Tag Archives: markov

Discrete-time Markov Chains

Much of how we interact with life could be described as transitions between states. These states could be weather conditions (whether we are in a state of “sunny” or “rainy”), the places we may visit (maybe “school”, “the mall”, “the park” and “home”), our moods (“happy”, “angry”, “sad”). There are a number of other ways to model states and even the possibility of infinitely many states.

Markov Chains are based on the principle that the future is only dependent on the immediate past. so for example, if I wished to predict tomorrow’s weather using a Markov Chain, I would need to only look at the weather for today, and can ignore all previous data. I would then compare the state of weather for today with historically how weather has changed in between states to determine the most likely next state (i.e what the weather will be like tomorrow). This greatly simplifies the construction of models.

To use Markov Chains to predict the future, we first need to compute a transition matrix which shows the probability (or frequency) that we will travel from one state to another based on how often we have done so historically. This transition matrix can be calculated by looking at each element of the history as an instance of a discrete state, counting the number of times each transition occurs and dividing each result by the number of times the origin state occurs. I’ll next give an example and then I’ll focus on explaining the Finite Discrete State Markov Chain tool I built using javascript.

Next, I want to consider an example of using Markov Chains to predict the weather for tomorrow. Suppose that we have observed the weather for the last two weeks. We could then use that data to build a model to predict tomorrow’s weather. To do this, lets first consider some states of weather. Suppose that a day can be classified in one of four different ways: {Sunny, Cloudy, Windy, Rainy}. Further, suppose that over the last two weeks we have seen the following pattern.

Day 1 Sunny
Day 2 Sunny
Day 3 Cloudy
Day 4 Rain
Day 5 Sunny
Day 6 Windy
Day 7 Rain
Day 8 Windy
Day 9 Rain
Day 10 Cloudy
Day 11 Windy
Day 12 Windy
Day 13 Windy
Day 14 Cloudy

We can look at this data and calculate the probability that we will transition from each state to each other state, which we see below:

Rain Cloudy Windy Sunny
Rain 0 1/3 1/3 1/3
Cloudy 1/2 0 1/2 0
Windy 2/5 1/5 2/5 0
Sunny 0 1/3 1/3 1/3

Given that the weather for today is cloudy, we can look at the transition matrix and see that historically the days that followed a cloudy day have been Rainy and Windy days each with probability of 1/5. We can see this more mathematically by multiplying the current state vector (cloudy) [0, 1, 0, 0] by the above matrix, where we obtain the result [1/2, 0, 1/2, 0].

In similar fashion, we could use this transition matrix (lets call it T) to predict the weather a number of days in the future by looking at Tn. For example, if we wanted to predict the weather two days in the future, we could begin with the state vector [1/2, 0, 1/2, 0] and multiply it by the matrix T to obtain [1/5, 4/15, 11/30, 1/6].

We can also obtain this by looknig at the original state vector [0, 1, 0, 0] and multiplying it by T2.

T2 =

1

3/10 8/45 37/90 1/9
1/5 4/15 11/30 1/6
13/50 16/75 59/150 2/15
3/10 8/45 37/90 1/9

When we multiply the original state vector by T2 we arrive at this same answer [1/5, 4/15, 11/30, 1/6]. This matrix T2 has an important property in that every state can reach every other state.

In general, if we have a transition matrix where for every cell in row i and column j, there is some power of the transition matrix such that the cell (i, j) in that matrx is nonzero, then we say that every state is reachable from every other state and we call the Markov Chain regular.

Regular Markov Chains are important because they converge to what’s called a steady state. These are state vectrs x = [x0, …, xn] such that xTn = x for very large values of n. The steady state tells us how the Markov Chain will perform over long periods of time. We can use algebra and systems of linear equations to solve for this steady state vector.

For the Javascript program I’ve written, I have generated a set of painting samples for a fictional artist. The states are the different colors and the transitions are the colors that the artist will use after other colors. as well as the starting and ending colors. Given this input, we can form a Markov Chain to understand the artist’s behavior. This Markov Chain can then be used to solve for the steady state vector or to generate random paintings according to the artist’s profile. Be sure to check it out and let me know what you think.

Hidden Markov Models: The Baum-Welch Algorithm

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the loaded die are the latent (hidden) variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

  • The N states of the model, defined by S = {S1, …, SN}
  • The M possible output symbols, defined by = {1, 2, …, M}
  • The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.
  • The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.
  • The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

    The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

    • How often does the casino change dice?
      • 0) Dealer Repeatedly Uses Same Dice
      • 1) Dealer Uniformly Changes Die
      • 2) Dealer Rarely Uses Same Dice
    • Which sides on the loaded dice are more likely?
      • 0) Larger Numbers Are More Likely
      • 1) Numbers Are Randomly Likely
      • 2) Smaller Numbers Are More Likely
    How often does the casino change dice?
    Which sides on
    the loaded dice
    are more likely?
    (0, 0) (0, 1) (0, 2)
    (1, 0) (1, 1) (1, 2)
    (2, 0) (2, 1) (2, 2)

    One of the interesting problems associated with Hidden Markov Models is called the Learning Problem, which asks the question “How can I improve a HMM so that it would be more likely to have generated the sequence O = o1, o2, …, oT?

    The Baum-Welch algorithm answers this question using an Expectation-Maximization approach. It creates two auxiliary variables t(i) and t(i, j). The variable t(i) represents the probability of being in state i at time t, given the entire observation sequence. Likewise t(i, j) represents the joint probability of being in state i at time t and of being in state j at time t+1, given the entire observation sequence. They can be calculated by

    t(i) =
    (t(i) * t(i) )
    j = 1 to N(t(j) * t(j))

    and

    t(i, j) =
    (t(i) * ai, j * t+1(j) * bj(ot+1) )
    i’ = 1 to Nj’ = 1 to N(t(i’) * ai’, j’ * t+1(j’) * bj’(ot+1) )

    As you can see, these are a direct result of calculations of from the Forward algorithm and from the Backwards algorithm. Once we have calculated these variables, we can update the parameters of the model as follows:

    i = 1(i)

    i,j = t = 1 to T-1(t(i))
    t = t to T-1 (t(i, j))

    // [b bar]_{j, k} = Sigma_{t = 1 to T, o_t = o_k} gamma_{t, j} / Sigma_{t = 1 to T} gamma_{t, j}, 1 <= j <= N, 1 <= k <= M

    j(ok) =
    t = 1 to T-1, ot = ok t(j)
    t = 1 to T-1 t(j)

    We can iterate this procedure a finite number of times or until it converges. This will generate a new model, = {N, , , , }.

    There is more on this example at LEARNINGlover.com: Hidden Marokv Models: The Baum-Welch Algorithm.

    Some further reading on Hidden Markov Models:

Hidden Markov Models: The Viterbi Algorithm

I just finished working on LEARNINGlover.com: Hidden Marokv Models: The Viterbi Algorithm. Here is an introduction to the script.

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

  • The N states of the model, defined by S = {S1, …, SN}
  • The M possible output symbols, defined by = {1, 2, …, M}
  • The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.
  • The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.
  • The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

    The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

    • How often does the casino change dice?
      • 0) Dealer Repeatedly Uses Same Dice
      • 1) Dealer Uniformly Changes Die
      • 2) Dealer Rarely Uses Same Dice
    • Which sides on the loaded dice are more likely?
      • 0) Larger Numbers Are More Likely
      • 1) All Numbers Are Equally Likely
      • 2) Smaller Numbers Are More Likely
    How often does the casino change dice?
    Which sides on
    the loaded dice
    are more likely?
    (0, 0) (0, 1) (0, 2)
    (1, 0) (1, 1) (1, 2)
    (2, 0) (2, 1) (2, 2)

    One of the interesting problems associated with Hidden Markov Models is called the Decoding Problem, which asks the question “What is the most likely sequence of states that the HMM would go through to generate the sequence O = o1, o2, …, oT?

    The Viterbi algorithm finds answers this question using Dynamic Programming. It creates an auxiliary variable t(i) which has the highest probability that the partial observation sequence o1, …, ot can have, given that the current state is i. This variable can be calculated by the following formula:

    t(i) = maxq1, …, qt-1 p{q1, …, qt-1, qt = i, o1, …, ot | }.

    1(j) = jbj(o1), for 1 j N.

    Once we have calculated t(j) we also keep a pointer to the max state. We can then find the optimal path by looking at arg max 1 j N T(j) and then backtrack the sequence of states using the pointer.

    There is more on this example at LEARNINGlover.com: Hidden Marokv Models: The Viterbi Algorithm.

    Some further reading on Hidden Markov Models:

Hidden Markov Models: The Backwards Algorithm

I just finished working on LEARNINGlover.com: Hidden Marokv Models: The Backwards Algorithm. Here is an introduction to the script.

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the number of loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

  • The N states of the model, defined by S = {S1, …, SN}
  • The M possible output symbols, defined by = {1, 2, …,M}
  • The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.
  • The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.
  • The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

  • How often does the casino change dice?
    • 0) Dealer Repeatedly Uses Same Dice
    • 1) Dealer Uniformly Changes Die
    • 2) Dealer Rarely Uses Same Dice
  • Which sides on the loaded dice are more likely?
    • 0) Larger Numbers Are More Likely
    • 1) All Numbers Are Randomly Likely
    • 2) Smaller Numbers Are More Likely
How often does the casino change dice?
Which sides on
the loaded dice
are more likely?
(0, 0) (0, 1) (0, 2)
(1, 0) (1, 1) (1, 2)
(2, 0) (2, 1) (2, 2)

One of the interesting problems associated with Hidden Markov Models is called the Evaluation Problem, which asks the question “What is the probability that the given sequence of observations O = o1, o2, …, oT are generated by the HMM . In general, this calculation, p{O | }, can be calculated by simple probability. However because of the complexity of that calculation, there are more efficient methods.

The backwards algorithm is one such method (as is the forward algorithm). It creates an auxiliary variable t(i) which is the probability that the model has generated the partially observed sequence ot+1, …, oT, where 1 t T. This variable can be calculated by the following formula:

t(i) = j = 1 to N(t+1(j) * aij * bj(ot+1))

We also need that T(i) = 1, for 1 i N.

Once we have calculated the t(j) variables, we can solve the evaluation problem by p{O | } i = 1 to N1(i)

There is more on this example at LEARNINGlover.com: Hidden Marokv Models: The Backwards Algorithm.

Some further reading on Hidden Markov Models:

Hidden Markov Models: The Forward Algorithm

I just finished working on LEARNINGlover.com: Hidden Marokv Models: The Forward Algorithm. Here is an introduction to the script.

Suppose you are at a table at a casino and notice that things don’t look quite right. Either the casino is extremely lucky, or things should have averaged out more than they have. You view this as a pattern recognition problem and would like to understand the number of ‘loaded’ dice that the casino is using and how these dice are loaded. To accomplish this you set up a number of Hidden Markov Models, where the number of loaded die are the latent variables, and would like to determine which of these, if any is more likely to be using.

First lets go over a few things.

We will call each roll of the dice an observation. The observations will be stored in variables o1, o2, …, oT, where T is the number of total observations.

To generate a hidden Markov Model (HMM) we need to determine 5 parameters:

  • The N states of the model, defined by S = {S1, …, SN}
  • The M possible output symbols, defined by = {1, 2, …, M}
  • The State transition probability distribution A = {aij}, where aij is the probability that the state at time t+1 is Sj, given that the state at time t is Si.
  • The Observation symbol probability distribution B = {bj(k)} where bj(k) is the probability that the symbol k is emitted in state Sj.
  • The initial state distribution = {i}, where i is the probability that the model is in state Si at time t = 0.

    The HMMs we’ve generated are based on two questions. For each question, you have provided 3 different answers which leads to 9 possible HMMs. Each of these models has its corresponding state transition and emission distributions.

    • How often does the casino change dice?
      • 0) Dealer Repeatedly Uses Same Dice
      • 1) Dealer Uniformly Changes Die
      • 2) Dealer Rarely Uses Same Dice
    • Which sides on the loaded dice are more likely?
      • 0) Larger Numbers Are More Likely
      • 1) All Numbers Are Equally Likely
      • 2) Smaller Numbers Are More Likely
    How often does the casino change dice?
    Which sides on
    the loaded dice
    are more likely?
    (0, 0) (0, 1) (0, 2)
    (1, 0) (1, 1) (1, 2)
    (2, 0) (2, 1) (2, 2)

    One of the interesting problems associated with Hidden Markov Models is called the Evaluation Problem, which asks the question “What is the probability that the given sequence of observations O = o1, o2, …, oT are generated by the HMM . In general, this calculation, p{O | }, can be calculated by simple probability. However because of the complexity of that calculation, there are more efficient methods.

    The forward algorithm is one such method. It creates an auxiliary variable t(i) which is the probability that the model has generated the partially observed sequence o1, …, ot, where 1 t T. This variable can be calculated by the following formula:

    t+1(j) = bj(ot+1)i = 1 to N(t(i)aij), where 1 j N, 1 t T-1.

    1(j) = jbj(o1), for 1 j N.

    Once we have calculated the t(j) variables, we can solve the evaluation problem by p{O | } = i = 1 to N T(i)

    There is more on this example at LEARNINGlover.com: Hidden Marokv Models: The Forward Algorithm.

    Some further reading on Hidden Markov Models: