<p>Is it the case that since you can rearrange this expression to $\Sigma (\frac{1}{y^2})(\hat{y}-y)^2$, that you are really weighting the standard L2 norm with  weights that are higher for lower absolute values of y? </p>

CC BY-SA 3.0
2015-01-30T16:38:38.390Z
Yes, check out weighted least squares. You could do this if you cared more about getting small absolute values of y predicted correctly or if you though that there was measurement error in y that was larger for large values of y.
BKay
CC BY-SA 3.0
2015-01-30T16:25:05.377Z
2015-01-31T16:33:40.203Z
2015-01-31T16:33:40.203Z
Hessian
<p>So I was reading about the consequences of various loss functions on what regressions result in, i.e. the L1 norm giving conditional median estimation and L2 giving conditional mean, etc.</p>

<p>What happens if you minimize the sum of $\left(\frac{\hat{y}}{y}-1 \right)^2$ over all observations? Provided no y != 0, I get Monte Carlo results that are very close to standard OLS results.</p>

<p>I SHOULD be doing my time series homework but instead I've been tinkering around.</p>

<p>Not a super serious question, just curiosity :)</p>

CC BY-SA 3.0
2015-01-31T20:16:50.210Z
This is probably better asked on stats.stackexchange.com
shadowtalker
<econometrics>
Consequences of a particular loss function
curiousregmonkey
# Consequences of a particular loss function

(Asked by: username_0 on 2015-01-29)

So I was reading about the consequences of various loss functions on what regressions result in, i.e. the L1 norm giving conditional median estimation and L2 giving conditional mean, etc.

What happens if you minimize the sum of $\left(\frac{\hat{y}}{y}-1 \right)^2$ over all observations? Provided no y != 0, I get Monte Carlo results that are very close to standard OLS results.

I SHOULD be doing my time series homework but instead I've been tinkering around.

Not a super serious question, just curiosity :)

**Tags:** econometrics
> **Comments:**
> **username_1**: This is probably better asked on stats.stackexchange.com  

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## 1 answers

### Answer by username_2 on 2015-01-30. Score: 5

Is it the case that since you can rearrange this expression to $\Sigma (\frac{1}{y^2})(\hat{y}-y)^2$, that you are really weighting the standard L2 norm with weights that are higher for lower absolute values of y?

> **Comments:**
> **username_3**: Yes, check out weighted least squares. You could do this if you cared more about getting small absolute values of y predicted correctly or if you though that there was measurement error in y that was larger for large values of y.  

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