关于二项式分布的众数

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Binomial distribution
PMF: $\displaystyle \binom {n}{k}p^{k}q^{n-k}$
Mode: $ ⌊ ( n + 1 ) p ⌋$ or $\displaystyle \lceil (n+1)p\rceil -1$

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Finding mode in Binomial distribution
Let $a_k=P(X=k)$, we have
$$a_k=\binom{n}{k}p^kq^{n-k}\qquad\text{and}\qquad a_{k+1}=\binom{n}{k+1}p^{k+1}q^{n-k-
1},$$
where as usual $q=1-p$ in binomial distribution.

We calculate the ratio $\dfrac{a_{k+1}}{a_k}$. Note that $\frac{\binom{n}{k+1}}{\binom{n}{k}}$ simplifies to $\frac{n-k}{k+1},$
and therefore
$$\frac{a_{k+1}}{a_k}=\frac{n-k}{k+1}\cdot\frac{p}{q}=\frac{n-k}{k+1}\cdot\frac{p}{1-p}.$$

From this equation we can follow:

\begin{align*}
k > (n+1)p-1 \implies a_{k+1} < a_k \\  
k = (n+1)p-1 \implies a_{k+1} = a_k \\
k < (n+1)p-1 \implies a_{k+1} > a_k
\end{align*}
  
The calculation (almost) says that we have equality of two consecutive probabilities precisely if $a_{k+1}=a_k$, that is, if $k=np+p-1$.  Note that $k=np+p-1$ implies that $np+p-1$ is an integer.


So if $k=np+p-1$ is not an integer, there is a single mode; and if $k=np+p-1$ is an integer, there are two modes, at $np+p-1$ and at $np+p$.  

Not quite! We have been a little casual in our algebra. We have not paid attention to whether we might be multiplying or dividing by $0$.  We also have casually accepted what the algebra seems to say, without doing a reality check.  

Suppose that $p=0$. Then $np+p-1$ is an integer, namely $-1$. But whatever $n$ is, there is a single mode, namely $k=0$. In all other situations where $np+p-1$ is an integer, the $k$ we have identified is non-negative.

However, suppose that $p=1$. Again, $np+p-1$ is an integer, and again there is no double mode. The largest $a_k$ occurs at one place only, namely $k=n$, since $np+p$ is in this case beyond our range.

That completes the analysis when $np+p-1$ is an integer. When it is not, the analysis is simple. There is a single mode, at $\lfloor np+p\rfloor$.

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回复 1# hbghlyj
a=5

1. Last /@ Last /@
2.   Table[MaximalBy[
3.     Table[{Coefficient[(5 x + 1)^n, x, i], i}, {i, 0, n}], First], {n,
4.      1, 100}]
5. {1, 2, 3, 4, 5, 5, 6, 7, 8, 9, 10, 10, 11, 12, 13, 14, 15, 15, 16, \
6. 17, 18, 19, 20, 20, 21, 22, 23, 24, 25, 25, 26, 27, 28, 29, 30, 30, \
7. 31, 32, 33, 34, 35, 35, 36, 37, 38, 39, 40, 40, 41, 42, 43, 44, 45, \
8. 45, 46, 47, 48, 49, 50, 50, 51, 52, 53, 54, 55, 55, 56, 57, 58, 59, \
9. 60, 60, 61, 62, 63, 64, 65, 65, 66, 67, 68, 69, 70, 70, 71, 72, 73, \
10. 74, 75, 75, 76, 77, 78, 79, 80, 80, 81, 82, 83, 84}

Copy the Code

规律是遇到5的倍数重复一次
a=6

1. Last /@ Last /@
2.   Table[MaximalBy[
3.     Table[{Coefficient[(6 x + 1)^n, x, i], i}, {i, 0, n}], First], {n,
4.      1, 100}]
5. {1, 2, 3, 4, 5, 6, 6, 7, 8, 9, 10, 11, 12, 12, 13, 14, 15, 16, 17, \
6. 18, 18, 19, 20, 21, 22, 23, 24, 24, 25, 26, 27, 28, 29, 30, 30, 31, \
7. 32, 33, 34, 35, 36, 36, 37, 38, 39, 40, 41, 42, 42, 43, 44, 45, 46, \
8. 47, 48, 48, 49, 50, 51, 52, 53, 54, 54, 55, 56, 57, 58, 59, 60, 60, \
9. 61, 62, 63, 64, 65, 66, 66, 67, 68, 69, 70, 71, 72, 72, 73, 74, 75, \
10. 76, 77, 78, 78, 79, 80, 81, 82, 83, 84, 84, 85, 86}

Copy the Code

规律是遇到6的倍数重复一次

mowxqq

这样解得的是必要条件。只能说明${{\text{a}}_{\text{k}}} \geqslant {a_{k - 1}}$与${{\text{a}}_{\text{ ...
hbghlyj 发表于 2019-9-25 00:04

根据这两个不等式解出 $k$ 的范围，两侧系数都是单调的，不就是只有一个极值吗

力工

这两个不等式说明了$k$的左边不减，右边不增，则$a_k$必是最大值，但觉得有风险，如果$k$恰好是$1或n$就有一个不等式无解了。

hbghlyj

$$\binom{n}{\lfloor n/2\rfloor}=\max_k\binom nk$$
刚才Erdős-Ko-Rado定理那篇帖子里又遇到这个了

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Hypergeometric distribution(超几何分布)
PMF(概率质量函数):$$p_{X}(k)=\Pr(X=k)={\frac {{\binom {K}{k}}{\binom {N-K}{n-k}}}{\binom {N}{n}}}$$
The pmf is positive when $\displaystyle \max(0,n+K-N)\leq k\leq \min(K,n)$.

Mode(众数): $\displaystyle \left\lceil {\frac {(n+1)(K+1)}{N+2}}\right\rceil -1,\left\lfloor {\frac {(n+1)(K+1)}{N+2}}\right\rfloor $.

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类似地计算:
$\frac{p_X(k)}{p_X(k-1)}=\frac{K-k+1}k\cdot\frac{n-k+1}{N-K-n+k}$
当$p_X(k)>0,p_X(k-1)>0$时$\frac{p_X(k)}{p_X(k-1)}>1\Leftrightarrow\frac{K-k+1}k>\frac{N-K-n+k}{n-k+1}\Leftrightarrow\frac{K+1}k>\frac{N-K+1}{n-k+1}\Leftrightarrow\frac{K+1}k>\frac{N+2}{n+1}\Leftrightarrow k<\frac{(n+1)(K+1)}{N+2}$
所以,对于$\max(0,n+K-N)+1\leq k\leq\left\lfloor {\frac {(n+1)(K+1)}{N+2}}\right\rfloor $有$p_X(k-1)<p_X(k)$.
同理,对于$\left\lceil {\frac {(n+1)(K+1)}{N+2}}\right\rceil -1\leq k\leq\min(K,n)-1$有$p_X(k)>p_X(k+1)$.

hbghlyj

hyperGeom.pdf
Examples of the Hypergeometric Distribution
The hypergeometric distribution, $h(N; k; n; x)$, arises in the following way:
Suppose we have $N$ balls $k$ of which are red and $N-k$ are blue. We draw a sample, without replacement, of $n$ balls. Let $X$ equal the number of red balls in our sample of size $n$.
Then$$\Pr(X=x)=h(N, k, n, x)=\frac{\left(\begin{array}{l}k \\ x\end{array}\right)\left(\begin{array}{l}N-k \\ n-x\end{array}\right)}{\left(\begin{array}{l}N \\ n\end{array}\right)}, \quad 0 \leq x \leq n$$
Example 1. (Exercise 5.1.36) A bin of 1000 turnbuckles has an unknown number $D$ of defectives. A sample of 100 turnbuckles has 2 defectives. The maximum likelihood estimate for $D$ is the number which gives the highest probability for obtaining the number of defectives observed in a sample. Find that value of $D$.
Solution: In this example $N = 1000, k = D$ (the defective turnbuckles are called red balls), and the sample size is $n=100$. What value of $D$ makes this the most probable event? We make a table of $h[1000; D; 100; 2]$ for varying values of $D$. Here's the result from Mathematica:
In[15]:= n1=Binomial[1000,100];
In[16]:= pr[d_]:=Binomial[d,2]*Binomial[1000-d,98]/n1;
In[17]:= Table[{d,N[pr[d]]},{d,2,40}]
Out[17]= {{2, 0.00990991}, {3, 0.0268104}, {4, 0.0483501}, {5, 0.0726546},
> {6, 0.098248}, {7, 0.123986}, {8, 0.149}, {9, 0.172646}, {10, 0.194466},
> {11, 0.214153}, {12, 0.231519}, {13, 0.246474}, {14, 0.259001},
> {15, 0.269145}, {16, 0.276991}, {17, 0.282658}, {18, 0.286288},
> {19, 0.288038}, {20, 0.28807}, {21, 0.286554}, {22, 0.283656},
> {23, 0.27954}, {24, 0.274364}, {25, 0.268278}, {26, 0.261422},
> {27, 0.253928}, {28, 0.245918}, {29, 0.237503}, {30, 0.228785},
> {31, 0.219855}, {32, 0.210795}, {33, 0.201677}, {34, 0.192565},
> {35, 0.183516}, {36, 0.174578}, {37, 0.165792}, {38, 0.157194},
> {39, 0.148812}, {40, 0.14067}}
So we see that the most likely value of $D$ is $20$.
Example 2. (Exercise 5.1.37) There are an unknown number of moose on Isle Royale. To estimate the number of moose, 50 moose are captured and tagged. Six months later 200 moose are captured and it is found that 8 of these are tagged. Estimate the number of moose on Isle Royale 1 from these data.
Solution: In the hypergeometric distribution let $N$ equal the total number of moose. We are told that $k$ of them are tagged. (The tagged moose are the red balls.) A sample of size $k = 200$ is drawn and we are told that 8 are tagged. We ask for what choice of $N$ maximizes the probability $h(N; 50; 200; 8]$, where $h(N; k; n; x)$ is given above in (1). If one makes a table as in the above example, one finds that $N = 1250$ maximizes the hypergeometric distribution for the above values of $n, k$ and $x$. Figure 1 shows a plot of $h(N; 50; 200; 8)$ as a function of $N$.

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So we see that the most likely value of $D$ is $20$.

$$\frac{h(N, k\phantom{-,1}, n, x)}{h(N, k-1, n, x)}=\frac k{k-x}\cdot\frac{N-k-n+x+1}{N-k+1}\ge1\Leftrightarrow k\le\frac{(N + 1) x}n$$
代入$N=1000,x=2,n=100$得$k=20$时取最大值.

one finds that $N = 1250$ maximizes the hypergeometric distribution for the above values of $n, k$ and $x$.

$$\frac{h(N\phantom{-,1}, k, n, x)}{h(N-1, k, n, x)}=\frac{\frac{N-k}{N-k-n+x}}{\frac{N}{N-n}}\ge1\Leftrightarrow N\le\frac{kn}x$$
代入$k=50,n=200,x=8$得$N=1250,1249$时取最大值.

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还可以计算Gamma函数的导数
对于$n=4,p=\frac23,q=\frac13,$
FindRoot[D[2^k/(k!(4-k)!),k]==0,{k,2}]
得到k≈2.84522
用1#得到$\lfloor5\cdot\frac23\rfloor=\lceil5\cdot\frac23\rceil-1=3$

对于$n=5,p=\frac23,q=\frac13,$
FindRoot[D[2^k/(k!(5-k)!),k]==0,{k,4}]
计算Gamma函数的导数得到k≈3.51004
用1#得到$\lfloor6\cdot\frac23\rfloor=4$ 与 $\lceil6\cdot\frac23\rceil-1=3$