I first give the notations of bias-variance tradeoff:
sample set: $D = \{(x_1,y_1),\cdots,(x_N,y_N)\}.$
relation between $y$ and $x:$ $y = f(x) + \epsilon.$ here $f(x)$ is a deterministic function; $\epsilon \sim (0,\sigma^2)$ is noise.
$\hat{f}(x;D):$ the estimation of $f(x)$ base on the samples $D.$
Then we have the decomposition of mean squared estimation (MSE)
:
$$E_D[(y-\hat{f}(x;D))^2] =\left(E_D[\hat{f}] - f\right)^2 + E_D[(E_D[\hat{f}] - f)^2] + \sigma^2$$ $$=\left(Bias_D(\hat{f})\right)^2 + Var_D(\hat{f}) + \sigma^2.$$
bias-variance tradeoff
usually means that we cannot reduce bias and variance simultaneously. But I confuse that
Here we only assume $y_i$ is random? i.e. all the randomness is from noise?
The tradeoff should be base on the assumption that MSE is fixed ($\sigma^2$ is irreducible). Then how do we understand this assumption? Does that mean when the sample set $D$ is given, there exists a set of optimal estimations $\{\hat{f}(x;D)\}$ minimizing MSE which is the fixed value we mentioned above? And the tradeoff happens among those optimal estimation $\{\hat{f}(x;D)\}?$
We know that the bagging usually reduces the variance without increasing the bias. Is it conflict to the tradeoff?
Here is my understanding after some discussions:
The equation of MSE decomposition is actually nothing about the trade off. It just illustrates that the MSE is from both Bias and Variance.
The trade off is a empirical conclusion that 'complex' algorithm usually has low bias but high variance vice versa.