Xrfgtjtk

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

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

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

form a Markov semigroup, i.e. for all $(x,B)inmathbb{R}timesmathcal{B}(mathbb{R})$ and for all $s,tin mathbb{R}_+$ 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{pisigma^2}{alpha}sqrt{1-e^{-2at}-e^{-2as}+e^{-2a(t+s)}}}\
&quadcdotint_mathbb{R}bigg(int_{B-ye^{-alpha t}}expbigg(frac{z^2}{frac{sigma^2}{alpha}(1-e^{-alpha t})}bigg)dzbigg)expbigg(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!

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.

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{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha t})}}
e^{- frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}}
dz
frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha s})}}
e^{- frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}}
dy\
& =
frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha t})}}
frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha s})}}
int_B
int_{mathbb R}
e^{- frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}
- frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha 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
$$
- frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}
- frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}
=
- frac{1-e^{-2alpha(s+t)}}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})(1-e^{-2alpha s})}
left(
y-frac{ze^{-alpha}(1-e^{-2alpha s}) + xe^{-alpha s}(1-e^{-2alpha t})}{1-e^{-2alpha(s+t)}}
right)^2
- frac{(z-xe^{-alpha(s+t)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)})}.
$$
So
begin{align}
int_{mathbb R}K_t(y,B)K_s(x,dy)
& =
frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)})}}
int_B
frac{sqrt{1-e^{-2alpha (s+t)}}}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha t})(1-e^{-2alpha s})}}
%%
int_{mathbb R}
e^{- frac{(z-ye^{-alpha t})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha t})}
- frac{(y-xe^{-alpha s})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha s})}
}
dy dz\
& =
frac{1}{sqrt{2pifrac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)})}}
int_B
e^{- frac{(z-xe^{-alpha(s+t)})^2}{2frac{sigma^2}{2alpha}(1-e^{-2alpha (s+t)})}}
dz\
& = K_{s+t}(x,B).
end{align}

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
$$
y sim mathcal{N}(y mid mu_0, sigma^2_0), qquad zmid y sim mathcal{N}(z mid Ay + b, sigma_1^2),
$$
and in particular for the linear Gaussian model you have the marginal distribution
$$
p(z) = mathcal{N}(z mid Amu_0 + b, sigma_1 + A^2sigma_0).
$$
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
$$
begin{align}
v(t) + e^{-2at}v(s) &=frac{sigma^2}{2a}left( 1-e^{-2at}right) + e^{-2at}frac{sigma^2}{2a}(1-e^{-2as}) \
&=frac{sigma^2}{2a}left(1 - e^{-2at} + e^{-2at}-e^{-2a(t+s)} right) \
&= frac{sigma^2}{2a}left(1-e^{-2(a+s)}right).
end{align}
$$