Timeline for Does an expectation for a Markov chain simplify like this?
Current License: CC BY-SA 4.0
10 events
| when toggle format | what | by | license | comment | |
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| Feb 2, 2023 at 17:55 | comment | added | jbowman | Ah, I see... that's actually kind of clever, if roundabout! | |
| Feb 2, 2023 at 16:56 | comment | added | mhdadk | @jbowman the reason I mentioned that it is a Markov chain is to justify the decomposition $$p(x_0,x_1,\dots,x_N) = p(x_N \mid x_{N-1})p(x_{N-1} \mid x_{N-2})\cdots p(x_1 \mid x_0) p(x_0)$$ and then to use this decomposition to prove my statement. I now see from @ThomasLumley’s answer that there a better way to prove this statement without mentioning that the sequence $\{x_k\}$ is a Markov chain. | |
| Feb 2, 2023 at 16:05 | comment | added | jbowman | It's deterministic. One could say that $x_{i+t} = x_i + 1$ is a Markov chain, but why? There's no randomness in the transitions; describing multiplication and addition as a Markov chain adds nothing useful to the discussion. | |
| Feb 2, 2023 at 14:56 | vote | accept | mhdadk | ||
| Feb 2, 2023 at 14:56 | comment | added | mhdadk | @jbowman What I mean by "the sequence $x_0 \to x_1 \to \cdots \to x_N$ forms a Markov chain" is that $x_{k}$ is conditionally independent of $x_{k-2}$ given $x_{k-1}$ for $k = 2,\dots,N-1$. Is there something wrong with saying so? | |
| Feb 2, 2023 at 5:10 | history | edited | User1865345 |
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| Feb 2, 2023 at 1:04 | comment | added | jbowman | As written, $x_k = A^kx_0$, so $\mathbb{E}x_k = A^k\mathbb{E}x_0$ | |
| Feb 2, 2023 at 0:59 | answer | added | Thomas Lumley | timeline score: 6 | |
| Feb 2, 2023 at 0:57 | comment | added | jbowman | Why do you think it forms a Markov chain instead of being deterministic? Imagine $A$ is a square matrix of all ones; then at each step, the first element of $x_{i+1}$ is just the sum of the elements of $x_i$, etc. It's a matrix multiplication, not a transition matrix. | |
| Feb 2, 2023 at 0:46 | history | asked | mhdadk | CC BY-SA 4.0 |