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:

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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:

enter image description here

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!


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:


and substitute the power series for the exponential term:


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:

$\lim_{a \to 0} L\left(y,\overset{\wedge}{y}\right)=\left(y-\overset{\wedge}{y}\right)^2$

| improve this answer | |
  • $\begingroup$ Hi Magic, thank you for your help! I totally did not think to use the Maclaurin's. $\endgroup$ – Kai Jun 14 at 10:21

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