![](/img/trans.png)
[英]How to calculate the transition probability matrix of a second order Markov Chain
[英]How to find the Markov Chain Probability?
我試圖找到鏈條從狀態跳躍的概率k-1
到狀態1
它擊中狀態之前k
。 誰能發現我的錯誤?
我試圖模擬馬爾可夫鏈,但我想編寫一個代碼,使我能夠找到k ={1, 2, 3, ........17}
概率。 但是我真的無法獲取代碼。
這是我總是收到的錯誤消息
Error in while (X[i] > 1 && X[i] < k) { :
missing value where TRUE/FALSE needed
這是我的代碼:
k <- 17
{ p <- 0.5
q <- 0.1
P <- matrix (0, nrow = k, ncol = k, byrow = TRUE)
for (i in 1:k)
{ for (j in 1:k)
{ if (i == 1 && i == j)
{ P[i,j] <- 1
}
else if (i == k && i == j)
{ P[i,j] <- 1
}
else if (i == j)
{ P[i,j] <- p*(1-q)
}
else if (j == k && i != 1)
{ P[i,j] <- q
}
else if (i == j+1 && i != k)
{ P[i,j] <- (1-p)*(1-q)
}
}
}
P
X <- (k-1)
trials <- 1000
hits <- 0 #counter for no. of hits
for (i in 1:trials)
{ i <- 1 #no. of steps
while(X[i] > 1 && X[i] < k)
{ Y <- runif(1) #uniform samples
p1 <- P[X[i],] #calculating the p-value
p1 <- cumsum(p1)
# changes in the chain
if(Y <= p1[1])
{ X[i+1] = 1}
else if(Y <= p1[2])
{ X[i+1] = 2}
else if(Y <= p1[3])
{ X[i+1] = 3}
else if(Y <= p1[4])
{ X[i+1] = 4}
else if(Y <= p1[5])
{ X[i+1] = 5}
else if(Y <= p1[6])
{ X[i+1] = 6}
else if(Y <= p1[7])
{ X[i+1] = 7}
else if(Y <= p1[8])
{ X[i+1] = 8}
else if(Y <= p1[9])
{ X[i+1] = 9}
else if(Y <= p1[10])
{ X[i+1] = 10}
else if(Y <= p1[11])
{ X[i+1] = 11}
else if(Y <= p1[12])
{ X[i+1] = 12}
else if(Y <= p1[13])
{ X[i+1] = 13}
else if(Y <= p1[14])
{ X[i+1] = 14}
else if(Y <= p1[15])
{ X[i+1] = 15}
else if(Y <= p1[16])
{ X[i+1] = 16}
else if(Y <= p1[17])
{ X[i+1] <= 17}
i <- i+1
}
if(X[i]==1)
{ hits <- hits+1}
else
{ hits <- hits+0}
}
Probability <- hits/trials
Probability
}
我認為線
i <- 1 #no. of steps
不應該在那里。 嘗試這個:
k <- 17
{ p <- 0.5
q <- 0.1
P <- matrix (0, nrow = k, ncol = k, byrow = TRUE)
for (i in 1:k)
{ for (j in 1:k)
{ if (i == 1 && i == j)
{ P[i,j] <- 1
}
else if (i == k && i == j)
{ P[i,j] <- 1
}
else if (i == j)
{ P[i,j] <- p*(1-q)
}
else if (j == k && i != 1)
{ P[i,j] <- q
}
else if (i == j+1 && i != k)
{ P[i,j] <- (1-p)*(1-q)
}
}
}
P
X <- (k-1)
trials <- 1000
hits <- 0 #counter for no. of hits
for (i in 1:trials)
{
while(X[i] > 1 && X[i] < k)
{ Y <- runif(1) #uniform samples
p1 <- P[X[i],] #calculating the p-value
p1 <- cumsum(p1)
# changes in the chain
if(Y <= p1[1])
{ X[i+1] = 1}
else if(Y <= p1[2])
{ X[i+1] = 2}
else if(Y <= p1[3])
{ X[i+1] = 3}
else if(Y <= p1[4])
{ X[i+1] = 4}
else if(Y <= p1[5])
{ X[i+1] = 5}
else if(Y <= p1[6])
{ X[i+1] = 6}
else if(Y <= p1[7])
{ X[i+1] = 7}
else if(Y <= p1[8])
{ X[i+1] = 8}
else if(Y <= p1[9])
{ X[i+1] = 9}
else if(Y <= p1[10])
{ X[i+1] = 10}
else if(Y <= p1[11])
{ X[i+1] = 11}
else if(Y <= p1[12])
{ X[i+1] = 12}
else if(Y <= p1[13])
{ X[i+1] = 13}
else if(Y <= p1[14])
{ X[i+1] = 14}
else if(Y <= p1[15])
{ X[i+1] = 15}
else if(Y <= p1[16])
{ X[i+1] = 16}
else if(Y <= p1[17])
{ X[i+1] <= 17}
i <- i+1
}
if(X[i]==1)
{ hits <- hits+1}
else
{ hits <- hits+0}
}
Probability <- hits/trials
Probability
}
您將X設置為k-1。 在R中,將其視為長度為1的向量。一旦i達到2,X [i]返回索引錯誤,因為X沒有第二個元素。
進一步說明:在兩個不同的嵌套級別中使用相同的索引是非常糟糕的形式。 同樣,當您開始擁有大量的if-then-else語句列表時,該重新考慮代碼了。 在這種情況下,您可以將p1 [i]> = Y上的子集1:17設為最小值,然后將X設置為該值。
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