# Expectation of power spectrum for nonorthogonal wavelets

I'm working through "A Practical Guide to Wavelet Analysis" by Torrence and Compo, and I am confused about section 3d ("Wavelet Power Spectrum").

Let $$x_n$$ denote the signal, sampled at increments $$\delta t$$; $$\hat{x}_k$$ its Fourier transform; $$\psi$$ the (normalized) mother wavelet, so that the amplitude of the wavelet transform at scale $$s$$ centered at $$n\delta t$$ is given by $$W_n(s) = \sum_k \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t}$$ There's some fiddling with constants here, but I understand this so far.

Then, they write that, for a white noise signal, $$\mathbb{E}|W_n(s)|^2 = N \mathbb{E}|\hat{x}_k|^2 = \sigma^2$$. Trying to work through this: \begin{align} \mathbb{E}|W_n(s)|^2 &= \mathbb{E}\left|\sum \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t}\right|^2 \end{align} I see that if the wavelets are orthogonal, this works out nicely, where you can write $$\mathbb{E}\left|\sum \hat{x}_k \hat{\psi}^*(s \omega_k)e^{i\omega_k n\delta t}\right|^2 = \sum |\hat{\psi}(s \omega_k)|^2\mathbb{E}|\hat{x}_k|^2 = \frac{\sigma^2}{N}\sum |\hat{\psi}(s \omega_k)|^2$$ However, the authors clearly state they are not working with orthogonal wavelets specifically here, so I am unsure how they get their white noise result.