Let's assume we are faced with showing that the following covariance between two functions $g(\cdot)$ and $f(\cdot)$ of the same set of $Z_1,...,Z_K$ i.i.d. standard normal random variables is equal to $0$ (we want to show that they are independent), such that:
$$Cov\left[f(Z_k),g(Z_k)\right]= Cov\left[\sqrt{\hat{\rho}_{s}}\hat{Y},\sum_{k=1}^K\left(\sqrt{\rho_{s}}\alpha_{s,k} - \sqrt{\hat{\rho}_{s}}\beta_{k}\right)Z_k\right] = 0 $$ where $S=K$ and $$\hat{Y} = \sum_{k=1}^K\beta_{k} Z_k \quad\;,\quad\; \sum_{k=1}^K\beta^2_{k} = 1 \quad\;,\quad\; \sum_{k=1}^K\alpha^2_{s,k}=1$$ Additionally, we know the following relationship between the parameters: $$\sqrt{\hat{\rho}_{s}} = \sqrt{\rho_{s}}\sqrt{\gamma_{s}} = \sqrt{\rho_{s}}\sum_{k=1}^{K}\alpha_{s,k}\beta_k$$ consider that, overall, $\rho_{s},\gamma_{s},\hat{\rho}_{s},\alpha_{s,k},\beta_k$ are just model parameters.
I know that the above boils down to needing to show that (as all other covariances are 0): $$Cov\left[f(Z_k)|g(Z_k)\right] = \left[\sqrt{\hat{\rho}_{s}}\sum_{k=1}^K\beta_{k} \times \sum_{k=1}^K\left(\sqrt{\rho_{s}}\alpha_{s,k} - \sqrt{\hat{\rho}_{s}}\beta_{k}\right)\right] \underbrace{V[Z_k]}_{=1} = 0 $$
I have tried running the algebra a couple of times to show that the part inside the squared brackets $=0$ but was unable to reach that conclusion. Is there some properties of sums I am missing?
Any help or suggestion would be much appreciated.
Recall $\textrm{Cov}[X,Y]=E[XY]-E[X]E[Y]$. Note that $E[f]=E[g]=0$. If I understood correctly all the notation, we then have $$\begin{aligned}E[fg]&=\sum_{k\leq K}\sum_{\ell\leq K}(\sqrt{\rho_s}\alpha_{s,k}-\sqrt{\hat{\rho}_s}\beta_k)\sqrt{\hat{\rho}_s}\beta_\ell E[Z_kZ_\ell]\\ &=\sum_{k\leq K}(\sqrt{\rho_s}\sqrt{\hat{\rho}_s}\alpha_{s,k}\beta_k-{\hat{\rho}_s}\beta_k^2)\\ &=\sqrt{\hat{\rho}_s}\bigg(\sqrt{\rho_s}\sum_{k\leq K}\alpha_{s,k}\beta_k\bigg)-\hat{\rho}_s\bigg(\sum_{k\leq K}\beta_k^2\bigg)\\ &=\hat{\rho}_s-\hat{\rho}_s\\\ &=0 \end{aligned}$$