sqrt(y/x) to get what you want.
 
What a nice and easy way. How stupid I am. Thanks a lot!
 

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