For $a\in\mathbb R$ and $\sigma^2>0$ let $\phi_{a,\sigma^2}$ be the Gauss distribution on $(\mathbb R, \mathcal B)$ with the expectation $a$ and variance $\sigma^2$. Show that $K\left((a,\sigma^2),A\right):=\phi_{a,\sigma^2}(A)$ is a Markov kernel from $(\mathbb R\times\mathbb R,\mathcal B_{\mathbb R\times\mathbb R_+})$ to $(\mathbb R, \mathcal B)$

First condition is trivial since we have Gauss distribution the function $A\mapsto K\left((a,\sigma^2),A\right)$ is a probability measure.

For the second condition, one has to show that the function $\displaystyle F_A:(a,\sigma^2)\mapsto\int_A\frac{e^{\frac{-(x-a)^2}{2\sigma^2}}}{\sqrt{2\pi\sigma^2}}dx$ is measurable on $\mathbb R\times\mathbb R_+$, but this function is continuous in both arguments for bounded $A$, or not, what if $A$ is not bounded ?

  • $\begingroup$ Why do you think it matters whether $A$ is bounded or unbounded? $\endgroup$ – Zoran Loncarevic Sep 25 '15 at 11:26
  • $\begingroup$ @ZoranLoncarevic can you also use then Dominated Convergence Theorem ? $\endgroup$ – ketum Sep 25 '15 at 11:27

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