If we have an AR(1) process, i.e: $X_{t+1} = \alpha X_t + e_{t+1}$ with $X_0=0$ then what is its Markov Chain transition density? We know that for a Markov chain, the following holds:

$P(X_{t+1}\leq x_{t+1} \mid X_0\leq x_0, X_1 \leq x_1, ... ) = P(X_{t+1}\leq x_{t+1} \mid X_t\leq x_t)$

For an AR(1) process this should then be

$P(\alpha X_{t} + e_{t+1}\leq x_{t+1} \mid X_t\leq x_t)$

From what I understand, the transition density $q(x_{t+1}\mid x_t)$ is the derivative of $P(\alpha X_{t} + e_{t+1}\leq x_{t+1} \mid X_t\leq x_t)$. But how should I take the derivative...?

  • 1
    $\begingroup$ To start with, your first equation does not hold for Markov chains in general: mathoverflow.net/questions/23478/… $\endgroup$
    – user940
    Aug 2 '15 at 23:18
  • $\begingroup$ So the above dosen't hold? That is given on my lecture notes thats why I take it for granted.... $\endgroup$
    – Elekko
    Aug 2 '15 at 23:28
  • $\begingroup$ That is correct. It doesn't hold, it is a common misconception. It may hold for some Markov chains, but generally it is false. $\endgroup$
    – user940
    Aug 2 '15 at 23:36
  • $\begingroup$ I edited the question, it has an initial value $X_0=0$ maybe it will then hold...? $\endgroup$
    – Elekko
    Aug 2 '15 at 23:44
  • 2
    $\begingroup$ That doesn't help. The counterexample from the MathOverflow link also has $X_0=0$. The issue is much more complicated than that. Did's answer here may help: math.stackexchange.com/questions/75791/… $\endgroup$
    – user940
    Aug 3 '15 at 0:14

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