Why Linex Loss Approaches Quadratic Loss as Asymmetry Vanishes
Summary
The document explains why the linear-exponential (Linex) loss function converges to squared-error loss as its asymmetry parameter approaches zero. It considers a forecast error and expands the exponential term in the Linex formula as a Maclaurin series. The constant and first-order terms cancel against the other terms in the formula, leaving the squared error as the leading term after scaling. Higher-order terms contain powers of the asymmetry parameter, so they vanish in the limit.
The derivation establishes the limiting relationship for a fixed forecast error and provides an intuitive way to see how an asymmetric loss becomes quadratic near zero asymmetry. A second answer suggests applying L’Hôpital’s rule twice, but does not show the steps. The discussion is a mathematical explanation rather than a trading application: it gives no empirical tests, forecast comparisons, or guidance on selecting the asymmetry parameter. Its result concerns the stated Linex formulation and normalization; other parameterizations may scale the limiting quadratic loss differently.
Key ideas
- A Maclaurin expansion of the exponential term exposes the Linex loss limit as its asymmetry parameter approaches zero.
- The constant and linear terms cancel, leaving squared forecast error as the leading term.
- Terms of higher order in the asymmetry parameter disappear in the limit.
- The resulting convergence is a mathematical property, not evidence about forecast performance.
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# How does Linear-Exponential Loss (Linex) function tend towards Quadratic Loss function?
# How does Linear-Exponential Loss (Linex) function tend towards Quadratic Loss function?
Thank you for your help everyone, and I apologise beforehand if this is a lousy or dumb question.
I am looking to read up more on Quadratic Loss & Linex Loss, and forecast optimality. In my university text, we were told that the quadratic loss function is essentially the squared-error of the loss function, or square of the difference between actual and forecasted value, as shown below:
We are then told that the Linear-exponential loss function is an asymmetric loss function that tends towards the Quadratic loss function as a tends towards zero, as shown below:
My question is how can the Linex loss function tend towards the quadratic loss function as a tends towards zero? I might be a bit clueless and missing the math here but I can't seem to prove that as a tends towards zero, Linex function tends to Quadratic function.
Any help would be greatly appreciated. Thank you!
## Answer by Magic is in the chain (score 4, accepted)
https://quant.stackexchange.com/a/54882
It just needs the power series(Maclaurin) expansion: $e^{x}=1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+\dots$
Take the Linex function:
$L\left(y,\overset{\wedge}{y}\right)=\frac{2}{a^2}\left[e^{a\left(y-\overset{\wedge}{y}\right)}-a\left(y-\overset{\wedge}{y}\right)-1\right]$
and substitute the power series for the exponential term:
$L\left(y,\overset{\wedge}{y}\right)=\frac{2}{a^2}\left[1+a\left(y-\overset{\wedge}{y}\right)+\frac{a^2\left(y-\overset{\wedge}{y}\right)^2}{2!}+\frac{a^3\left(y-\overset{\wedge}{y}\right)^3}{3!}+\dots-a\left(y-\overset{\wedge}{y}\right)-1\right]$
And then simplify:
$L\left(y,\overset{\wedge}{y}\right)=\frac{2}{a^2}\left[\frac{a^2\left(y-\overset{\wedge}{y}\right)^2}{2!}+\frac{a^3\left(y-\overset{\wedge}{y}\right)^3}{3!}+\dots\right]=\left(y-\overset{\wedge}{y}\right)^2 +\frac{2a\left(y-\overset{\wedge}{y}\right)^3}{3!}+\dots$
And then the limit is easily seen to be the quadratic:
$\lim_{a \to 0} L\left(y,\overset{\wedge}{y}\right)=\left(y-\overset{\wedge}{y}\right)^2$
## Answer by Raju Dey (score 0)
https://quant.stackexchange.com/a/68748
Just apply L Hospital's rule two times on the Linex loss function, you will obtain the required result.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.