对概率中随机过程支持的改进 

在给定时间点前后的过程切片，假设一个特定值，求泊松过程切片 的概率.

显示完整的 Wolfram 语言输入隐藏输入

In[1]:=

X assumps = Element[{k1, k, n}, Integers] && t1 < t2 < t3 && k1 >= 0 && 0 <= k <= n;

In[2]:=

X pdf = Probability[ x[t2] == k1 + k \[Conditioned] {x[t1] == k1, x[t3] == k1 + n}, x \[Distributed] PoissonProcess[\[Mu]], Assumptions -> assumps]

Out[2]=

In[3]:=

X FullSimplify[pdf, assumps]

Out[3]=

对概率和二项分布的概率分布函数（PDF）进行比较.

In[4]:=

X PDF[BinomialDistribution[n, (t2 - t1)/(t3 - t1)], k]

Out[4]=

In[5]:=

X PDF[BinomialDistribution[n, (t2 - t1)/(t3 - t1)], k]; FullSimplify[% == pdf, assumps]

Out[5]=

在给定时间以后的过程切片，假设一个特定值，求二项过程切片 的概率.

显示完整的 Wolfram 语言输入隐藏输入

In[6]:=

X assumps2 = 0 <= k <= n && 0 <= t1 < t2 && n <= t2 && k <= t1;

In[7]:=

X pdf2 = Probability[x[t1] == k \[Conditioned] x[t2] == n, x \[Distributed] BinomialProcess[p], Assumptions -> assumps2] // FullSimplify

Out[7]=

对概率和超几何分布的概率分布函数（PDF）进行比较.

In[8]:=

X PDF[HypergeometricDistribution[n, t1, t2], k]

Out[8]=

In[9]:=

X PDF[HypergeometricDistribution[n, t1, t2], k]; FullSimplify[% == pdf2, assumps2]

Out[9]=