I was wondering how to calculate the autocorrelation of a deterministic signal $x(t)$ multiplied by a stochastic process $M(t)$, whose autocorellation $R_M(\tau)$ is known a priori. In my case, $x(t)$ is a truncated monolateral exponentially decaying function.
I suppose that the result of such multiplication $y(t) = x(t) \cdot M(t)$ is again a stochastic process, but when approaching the calculation of the autocorrelation of $y(t)$ I obtain something that is not even, therefore I suppose I am making some mistakes. I know that the definition of autocorrelation for deterministic signals is different from the one of stochastic processes, but I do not know how to connect the two of them.
Answer
As correctly pointed out in the comments, in general the process $Y(t)=x(t)M(t)$ is not wide-sense stationary (WSS), i.e. its autocorrelation function depends not only on the time difference parameter $\tau$, but also on the absolute time $t$:
$$R_Y(\tau,t)=E[Y(t+\tau)Y^*(t)]=E[x(t+\tau)M(t+\tau)x^*(t)M^*(t)]=\\ =x(t+\tau)x^*(t)E[M(t+\tau)M^*(t)]=x(t+\tau)x^*(t)R_M(\tau)\tag{1}$$
In your case you just need to evaluate (1) with the given function $x(t)$. However, as expected you will end up with a function of two variables because $Y(t)$ is not WSS.
There are a few special cases in which the resulting process is indeed WSS or can be easily made WSS. The first case is modulation by a complex exponential:
$$x(t)=e^{j\omega_ct}$$
in which case
$$x(t+\tau)x^*(t)=e^{j\omega_c(t+\tau)}e^{-j\omega_ct}=e^{j\omega_c\tau}$$
only depends on $\tau$ and not on $t$. Another case of interest is the case where $x(t)$ is $T$-periodic. In this case the process $x(t)M(t)$ can be made WSS by introducing a random phase epoch $\Theta$ with uniform distribution in the interval $[0,T]$ which is independent of $M(t)$:
$$Y(t)=x(t+\Theta)M(t)$$
The autocorrelation function of $Y(t)$ is then given by
$$R_Y(\tau)=R_M(\tau)\frac{1}{T}\int_0^Tx(\alpha+\tau)x^*(\alpha)d\alpha$$
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