Understanding the H1 and H2 estimators
Proving that the H1 estimator ignores noise on the output:
If we assume that the output is the input processed by the transfer function plus some output noise, we have $Y = H(f)X(f) + N$, where everything is in the freq domain.
So, $S_{xy}(f) = E[X(f)^*Y(f)] = E[X(f)^*(H(f)X(f) + N(f))]$. where E[] denotes expectation and * denotes complex conjugate.
If you multiply this out, you get
$S_{xy}(f) = H(f)E[X(f)^*X(f)] + E[X(f)^*N(f)]  = H(f)S_{xx}(f)$
Crucially, we assume that the input is uncorrelated with the noise so the expectation of the noise term cancels to zero.
Rearranging, we get the H1 estimator, which is not affected by noise on the output:
$H1 = \frac{S_{xy}}{S_{xx}}$
If you try this with noise on the input, the answer isn't as clean - this is because the H1 estimator is not immune to noise on the input. If you repeat a similar proof for the H2 estimator you will be pleasantly surprised :-)