Markov semigroup for normal distributed kernels

Question

Let $\alpha,\sigma^2>0$. I want to show that the kernels defined by

$$K_t(x,\cdot):=\mathcal{N}\big(xe^{-\alpha t},\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})\big)\quad\text{for}\quad t>0$$

$$K_0(x,\cdot)=\varepsilon_x$$

form a Markov semigroup, i.e. for all $(x,B)\in\mathbb{R}\times\mathcal{B}(\mathbb{R})$ and for all $s,t\in \mathbb{R}_+$ it holds

$$K_{s+t}(x,B)=\int_\mathbb{R}K_t(y,B)K_s(x,dy)$$

Clearly we have

$$K_{0+t}(x,B)=\int_\mathbb{R}K_t(y,B) \varepsilon_{x}(dy)=K_t(x,B)$$

and also

$$K_{s+0}(x,B)=\int_\mathbb{R}\varepsilon_y(B)K_s(x,dy)=K_s(x,B)$$

For both $s,t>0$ my attempt is following:

\begin{align}\int_\mathbb{R}K_t(y,B)K_s(x,dy)&=\frac{1}{\frac{\pi\sigma^2}{\alpha}\sqrt{1-e^{-2at}-e^{-2as}+e^{-2a(t+s)}}}\\ &\quad\cdot\int_\mathbb{R}\bigg(\int_{B-ye^{-\alpha t}}\exp\bigg(\frac{z^2}{\frac{\sigma^2}{\alpha}(1-e^{-\alpha t})}\bigg)dz\bigg)\exp\bigg(\frac{(y-xe^{-\alpha s})^2}{\frac{\sigma^2}{\alpha}(1-e^{-\alpha s})}\bigg)dy \end{align}

but I do not know where I have to go from here... I am grateful for any advice or help. Thanks in advance!

Answer 1

Addendum. If we just want to see it's true, then of course it's true because it's the transition pdf of an OU process. I thought the purpose of the exercise was to "double check" it by direct integration.

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We want to show (by direct integration) $$ \int_{\mathbb R}K_t(y,B)K_s(x,dy)=K_{s+t}(x,B). $$ Note that \begin{align} \int_{\mathbb R}K_t(y,B)K_s(x,dy) & = \int_{\mathbb R}\int_B \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}} e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}} dz \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}} e^{- \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}} dy\\ & = \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})}} \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})}} \int_B \int_{\mathbb R} e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})} - \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})} } dy dz \end{align} where in the second equality I changed the order of integration. And from this point on, it's mechanical. First off, the exponent can be rewritten as $$ - \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})} - \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})} = - \frac{1-e^{-2\alpha(s+t)}}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})(1-e^{-2\alpha s})} \left( y-\frac{ze^{-\alpha}(1-e^{-2\alpha s}) + xe^{-\alpha s}(1-e^{-2\alpha t})}{1-e^{-2\alpha(s+t)}} \right)^2 - \frac{(z-xe^{-\alpha(s+t)})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}. $$ So \begin{align} \int_{\mathbb R}K_t(y,B)K_s(x,dy) & = \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}} \int_B \frac{\sqrt{1-e^{-2\alpha (s+t)}}}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})(1-e^{-2\alpha s})}} %% \int_{\mathbb R} e^{- \frac{(z-ye^{-\alpha t})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha t})} - \frac{(y-xe^{-\alpha s})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha s})} } dy dz\\ & = \frac{1}{\sqrt{2\pi\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}} \int_B e^{- \frac{(z-xe^{-\alpha(s+t)})^2}{2\frac{\sigma^2}{2\alpha}(1-e^{-2\alpha (s+t)})}} dz\\ & = K_{s+t}(x,B). \end{align}

Answer 2

The notation makes everything look much more grim than it actually is, so lets quickly define $$ m(t, x) = e^{-at}x, \qquad v(t) = \frac{\sigma^2}{2a}\left(1-e^{-2at}\right). $$ Then if you form the double integral, switch the order of integration and then inspect the inner integral $$ \int_{\mathbb{R}}\mathcal{N}(z \mid m(t, y), v(t))\mathcal{N}(y \mid m(s, x) , v(s) )\mathrm{d}y, $$ you see this is exactly the same as what we would have in the linear Gaussian model $$ y \sim \mathcal{N}(y \mid \mu_0, \sigma^2_0), \qquad z\mid y \sim \mathcal{N}(z \mid Ay + b, \sigma_1^2), $$ and in particular for the linear Gaussian model you have the marginal distribution $$ p(z) = \mathcal{N}(z \mid A\mu_0 + b, \sigma_1 + A^2\sigma_0). $$ Applying that in this case you have $$ \begin{align} p(z) &= \mathcal{N}(z \mid e^{-at}m(s, x), v(t) + e^{-2at}v(s)) \\ \end{align} $$ then you have $$ e^{-at}m(s, x) = e^{-at}e^{-as}x =e^{-a(t+s)}x $$ and $$ \begin{align} v(t) + e^{-2at}v(s) &=\frac{\sigma^2}{2a}\left( 1-e^{-2at}\right) + e^{-2at}\frac{\sigma^2}{2a}(1-e^{-2as}) \\ &=\frac{\sigma^2}{2a}\left(1 - e^{-2at} + e^{-2at}-e^{-2a(t+s)} \right) \\ &= \frac{\sigma^2}{2a}\left(1-e^{-2(a+s)}\right). \end{align} $$