Question: The Markov property does not imply that the past and the future are independent given *any* information concerning the present. Find an example of a homogeneous Markov chain $\left\{X\_n\right\}\_{n\ge0}$ with state space $E=\left\{1,2,3,4,5,6\right\}$ such that 

$P(X\_2=6\mid X\_1\in\{3,4\},X\_0=2)\neq P(X\_2=6\mid X\_1\in\{3,4\})$.

I don't understand the phrasing of this question. How should I approach this?

Could this work? [![enter image description here](https://i.stack.imgur.com/SLR2e.jpg)](https://i.stack.imgur.com/SLR2e.jpg)

$P(X\_2=6\mid X\_1\in\{3,4\}) = 1$

But $P(X\_2=6\mid X\_1\in\{3,4\},X\_0=2) =0$

Answer: 
**Hint $1$**: The fundamental theorem of calculus states that
$$\int\_a^b g(x)\,\mathrm dx = G(b) - G(a)$$
for a primitive $G$ of $g$. Use it to find $f'(x)$. What can you conclude now?

**Hint $2$**: To find the asymptotes, visualize the graph of the integrand function. This should give you an idea for $x \to -\infty$. For the other side, it boils down to the improper integral $\lim\limits\_{x \to +\infty} f(x)$. Does it converge?
The derivative is $f'(x)=e^{-x^3}(x+1)^2(x-1)$ so it only vanishes at $1$ and $-1$.

However $f$ is decreasing on $(-\infty,1]$ and increasing on $[1,\infty)$, because the derivative is positive in a neighborhood of $-1$ (except only at $-1$).

For the question about asymptotes, note that
$$
\lim\_{x\to\infty}f'(x)=0
$$
and
$$
\lim\_{x\to-\infty}f'(x)=-\infty
$$