Cross Power Spectral Density from Individual Power Spectral Densities Let $X$ and $Y$ be two zero-mean, wide-sense stationary random processes.
The power spectral density of a process is the Fourier transform of the process's auto-correlation function. The cross power spectral density of two processes is the Fourier transform of the cross-correlation function of two processes. 
Given the power spectral density of $X$ and $Y$, $S_X(f)$ and $S_Y(f)$, how would I find the cross power spectral density of $X$ and $Y$? For this question to be doable, would I need to assume independence between $X$ and $Y$?
 A: There are two parts to your question. For the second part, assuming X(t) and Y(t') are independent, requires the cross spectral density to be zero. The cross spectral density is the Fourier transform of the cross correlation function.   The cross correlation is the ensemble average of the time-shifted product of X(t) and Y(t'), and if these are independent zero-mean processes than the ensemble average is the product of the two means is zero, thus making the cross spectral density zero.
To find an expression for the cross spectral density start by showing the relationship of $X(t)$ to $S_X(f)$.  A WSS random function $X(t)$ can be represented by a Fourier series over the interval $T$ $$X(t) = \sum_{m=-\infty}^{+\infty}A_me^{i2\pi mt/T + i\zeta _m}$$ where $A_m$ is the Fourier coefficient related to the spectrum $S_X(f_m)$, the frequency is given by $f_m = m/T$, and the random behavior of $X(t)$ is carried in the random phase $\zeta_m$.  The phase $\zeta_m$ is a uniformly distributed random variable between $0$ and $2\pi$; a random value of $\zeta_m$ is assigned to each term in the Fourier series.  An infinite number of ensembles of $X(t)$ can be created by choosing new values of the phases $\zeta_m$ from the distribution.  The mean value of $X(t)$ is given by $A_0$ with $\zeta_0 = 0$, and for a zero-mean process $A_0 = 0$.  (Note: most of the literature that I have seen erroneously treats $\eta_0$ as a random variable; this is wrong because this makes the mean value a member of a distribution instead of a constant as required by stationarity) 
The autocorrelation $R(\tau)$ is given by the temporal average over $t$ or the ensemble averages over $\zeta_m$; here we use the ensemble averages $$R(\tau) = \langle X(t)X^*(t+\tau)\rangle = \sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}A_m A_n e^{i2\pi (m-n)t/T -i2\pi n\tau/T} \langle e^{i(\zeta _m-\zeta_n)}\rangle $$ The ensemble averages are zero everywhere except when $m = n$ so that the autocorrelation reduces to $$R(\tau) = \sum_{m=-\infty}^{+\infty}A_m^2 e^{-i2\pi m\tau/T} $$  
An estimate of the $m^{th}$ component of the spectrum of $X(t)$ on the finite interval $T$ is given by taking the Fourier transform of the autocorrelation estimate$$S_m = \int_{-T/2}^{T/2}R(\tau) e^{i2\pi p\tau/T}d\tau = \sum_{m=-\infty}^{+\infty}A_m^2 \int_{-T/2}^{T/2} e^{-i2\pi (m-p)\tau/T}d\tau = A_m^2T $$ where the integral is zero except when $p = m$. If we think of $S_m$ as a sample of $S_X(f_m)$ then the Fourier coefficient $A_m$ can be related to the power spectral density$$A_m = \sqrt\frac{S_m}T = \sqrt\frac{S_X(f_m)}T$$ Substituting this value for $A_m$ into the autocorrelation function yields $$R(\tau) = \sum_{m=-\infty}^{+\infty}\frac{S_m}T e^{-i2\pi m\tau/T} $$ Define $2\pi /T = \Delta\omega$ and after substituting this into the expression for autocorrelation take the limit as $T$ goes to infinity $$R(\tau) = \lim_{T \to \infty}\sum_{m=-\infty}^{+\infty} S_m e^{-i\Delta\omega \tau} \Delta\omega = \frac{1}{2\pi}\int_{-\infty}^{+\infty}S(\omega) e^{-i\omega\tau}d\omega  $$ where the Reimann sum has been converted to an integral. Assume all continuity and convergence requirements are met.
All of the foregoing was just to establish that $X(t)$ and $Y(t+\tau)$ can be written in terms of their spectra as $$X(t) = \sum_{m=-\infty}^{+\infty}\sqrt\frac{S_X(f_m)}T e^{i2\pi mt/T + i\zeta _m}$$ $$Y(t+\tau) = \sum_{n=-\infty}^{+\infty}\sqrt\frac{S_Y(f_n)}T e^{i2\pi n(t+\tau)/T + i\mu_n}$$  The cross correlation is the ensemble average of the product of $X(t)$ and the complex conjugate of $Y(t+\tau)$  $$R_{XY}(\tau) = \sum_{m=-\infty}^{+\infty}\sum_{n=-\infty}^{+\infty}\frac{\sqrt{S_X(f_m)S_Y(f_n)}}T e^{i2\pi (m-n)t/T -i2\pi n\tau/T} \langle e^{i(\zeta _m-\mu_n)}\rangle $$ 
To achieve a non-zero cross spectral density the ensemble average of the cross correlation function cannot be zero; this means that phases between the spectral components must be related. Otherwise, if the phases $\zeta_m$ and $\mu_n$ are independent the ensemble average will always be zero. So if the phases between components with the same frequency are assumed to be related by $\zeta_m - \mu_m = \delta_m$ and $\zeta_m - \mu_n = 0$ the cross correlation can be written as $$R_{XY}(\tau) = \sum_{m=-\infty}^{+\infty}\frac{\sqrt{S_X(f_m)S_Y(f_m)}}T e^{-i2\pi m\tau/T+i\delta_m} $$ 
Again, define $2\pi /T = \Delta\omega$ and after substituting this into the expression for autocorrelation take the limit as $T$ goes to infinity $$R_{XY}(\tau) = \lim_{T \to \infty}\sum_{m=-\infty}^{+\infty} \sqrt{S_X(f_m)S_Y(f_m)} e^{-i\Delta\omega \tau+i\delta(f_m)} \Delta\omega $$ $$ = \frac{1}{2\pi}\int_{-\infty}^{+\infty} \sqrt{S_X(\omega)S_Y(\omega)} e^{-i\omega\tau+i\delta(\omega)}d\omega  $$ where the Reimann sum has been converted to an integral.
The cross power spectral density is just the Fourier transform of the cross correlation and it can be obtained by visual inspection of the above expression $$ S_{XY}(\omega) = \sqrt{S_X(\omega)S_Y(\omega)} e^{i\delta(\omega)} $$
A few observations on the above result. Although the question pertained to the spectra of the two processes, the more crucial parameter is the phase relationship $\delta(\omega)$. Also, consider relaxing the WSS requirement. One common mechanism that produces correlated processes is for one of them to be the time evolved version of the other; this only happens when WSS does not apply.      
