Estimating Transition Probabilities with Repeated States
Summary
The document examines a three-state price process and asks whether repeated observations of the same state should be removed before estimating transition probabilities. Counting every time step estimates the chance of moving to another state at the next observation, including the possibility of staying put. Collapsing consecutive repetitions instead focuses on what state follows when the current run ends, and gives equal weight to runs of different lengths.
The response distinguishes these quantities using conditional probability. The probability of moving to a particular state given that a move away from the current state occurs is the corresponding transition probability divided by the total probability of leaving. Thus, the collapsed estimate is conditional on departure; it is not the unconditional one-step transition probability. Which quantity is useful depends on the trading question, and the brief answer does not discuss estimation uncertainty or validate a trading strategy.
Key ideas
- Counting every time step estimates unconditional one-step transition probabilities, including the chance of staying in place.
- Removing repeated states estimates transitions conditional on leaving the current state.
- The conditional departure probability is obtained by normalizing each exit probability by the total exit probability.
- Long runs receive more weight in the unreduced sequence than in the sequence of runs.
- The document does not assess statistical uncertainty or trading performance.
Tags
Full text
# To lump or not to lump
# To lump or not to lump
Suppose I have a very simple asset whose price takes only three possible values: $X_t\in \{-1,0,1\}$. I also got some discrete time series $X = (X_t)_{t\geq 0}$ and I would like to come up with a trading rule based on these observations. Let's focus on the following naive approach: given the current level of the asset, I would like to estimate of what the next change would be. Thus, I am fitting this time series into Markov Chain where I disregard transition that do not change the state. For example, based on the following sample from the time series: $$ \dots0,0,0,0,0,0,0,1,1,1,0,0,0,-1,-1\dots \tag{1} $$ and suppose that there are no appearances of $0$ anymore. I can conclude that out of $10$ appearances of $0$, $7$ are followed by $1$ and $3$ are followed by $-1$, hence a naive algorithm would say that transition probabilities are $p(1|0) = 0.7$ and $p(-1|0) = 0.3$.
Now, I wanted to make it faster, so I preprocessed the data to get rid of repetitions since I thought that it shall not affect the end result. For example, the sample $(1)$ transforms into $$ \dots0,1,0,-1\dots \tag{2} $$ but now $p(1|0) = 0.5$ and $p(-1|0) = 0.5$ which is quite different from the previous estimation. Of course, that's just a simple example, but it gives a general impression: the lumping procedure $(1)\to(2)$ changes the estimates of transition probabilities. It surprised me first, but now it seems very natural: in $(2)$ when sampling I give equal weight to each interval of consecutive $0$'s whereas in $(1)$ more weight goes to a longer interval of consecutive $0$'s.
The question is: given my purpose, what would be the correct method to estimate probabilities? Note that I am trying to predict the price move without taking into account for how long have I stayed at the current price of the asset, only the price level itself. From that perspective, I guess the second approach is more appropriate: if I observe new prices and I see the price is $0$, I'd say I'd rather rely on that the next price change happens with probabilities $(0.5,0.5)$ rather than $(0.7, 0.3)$. At the same time, here I do not feel confident since my background in statistics is rather weak, so any feedback on this topic is highly appreciated. Namely: what is formally more correct, to use probabilities from $(1)$ or from $(2)$, or if both are correct depending on how they are used, what is a proper way of using them. Practical comment on that approach is also welcome.
## Answer by Xodarap (score 1)
https://quant.stackexchange.com/a/15810
It sounds to me like you have a Markov model that is not "lumped", it's just that certain transitions don't provide you with any payout. I would model the true transition probabilities.
Now, let's ask what the probability of getting a one is, assuming that we won't stay at zero, i.e. $P(X_1=1 | X_0=0, X_1 \not = 0)$. We recall that $$P(A|B)=P(A,B) / P(B)$$ and therefore get
$$ \begin{align*} P(X_1=1 | X_0=0, X_1 \not = 0) &=\frac{P(X_1=1, X_1 \not = 0 | X_0=0)}{P(X_1 \not = 0| X_0=0)}\\ & =\frac{P(X_1=1 | X_0=0)}{P(X_1 = 1| X_0=0) + P(X_1 = -1| X_0=0)} \\ & =\frac{P(1|0)}{P(1|0) + P(-1|0)} \end{align*} $$
Which is just your second definition where you found it to be $(.5,.5)$ as indeed half the times it transitioned away from zero it transitioned to one.
Importantly, this is not $P(1|0)$! I think this is where you were getting confused before.Shown in full with attribution under the source's licence. Licence: CC BY-SA 4.0 (Stack Exchange)
This summary was written by Stratmill's research agent from the original; it is not a copy of the source.