$Z(t)=Xt+Y$ is a random process find the following

Question

$X$ and $Y$ are jointly Gaussian random variables. $$E[X]=\nu_x=0$$ $$E[Y]=\nu_y=0$$ where $\nu$ is the mean. $$\sigma_x=1$$ $$\sigma_y=1$$ where $\sigma $ is variance and $$\rho_{xy}=0.75$$ where $\rho$ is correlation coefficient. Find $Z(t)$, $R_{zz}(t_1,t_2)$, $C_{zz}(t_1,t_2)$ and joint pdf of $Z(0)$ and $Z(t_1)$ where $R_{zz}(t_1,t_2)$ is autocorrelation function and $C_{zz}(t_1,t_2)$ is autocovariance function. $$E[Z(t)]=E[Xt+Y]$$ and cannot separate X and Y because they are not specified to be independent. $$R_{zz}(t_1,t_2)=E[(Xt_1+Y)(Xt_2+Y)]=E[X^2t_1t_2+XYt_1+XYt_2+Y^2]=E[X^2t_1t_2+XY(t_1+t_2)+Y^2]$$ Same problem as for $E[Z(t)]$ $$C_{zz}(t_1,t_2)=R_{xx}(t_1,t_2)-E[Z(t_1)]E[Z(t_2)]=E[X^2t_1t_2+XY(t_1+t_2)+Y^2]-E[Z(t_1)]E[Z(t_2)]$$

Answer 1

Due to linearity of $E$ (no need for independence): $$E[Z(t)]=E[Xt+Y]=tE[X]+E[Y]=0$$

then

$$R_{zz}(t_1,t_2)=E[(Xt_1+Y)(Xt_2+Y)]=E[X^2t_1t_2+XYt_1+XYt_2+Y^2]=E[X^2t_1t_2+XY(t_1+t_2)+Y^2]=t_1t_2E[X^2]+(t_1+t_2)E[XY]+E[Y^2]\\=t_1t_2\sigma^2_X+(t_1+t_2)\rho\sigma_X\sigma_Y+\sigma^2_Y$$

$Z(t)$ is also a normal random variable with mean zero and variance $$\sigma^2_{Z(t)}=t^2\sigma^2_X+2t\rho\sigma_X\sigma_Y+\sigma_Y^2$$ Also since $Z(t)$ is a linear combination of joint Gaussian random variables at different $t$ values, each two random variables from $Z(t)$ at different $t$ are jointly Gaussian as well. You can use the above $R_{zz}$ and mean to write the joint pdf 0f $Z(0)$ and $Z(t_1)$ using the pdf of bivariate Normal density:

$$\mathbf{\mu}=\begin{bmatrix} E[Z(0)] \\ E[Z(t_1)] \end{bmatrix}=\begin{bmatrix} 0 \\ 0 \end{bmatrix}$$

$$\mathbf{\Sigma}=\begin{bmatrix} E[Z(0)]E[Z(0)] & E[Z(0)]E[Z(t_1)]\\ E[Z(t_1)]E[Z(0)] & E[Z(t_1)]E[Z(t_1)] \end{bmatrix}=\begin{bmatrix} \sigma^2_{Z(0)} & R_{zz}(0,t_1)\\ R_{zz}(0,t_1) & \sigma^2_{Z(t_1)} \end{bmatrix}$$ Now with $\mathbf{Z}=\begin{bmatrix} Z(0)\\Z(t_1) \end{bmatrix}$ $$f_{Z_0,Z_{t_1}}(z_0,z_{t_1})=\frac{1}{\sqrt{(2\pi)^2|\mathbf{\Sigma}|}}\exp\left(-\frac{1}{2}(\mathbf{z}-\mathbf{\mu})^T\mathbf{\Sigma}^{-1}(\mathbf{z}-\mathbf{\mu})\right)$$