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我已经实现了 Baum-Welch 算法，并且正在使用一些玩具数据，这些数据是用已知分布生成的。数据呈正态分布，根据隐藏状态具有不同的均值和标准差。有2个状态。除了隐藏状态的初始分布之外，该算法似乎对大多数参数都收敛，根据随机数据，隐藏状态总是收敛到 (0; 1) 或 (1; 0)。

这个算法正常吗？如果是这样，我将不胜感激一些参考，如果不是一些提示如何找到错误。

代码 (F#)。首先是一个辅助模块：

module MyMath

let sqr (x:float) = x*x

let inline (./) (array:float[]) (d:float) =
  Array.map (fun x -> x/d) array

let inline (.*) (array:float[]) (d:float) =
  Array.map (fun x -> x*d) array

let map f s =
  s |> Seq.map f |> Seq.toArray

let normalize v = 
  let sum = Seq.sum v
  map (fun x -> x/sum) v

let row i array = seq { for j in 0 .. (Array2D.length2 array)-1 do yield array.[i,j]}
let column j array = seq { for i in 0 .. (Array2D.length1 array)-1 do yield array.[i,j]}

let sum (v:float[]) = v |> Array.sum
let sumTo N (f:int->float) = Seq.init N f |> Seq.sum
let sum_column j (array:float[,]) = column j array |> Seq.sum
let sum_row i (array:float[,]) = row i array |> Seq.sum

let mean data = (sum data)/(float (Array.length data))
let var data = 
    let m=mean data
    let N=Array.length data
    let sum=Seq.sumBy (fun x -> sqr(x)) data
    sum/(float N)

let induction start T nextRow =
  let result =  Array.zeroCreate T
  result.[0] <- start
  for t=1 to T-1 do
    result.[t] <- nextRow t result.[t-1]
  result

let backInduction last T previousRow =
  let result =  Array.zeroCreate T
  result.[T-1] <- last
  for t=T-2 downto 0 do
    result.[t] <- previousRow t result.[t+1]
  result

let inductionNormalized start T nextRow =
  let result =  Array.zeroCreate T
  let norm = Array.zeroCreate T
  norm.[0] <- sum start
  result.[0] <- start./norm.[0]
  for t=1 to T-1 do
    result.[t] <- nextRow t result.[t-1]
    norm.[t] <- sum result.[t]
    result.[t] <- result.[t]./norm.[t]
  (result, norm)

主要模块：

module BaumWelch

open System
open MyMath

let mu (theta : float[,]) q = theta.[q,0]
let sigma (theta : float[,]) q = theta.[q,1]

let likelihood getDrift getVol dt parameters state observation  =
    let mu = getDrift parameters state
    let sigma =  Math.Abs (getVol parameters state:float)
    let sqrt_dt = Math.Sqrt dt
    let residueSquared = 
        let r = Likelihood.normalizedResidue mu sigma dt sqrt_dt observation in r*r
    let result = (Math.Exp (-0.5*residueSquared))/(sigma * (Math.Sqrt (2.0*Math.PI*dt)))
    if result<0.0 then failwith "Negative density, it certainly shouldn't have happened"
    else result

let alphaBeta b (initialPi:float[]) initialA observations= //notation in comments from the Erratum for Rabiner
    let T = Array.length observations
    let N = Array2D.length1 initialA
    let alphaStart = Array.init N (fun i -> initialPi.[i] * (b i observations.[0]))  //this contains \bar{\alpha}
    let alpha_j_t (previousRow:float[]) t j  = (sumTo N (fun i -> previousRow.[i]*initialA.[i, j]))* (b j observations.[t]) //this contains \bar{\alpha}
    let alphaInductionStep t previousRow = Array.init N (alpha_j_t previousRow t)
    let (alpha, norm) = inductionNormalized alphaStart T alphaInductionStep
    let betaStart = Array.init N (fun i -> 1.0/norm.[T-1])
    let beta_j_t (nextRow:float[]) t j = (sumTo N (fun i -> initialA.[j, i]*nextRow.[i]*(b i observations.[t+1])))/norm.[t]
    let betaInductionStep t nextRow = Array.init N (beta_j_t nextRow t)
    let beta = backInduction betaStart T betaInductionStep
    (alpha, beta, norm) //c_t = 1/norm_t


let log_P_O norm = 
  let result = norm |> Seq.sumBy (fun norm_t -> Math.Log norm_t)//c_t = 1/norm_t
  if Double.IsNaN result then failwith "log likelihood is NaN"
  else result

let gamma (alpha:float[][], beta:float[][], norm:float[])  i t = 
    alpha.[t].[i]*beta.[t].[i]*norm.[t]

let xi b (initialA:float[,]) (alpha:float[][]) (beta:float[][]) (observations:float[]) i j t = 
    alpha.[t].[i]*initialA.[i,j]*(b j observations.[t+1])*beta.[t+1].[j]


let oneStep llFunction dt (initialPi, initialA, initialTheta) observations =
  let T = Array.length observations
  let N = Array2D.length1 initialA
  let b = llFunction dt initialTheta
  let (alpha, beta, norm) = alphaBeta b initialPi initialA observations
  let gamma = gamma (alpha, beta, norm)
  let xi = xi b initialA alpha beta observations
  let pi = Array.init N (fun i -> gamma i 0) //Rabiner (40a)
  let A =  //Rabiner (40b)
    let A_func i j = (sumTo (T-1) (xi i j))/(sumTo (T-1) (gamma i))
    Array2D.init N N A_func
  let mean i = (sumTo T (fun t -> (gamma i t) * observations.[t]))/(sumTo T (gamma i))//Rabiner (53)
  let var i = 
    let numerator = sumTo T (fun t -> (gamma i t) * (sqr (observations.[t]-(mean i))))
    let denumerator = sumTo T (gamma i)
    numerator/denumerator
  let mu i = ((mean i) + 0.5*(var i))/dt
  let sigma i = Math.Sqrt ((var i)/dt)
  let theta = Array2D.init N 2 (fun i k -> if k=0 then mu i else sigma i) 
  let logLikelihood = log_P_O norm //Rabiner (103)
  (logLikelihood, (pi, A, theta))

let print (ll, (pi, A, theta)) = 
  printfn "pi = %A" pi
  printfn "A = %A" A
  printfn "theta = %A" theta
  printfn "logLikelihood = %f" ll

let baumWelch likelihood dt initialParams observations =
  let tolerance = 10e-5
  let rec doStep parameters previousLL =
    //print (previousLL, parameters)
    let (logLikelihood, parameters) = oneStep likelihood dt parameters observations
    if Math.Abs(previousLL - logLikelihood) < tolerance then (logLikelihood, parameters)
    else doStep parameters logLikelihood
  doStep initialParams -10e100
4

3 回答 3

2

我没有尝试通过 F# 猜测我的方式，但这里有一些观察：

1）你有多少初始状态的观察？如果答案是“只有一个”，那么观察的概率可以写成 P(state 0) P(obs | state is 0) + P(state 1) P(obs | state 1)。根据两个 P(obs | state is X) 中的哪一个更高，最大似然解将具有 P(state 0) = 1 或 P(state 1) = 1。我只希望看到中间概率当您可能正在观察从许多不同的初始状态派生的观察结果时的初始状态 - 例如，如果您有多个玩具数据要同时分析。

2）在寻找错误时，它可以帮助生成答案完全显而易见的玩具数据。如果我有 n 条 {0, 0, 0, 0...} 形式的数据和 m 条 {1, 1, 1, 1...} 形式的数据，我可能希望看到分配的状态 0初始概率 n/(m +n) - 或者当然是 m/(n + m)，因为程序不知道我希望与哪个序列关联哪个状态。

3) 检查程序的另一种方法是寻找某种一致性或保护性检查。由于两个初始状态的模型可以与只有一个初始状态的模型相同，第一次观察的一组特殊转换概率，可能还有一个特殊的虚拟第一次观察，你可以检查它在两个初始状态下的行为它的行为只有一个初始状态和一些捏造。

于 2012-08-08T17:37:34.027 回答
1

我的猜测是，仅使用一个观察序列几乎总是会导致概率收敛到 0/1。

于 2013-06-27T09:13:44.457 回答
0

经过一段时间的思考，我认为我描述的行为实际上可能是正确的。原因是在观察到的数据中，这个分布只“使用”了一次，所以直观上没有足够的统计数据来推断分布。话虽如此，我认为算法应该能够（以合理的准确度）恢复隐藏状态变量在时间 0 的实际值——因为这对整个时间序列产生了影响。

于 2012-08-09T09:15:16.087 回答