0
$\begingroup$

Markov chains are typically described as memoryless as the next state depends only on the current state but not any of the past states.

But wouldn't true memorylessness mean that the next state does not depend on any state?

The way I view it, if the next state depends on the current state, then the system has a memory of size 1 - making it not memoryless.

In other words:

  • X_t+1 = f(X_t) -> Markov Chain
  • X_t+1 = f() -> What is this called?

Put differently: If the Markov process as such is described as memoryless, then what would you call a process that can jump randomly to any state regardless of where it currently is (which the Markov process can not do)? Memorylesser?

$\endgroup$
11
  • 7
    $\begingroup$ You're standing in a room. There are doors in front of you. You don't remember how you got to this room, but you can walk through one of the doors you see. That's memorylessness. $\endgroup$ Oct 12, 2023 at 16:26
  • 4
    $\begingroup$ The use of the word "memoryless" in this context means exactly what the technical definition says. You might have a different meaning of the word in mind based on its common usage, but that doesn't relate to the technical definition. Mathematics and statistics have specialized terminologies which don't really have more than a suggestive relationship to the common meaning of the words, e.g. "imaginary," "real," "sheaf," "ring," "unscented." $\endgroup$
    – Sycorax
    Oct 12, 2023 at 16:26
  • 1
    $\begingroup$ I think you are confusing a computer science definition of memory capacity with the common definition of memory, being about the past. $\endgroup$
    – seanv507
    Oct 12, 2023 at 16:51
  • 5
    $\begingroup$ I think you aren't using these words even in their standard English sense. Here are some hypothetical dialogs to help appreciate what they mean. Memoryless: The triage nurse in a hospital emergency room asks a patient, "do you know where you are?" If the answer is yes, the follow-up is "do you know how you got here?" When the answer is no, this patient has lost their memory. Totally unaware: "do you know where you are?" "No." Nobody ever said a Markov chain is "totally unaware." $\endgroup$
    – whuber
    Oct 12, 2023 at 19:08
  • 1
    $\begingroup$ 1. With respect to: " If the only way to the next room is via the current room, then in the next room I have the memory that i came via this particular room." - not so, with a memoryless process. 2. You are confusing having a process/method for selecting the next room with having a memory of the past rooms. Having a process that selects the next room from a collection of possibilities with unequal probabilities is not equivalent to remembering how you got to the current room. $\endgroup$
    – jbowman
    Oct 12, 2023 at 21:07

2 Answers 2

2
$\begingroup$

A process that can jump to any state regardless of the state it is currently in would also be a Markov process and would also be memoryless.

For example, take the coin-flip process where $X_t$ is Heads or Tails with 50% probability each, independently for each time $t$. That's a Markov process -- you can easily write down its transition probabilities, and $X_{t+1}$ is independent of $X_{t-1}$ given $X_t$.

If the term 'memoryless' bothers you, just don't use it. I don't especially like it either. But it has a straightforward and unambiguous definition and there are plenty of worse names in statistics and probability.

$\endgroup$
2
  • $\begingroup$ > you can easily write down its transition probabilities, and Xt+1 is independent of Xt−1 given Xt. - I totally get that but as you say, X_t+1 is still dependent on X_t. Is there such a process where X_t+1 is not even dependent on X_t, nor X_t-1, nor on any other X, but just ... completely random? Kind of like white noise? I guess the only such case would be a "fully connected" Markov chain with uniform transition probabilities at every state .. $\endgroup$ Oct 12, 2023 at 22:14
  • $\begingroup$ No, $X_{t+1})$ is not dependent on $X_t$ or on anything else. It's a sequence of iid binary variables. $\endgroup$ Oct 12, 2023 at 22:39
0
$\begingroup$

I think what you are looking for is just a random variable. f() with no inputs is either deterministic, in which case it's constant, or random, in which case it's a random variable independent of the value at the previous time. In this latter case, it has lost any time-series structure, any sense of dependence on previous states. I would use the term "time-independent random variable".

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.