Cross-correlation of filtered random processes
As an addition to Dilip's answer I'll show you how to derive that result:

$$\begin{align}R_{y_1,y_2}(\tau)&amp;=E[y_1(t+\tau)y_2(t)]\\&amp;=E\left[\int_{-\infty}^{\infty}x(\alpha)h_1(t+\tau-\alpha)d\alpha\int_{-\infty}^{\infty}x(\beta)h_2(t-\beta)d\beta\right]\\&amp;=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}E[x(\alpha)x(\beta)]h_1(t+\tau-\alpha)h_2(t-\beta)d\alpha d\beta\\&amp;=\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}R_x(\alpha-\beta)h_1(t+\tau-\alpha)h_2(t-\beta)d\alpha d\beta\\&amp;\stackrel{\gamma=\alpha-\beta}{=}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}R_x(\gamma)h_1(t+\tau-\alpha)h_2(t-\alpha+\gamma)d\alpha d\gamma\\&amp;\stackrel{\zeta=\alpha-t}{=}\int_{-\infty}^{\infty}\underbrace{\int_{-\infty}^{\infty}R_x(\gamma)h_2(\gamma-\zeta)d\gamma}_{(R_x\star h_2^-)(\zeta)}\; h_1(\tau-\zeta) d\zeta\\&amp;=(R_x\star h_1\star h_2^-)(\tau)\qquad\qquad\qquad (1)\end{align}$$

with $h_2^-(t)=h_2(-t)$.

The cross-spectral density $S_{y_1,y_2}(\omega)$ is the Fourier transform of the cross-correlation function:

$$S_{y_1,y_2}(j\omega)=S_x(j\omega)H_1(j\omega)H_2^*(j\omega)\tag{2}$$

where $S_x(j\omega)$ is the power spectral density of $x(t)$, and $H_1(j\omega)$ and $H_2(j\omega)$ are the frequency responses of the two filters.

Using $(2)$ it is straightforward to define a condition on $H_1(j\omega)$ and $H_2(j\omega)$ such that the cross-spectral density, and, consequently, the cross-correlation function become zero.