Determining Probability from Specific Markov Transition Function I am trying to solve the following:
Consider a Markov chain whose state space is $\mathbb{R}$. Let $P(x,A), x ∈ \mathbb{R},A ∈ \mathcal{B}(\mathbb{R})$, be the following Markov transition function, $P(x,A) = \lambda([x − 1/2,x + 1/2] \cap A)$,where $\lambda$ is the Lebesgue measure. 
Assuming that the initial distribution is
concentrated at the origin, find $P(|\omega_{2}| ≤ 1/4).$
I have no idea how to approach this problem.
 A: The law of total probability and the Markov property imply that
\begin{equation}
P(\omega_2 \in A \mid \omega_0 = 0)
 =
\int P(\omega_2 \in A \mid \omega_1 = x) f_1(x) \; dx,
\end{equation}
where $f_1$ is the conditional probability density function of $\omega_1$ given $\omega_0 = 0$ (assume for now that such a density exists and try to find it later). To explain my earlier comment, define the function $F(x) = P(\omega_2 \in A \mid \omega_1 = x)$. Then by the definition of the expectation of a function of a random variable with a density, the right-hand side of the equation above is simply $E[F(\omega_1)]$.
Now to evaluate the integral (which we've seen is an expectation), we need to find the density $f_1$ and the function $F$ defined above. But the function $F$ is just the transition function $P(x, A)$, which you've been given.
To find the conditional density, follow the "usual" procedure: First, find the conditional cumulative distribution function $F_1(x) = P(\omega_1 \le x \mid \omega_0 = 0)$. Then the conditional density is the derivative: $f_1(x) = F'_1(x)$.
