Question

我已经实现了Baum-Welch算法，我正在使用已知发行版生成的一些玩具数据。数据是正态分布的，根据隐藏状态具有不同的均值和标准差。有2个州。除了隐藏状态的初始分布之外，该算法似乎收敛于大多数参数，隐藏状态总是收敛于（0; 1）或（1; 0），具体取决于随机数据。

这个算法是正常的吗？如果是这样，我会很感激一些参考，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

Answer 1

我没有试过通过F＃猜测我的方式，但这里有一些观察结果：

1）你有多少初始状态观察？如果答案是“仅一个”那么观察的概率可以写为P（状态0）P（状态为0）+ P（状态1）P（状态1）。根据两个P（obs | state是X）中的哪一个更高，最大似然解将具有P（状态0）= 1或P（状态1）= 1.我只期望看到中间概率初始状态，当你有可能观察到来自许多不同初始状态的观察时 - 例如，如果你有多个玩具数据同时进行分析。

2）在寻找错误时，它可以帮助生成玩具数据，其答案是完全明显的。如果我有n个{0,0,0,0 ...}形式的数据和m个{1,1,1,1 ...}形式的数据，我可能希望看到状态0分配初始概率n /（m + n） - 当然还有m /（n + m），因为程序不知道我希望哪个状态与哪个序列相关联。

3）检查程序的另一种方法是寻找某种一致性或保护检查。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，第一个观察的一组特殊的转移概率，可能是一个特殊的虚拟第一个观察，你可以用两个初始状态来检查它的行为。只有一个初始状态和一些捏造的行为。

Answer 2

我的猜测是，仅使用一个观察序列几乎总会导致probs收敛到0/1。

Answer 3

在考虑了一段时间之后，我认为我描述的行为可能实际上是正确的。原因是在观察到的数据中，这种分布仅被“使用”一次，因此直观地说，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恢复（以合理的准确度）隐藏状态变量在0时刻的实际值 - 因为这会对整个时间序列产生影响。

是否可以使用Baum-Welch算法估计（隐藏）状态的初始分布？

3 个答案: