Auto-covariance of Multiply Process

Question

Given the process $I=XY$, where $X$ and $Y$ are:

• independent
• WSS
• with auto-covariances $B_X(\tau)=E\left\lbrace x(t)x(t+\tau)\right\rbrace-E^2\left\lbrace X\right\rbrace$ and $B_Y(\tau)$
• $E\left\lbrace X\right\rbrace >0$, $E\left\lbrace Y\right\rbrace >0$
• positive-axis distributed (non-negative)

The question is, what is $B_I(\tau)$ as a function of $B_X(\tau)$ and $B_Y(\tau)$ ?

I've found in literature, that for special normalized case of $\hat{I}=I/{E\left\lbrace I\right\rbrace}$, the resulting auto-covariance is

$B_I(\tau) = B_X(\tau)+B_Y(\tau)+B_X(\tau)B_Y(\tau)$.

Unfortunately, I could not even prove the relation above.

Answer 1

We have to compute $\mathbb E\left[I(0)I(t)\right]-\left(\mathbb E\left[I(0)\right]\right)^2$. Notice that by independence of $X$ and $Y$, $$\mathbb E\left[I(0)\right]=\mathbb E\left[X(0)\right]\mathbb E\left[Y(0)\right]$$ and by independence of $X(0)X(t)$ with $Y(0)Y(t)$, we have $$\mathbb E\left[I(0)I(t)\right]=\mathbb E\left[X(0)Y(0)X(t)Y(t)\right] =\mathbb E\left[X(0)X(t)\right]\mathbb E\left[Y(0)Y(t)\right]= \left(B_X(t)+\left(\mathbb E\left[X(0)\right]\right)^2\right)\left(B_Y(t)+\left(\mathbb E\left[Y(0)\right]\right)^2\right),$$ which gives $$B_I(t)=B_X(t)B_Y(t)+B_X(t)\left(\mathbb E\left[Y(0)\right]\right)^2+ B_Y(t)\left(\mathbb E\left[X(0)\right]\right)^2.$$