Mathematical intuition of Bias-Variance equation I recently asked a question seeking a mathematical interpretation/intuition behind the elementary equation relating sample mean and variance: $ E[X^2] = Var(X) +(E[X])^2$, geometric or otherwise.
But now I'm curious about the superficially similar bias-variance tradeoff equation.
$$
\begin{eqnarray}
\text{MSE}(\hat{\theta}) = E [(\hat{\theta}-\theta)^2 ] 
      &=&  E[(\hat{\theta} - E[\hat\theta])^2] + (E[\hat\theta] - \theta)^2\\ 
      &=& \text{Var}(\hat\theta)               + \text{Bias}(\hat\theta,\theta)^2
\\
\end{eqnarray}
$$
(formulas from Wikipedia)
To me there is a superficial similarity with the bias-variance tradeoff equation for regression: three terms with squares and two adding to the other. Very Pythagorean looking. Is there a similar vector relationship including orthogonality for all these items? Or is there some other related mathematical interpretation that applies?
I am seeking a mathematical analogy with some other mathematical objects that might shed light. I am not looking for the accuracy-precision analogy which is well covered here. But if there are non-technical analogies that people can give between the bias-variance tradeoff and the much more basic mean-variance relation, that would be great too.
 A: This is a way to think visually about accuracy and the variance bias trade-off.  Suppose you are looking at a target and you make many shots that are all scattered close to the center of the target in such a way that there is no bias. Then accuracy is solely determined by variance and when the variance is small the shooter is accurate.
Now let us consider a case where there is great precision but a large bias.  In this case, the shots are scattered around a point far from the center.  Something is messing up the aim point but around this aim point, every shot is close to that new aim point.  The shooter is precise but very inaccurate because of the bias.
There are other situations where the shots are accurate because of small bias and high precision. What we want is no bias and small variance or small variance with small bias.  In some statistical problems, you can't have both.  So MSE becomes the measure of accuracy that you want to use that plays off the variance bias trade-off and minimizing MSE should be the goal.
A: The similarity is more than superficial. 
The "bias-variance tradeoff" can be interpreted as the Pythagorean Theorem applied to two perpendicular Euclidean vectors: the length of one is the standard deviation and the length of the other is the bias.  The length of the hypotenuse is the root mean squared error. 
A fundamental relationship
As a point of departure, consider this revealing calculation, valid for any random variable $X$ with a finite second moment and any real number $a$.  Since the second moment is finite, $X$ has a finite mean $\mu=\mathbb{E}(X)$ for which $\mathbb{E}(X-\mu)=0$, whence
$$\eqalign{
\mathbb{E}((X-a)^2) &= \mathbb{E}((X-\mu\,+\,\mu-a)^2) \\
&= \mathbb{E}((X-\mu)^2) + 2 \mathbb{E}(X-\mu)(\mu-a) + (\mu-a)^2 \\
&= \operatorname{Var}(X) + (\mu-a)^2.\tag{1}
}$$
This shows how the mean squared deviation between $X$ and any "baseline" value $a$ varies with $a$: it is a quadratic function of $a$ with a minimum at $\mu$, where the mean squared deviation is the variance of $X$.
The connection with estimators and bias
Any estimator $\hat \theta$ is a random variable because (by definition) it is a (measurable) function of random variables.  Letting it play the role of $X$ in the preceding, and letting the estimand (the thing $\hat\theta$ is supposed to estimate) be $\theta$, we have
$$\operatorname{MSE}(\hat\theta) = \mathbb{E}((\hat\theta-\theta)^2) = \operatorname{Var}(\hat\theta) + (\mathbb{E}(\hat\theta)-\theta)^2.$$
Let's return to $(1)$ now that we have seen how the statement about bias+variance for an estimator is literally a case of $(1)$.  The question seeks "mathematical analogies with mathematical objects."  We can do more than that by showing that square-integrable random variables can naturally be made into a Euclidean space.
Mathematical background
In a very general sense, a random variable is a (measurable) real-valued function on a probability space $(\Omega, \mathfrak{S}, \mathbb{P})$.  The set of such functions that are square integrable, which is often written $\mathcal{L}^2(\Omega)$ (with the given probability structure understood), almost is a Hilbert space.  To make it into one, we have to conflate any two random variables $X$ and $Y$ which don't really differ in terms of integration: that is, we say $X$ and $Y$ are equivalent whenever
$$\mathbb{E}(|X-Y|^2) = \int_\Omega |X(\omega)-Y(\omega)|^2 d\mathbb{P}(\omega) = 0.$$
It's straightforward to check that this is a true equivalence relation: most importantly, when $X$ is equivalent to $Y$ and $Y$ is equivalent to $Z$, then necessarily $X$ will be equivalent to $Z$.  We may therefore partition all square-integrable random variables into equivalence classes.  These classes form the set $L^2(\Omega)$.  Moreover, $L^2$ inherits the vector space structure of $\mathcal{L}^2$ defined by pointwise addition of values and pointwise scalar multiplication.  On this vector space, the function
$$X \to \left(\int_\Omega |X(\omega)|^2 d\mathbb{P}(\omega)\right)^{1/2}=\sqrt{\mathbb{E}(|X|^2)}$$
is a norm, often written $||X||_2$.  This norm makes $L^2(\Omega)$ into a Hilbert space.  Think of a Hilbert space $\mathcal{H}$ as an "infinite dimensional Euclidean space."  Any finite-dimensional subspace $V\subset \mathcal{H}$ inherits the norm from $\mathcal{H}$ and $V$, with this norm, is a Euclidean space: we can do Euclidean geometry in it.
Finally, we need one fact that is special to probability spaces (rather than general measure spaces): because $\mathbb{P}$ is a probability, it is bounded (by $1$), whence the constant functions $\omega\to a$ (for any fixed real number $a$) are square integrable random variables with finite norms.
A geometric interpretation
Consider any square-integrable random variable $X$, thought of as a representative of its equivalence class in $L^2(\Omega)$.  It has a mean $\mu=\mathbb{E}(X)$ which (as one can check) depends only on the equivalence class of $X$.  Let $\mathbf{1}:\omega\to 1$ be the class of the constant random variable.
$X$ and $\mathbf{1}$ generate a Euclidean subspace $V\subset L^2(\Omega)$ whose dimension is at most $2$.  In this subspace, $||X||_2^2 = \mathbb{E}(X^2)$ is the squared length of $X$ and $||a\,\mathbf{1}||_2^2 = a^2$ is the squared length of the constant random variable $\omega\to a$.  It is fundamental that $X-\mu\mathbf{1}$ is perpendicular to $\mathbf{1}$.  (One definition of $\mu$ is that it's the unique number for which this is the case.) Relation $(1)$ may be written
$$||X - a\mathbf{1}||_2^2 = ||X - \mu\mathbf{1}||_2^2 + ||(a-\mu)\mathbf{1}||_2^2.$$
It indeed is precisely the Pythagorean Theorem, in essentially the same form known 2500 years ago.  The object $$X-a\mathbf{1} = (X-\mu\mathbf{1})-(a-\mu)\mathbf{1}$$ is the hypotenuse of a right triangle with legs $X-\mu\mathbf{1}$ and $(a-\mu)\mathbf{1}$.
If you would like mathematical analogies, then, you may use anything that can be expressed in terms of the hypotenuse of a right triangle in a Euclidean space.  The hypotenuse will represent the "error" and the legs will represent the bias and the deviations from the mean.
