Why is the variance of the error term (a.k.a., the “irreducible error”) always 1 in examples of the bias-variance tradeoff?

I'm reading Introduction to Statistical Learning. The relevant part is referenced here: Proof/Derivation of Residual Sum of Squares (Based on Introduction to Statistical Learning).

When the author shows graphs that illustrate "Bias vs Variance Tradeoff" (as in Figure 2.12), the $${\rm Var}(\varepsilon)$$ is always $$1$$ (note the dashed lines in the figures):

The conditions of $$\varepsilon$$ are clarified elsewhere, as on page 16:

$$\varepsilon$$ is a random error term, which is independent of $$X$$ and has mean zero.

... and there is some explanation about going from "random error term" to "irreducible error":

However, even if it were possible to form a perfect estimate for $$f$$, so that our estimated response took the form $$\hat{Y} = f(X)$$, our prediction would still have some error in it! This is because $$Y$$ is also a function of $$\varepsilon$$, which, by definition, cannot be predicted using $$X$$. Therefore, variability associated with $$\varepsilon$$ also affects the accuracy of our predictions.

But I don't see anywhere in the other SO questions, nor in the book: why is $$Var(\varepsilon)$$ always at 1?

• Is it because the "mean is zero"? I don't think so; I could describe a dataset with mean of zero but a variance of $$\ne 1$$.
• Is it because, as described elsewhere, the "the error term $$\varepsilon$$ is normally distributed"? I don't know enough about the normal distribution; is the variance of a normal distribution is always equal to some value?

EDIT

In looking for help in Wikipedia's MSE article, I expected to find a consistent formula with the "three fundamental quantities" (i.e., the variance, the bias, and the variance of the error terms), but I didn't. Can someone tell me why the Wikipedia doesn't list the variance of error terms:

$$\operatorname{MSE}(\hat{\theta})=\operatorname{Var}(\hat{\theta})+ \left(\operatorname{Bias}(\hat{\theta},\theta)\right)^2$$

• When doing examples, people will often just set the variance of the error term $\epsilon$ equal to 1. If they didn't want to fix a value for $Var(\epsilon) = \sigma^2$ then they wouldn't have been able to make those plots with particular numbers on the side. – jld Aug 9 '16 at 1:23
• So is it just convention? Why 1? 1 is convenient? No other assumptions about error term can lead us to this effect? ( I realize I am probably over thinking something that is, by definition, unknowable...) – The Red Pea Aug 9 '16 at 1:25
• Author does say, "...the irreducible error will always provide an upper bound on the accuracy of our prediction for $Y$. This bound is almost always unknown in practice" Is this as good as admitting that 1 is an arbitrary choice? – The Red Pea Aug 9 '16 at 1:27
• also this has absolutely no bearing on real data. This is purely for the sake of the examples that they're doing. In real life $Var(\epsilon)$ could be anything – jld Aug 9 '16 at 1:29
• Thanks @Chaconne, you should've answered so I could upvote a tortoise. Can we calculate $Var(\epsilon)$ in real life? Or is that the "unknown" author describes in practice – The Red Pea Aug 9 '16 at 2:25

It isn't because the mean is $0$ or because the error term is normally distributed. In fact, the normal distribution is the only 'named' distribution where the mean and the variance are independent of each other (see: What is the most surprising characterization of the Gaussian (normal) distribution?).
More generally, my strong guess is that the purpose of setting the variance of the errors equal to $1$ is pedagogical. Everything in the figures can be related to the variance of the error term because the unit of measurement in the figures is $1$ and that was set as the variance of the error term.
Regarding the Wikipedia article, be aware that the variance of theta is a function of the variance of the error term, so ${\rm Var}(\hat\theta)$ does include ${\rm Var}(\varepsilon)$ (it's just out of sight).