I am trying to understand the function of a Wiener filter. I get that the Wiener filter minimizes the mean square error between the estimated random process and the desired process.
Does this mean that when you have a signal with low and high frequencies and the noise is only in in the high frequencies that the resulting signal only contains low frequencies? What would the transfer function look like in this case?
In general, an ideal Wiener filter (i.e. a non-causal filter with an infinitely long impulse response) has the following frequency response
$$W(\omega)=\frac{S_{dx}(\omega)}{S_x(\omega)}\tag{1}$$
where $S_{dx}(\omega)$ is the cross-power spectral density of the desired signal $d[n]$ and the input of the Wiener filter $x[n]$, and $S_x(\omega)$ is the power spectral density (PSD) of the input process $x[n]$. If the input is just a noisy version of the desired signal, and if the desired signal and the noise are uncorrelated, then we have
$$S_x(\omega)=S_d(\omega)+S_n(\omega)\tag{2}$$
where $S_n(\omega)$ is the PSD of the additive noise, and $S_d(\omega)$ is the PSD of the desired signal. Furthermore, we have
$$S_{dx}(\omega)=S_d(\omega)\tag{3}$$
because the noise and the desired signal are uncorrelated. Plugging (2) and (3) into (1) gives
$$W(\omega)=\frac{S_d(\omega)}{S_d(\omega)+S_n(\omega)}\tag{4}$$
This expression is very easily interpreted: for frequencies where the noise is much stronger than the signal (i.e. $S_n(\omega)\gg S_d(\omega)$ the Wiener filter strongly attenuates the input signal, whereas for frequencies with a very good signal-to-noise ratio (SNR) (i.e. $S_d(\omega)\gg S_n(\omega)$) the filter passes the input with a gain of approximately $1$.
So your guess is correct: if there's a lot of noise in the high frequencies (relative to the desired signal), then these frequencies are attenuated, and the low frequencies, where the SNR is high, will pass the filter with a gain of approximately $1$.
