Calculate the correlation function of $n^{\prime}(t)=\int_{0}^{T} k(t, \tau) n(\tau) d \tau$

Question

Let us consider the following signal: $$n^{\prime}(t)=\int_{0}^{T} k(t, \tau) n(\tau) d \tau$$ prove that the correlation function is $$R_{n^\prime}(t, s)=\int_{0}^{T} \int_{0}^{T} k(t, \tau) k^{*}(s, u) R_{n}(\tau, u)d\tau du$$ It is the equation (6.33) of the following report (page 104). I can't figure out how to come up with this expression. I tried with $R_{n^\prime}(s)=\int_{\mathbb R} \left(n^{\prime}(t)\right)^* n^{\prime}(t+s) dt$ but I can't get an expression in t and s.

Answer 1

Your attempted solution does not use the correct formula for the correlation function of a stochastic process (time-series), which is $$ R_x(t,s)=E[x(t)x(s)^*] $$ Substituting $x(t)=n'(t)=\int_0^T k(t,\tau)n(\tau)d\tau$ into the above and using the linearity of the expectation operator gives the desired result: $$ R_n'(t,s)=E\left[\left(\int_0^T k(t,a)n(a)da\right)\cdot \left(\int_0^T k(s,b)^*n(b)^* db\right)\right]= $$ $$ \iint_{[0,T]^2} k(t,a)k(s,b)^*R_n(a,b) \, da\,db $$