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Use the Itô product rule with the deterministic factor $e^{-\lambda t}$. Its quadratic covariation with $X$ is zero, givingMultiplication by the integrating factor $e^{\lambda t}$ solves the Ornstein-Uhlenbeck stochastic differential equation:The initial value $B_1$ is standard normal and independent of the future driving Brownian motion. Thus $Y$ is a centered Gaussian process with variance one. For $s\le t$, its independent future noise gives $\mathbb{E}(Y_s{Y}_t)=e^{-\lambda (t-s)}$. Its Gaussian finite-dimensional distributions consequently depend only on time differences, so it is a stationary process. This verifies the exponential Brownian time change to a stationary Ornstein-Uhlenbeck process directly.

For an adapted stationary process with the displayed bounds, the Itô formula and stationarity give ${\int }_0^t(\mathbb{E}[Z_s{g}_s]+\lambda )ds=0$. The Cauchy-Schwarz inequality makes each integrand nonnegative, so equality holds almost everywhere. Then $\mathbb{E}(g_s+\lambda Z_s{)}^2\le 0$, proving $g=-\lambda Z$ for almost every time and outcome. Thus the process solves the Ornstein-Uhlenbeck stochastic differential equation and its initial value, rather than stationarity alone, determines its coupling with W.

In the long-time growing horizontal-gradient branch, independent vertical shear noises combine as $\cos \varphi d{W}_1+\sin \varphi d{W}_2=dW$. The signed inclination $\beta =k_3{e}^{-ct}/k_0$ satisfies the displayed Ornstein-Uhlenbeck stochastic differential equation. For $c>0$ its stationary density is $\sqrt{c/(\pi g^2)}{e}^{-c{B}^2/g^2}$, with variance $g^2/(2c)$. The Fokker-Planck probability current vanishes at infinity and gives this Gaussian by one integration. At $c=0$ and nonzero noise there is no normalizable stationary density.