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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...?

Elekko
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    To start with, your first equation does not hold for Markov chains in general: http://mathoverflow.net/questions/23478/examples-of-common-false-beliefs-in-mathematics/23933#23933 –  Aug 02 '15 at 23:18
  • So the above dosen't hold? That is given on my lecture notes thats why I take it for granted.... – Elekko Aug 02 '15 at 23:28
  • That is correct. It doesn't hold, it is a common misconception. It may hold for some Markov chains, but generally it is false. –  Aug 02 '15 at 23:36
  • I edited the question, it has an initial value $X_0=0$ maybe it will then hold...? – Elekko Aug 02 '15 at 23:44
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    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: http://math.stackexchange.com/questions/75791/from-a-deterministic-discrete-process-to-a-markov-chain-conditions –  Aug 03 '15 at 00:14
  • Are you sure the formula $P(X_{t+1}\leq x_{t+1} \mid X_0\leq x_0, X_1 \leq x_1, ... ) = P(X_{t+1}\leq x_{t+1} \mid X_t\leq x_t)$ is indeed in your lecture notes? 2. If $(e_t)$ is i.i.d. and independent of $X_0$ and if the distribution of $e_t$ has density $f$ then the transition density $q$ of the Markov process $(X_t)$ is such that $$q(y\mid x)=f(y-\alpha x).$$
    • – Did Aug 17 '15 at 11:39