 Solution

For $a<0$, set $v_{\infty }=-{\sigma }^2/(2a)>0$. The candidate invariant probability law of a Markov process isTo prove invariance rather than merely guess a limiting probability distribution, initialize $X_t\sim \pi$ independently of future Brownian increments. Integrating the conditional characteristic function from part (i) against this centered normal distribution givesbecause $v_{\Delta }=v_{\infty }(1-e^{2a\Delta })$. The uniqueness theorem for characteristic functions implies $X_{t+\Delta }\sim \pi$ for every $\Delta \ge 0$. This proves $\pi P_{\Delta }=\pi$, as required. The time-homogeneous Markov property then also makes the process with this Ornstein-Uhlenbeck stationary initialization a strictly stationary process.