统计代写|Story proofs stat代写
统计代考
A story proof is a proof by interpretation. For counting problems, this often means counting the same thing in two different ways, rather than doing tedious algebra. A story proof often avoids messy calculations and goes further than an algebraic proof toward explaining why the result is true. The word “story” has several meanings, some more mathematical than others, but a story proof (in the sense in which we’re using the term) is a fully valid mathematical proof. Here are some examples of story proofs, which also serve as further examples of counting.
Example 1.5.1 (Choosing the complement). For any nonnegative integers $n$ and $k$ with $k \leq n$, we have
$$
\left(\begin{array}{l}
n \
k
\end{array}\right)=\left(\begin{array}{c}
n \
n-k
\end{array}\right)
$$
This is easy to check algebraically (by writing the binomial coefficients in terms of factorials), but a story proof makes the result easier to understand intuitively.
Story proof: Consider choosing a committee of size $k$ in a group of $n$ people. We know that there are $\left(\begin{array}{l}n \ k\end{array}\right)$ possibilities. But another way to choose the committee is to specify which $n-k$ people are not on the committee; specifying who is on the committee determines who is not on the committee, and vice versa. So the two sides are equal, as they are two ways of counting the same thing.
Example 1.5.2 (The team captain). For any positive integers $n$ and $k$ with $k \leq n$,
$$
n\left(\begin{array}{l}
n-1 \
k-1
\end{array}\right)=k\left(\begin{array}{l}
n \
k
\end{array}\right)
$$
This is again easy to check algebraically (using the fact that $m !=m(m-1) !$ for any positive integer $m$ ), but a story proof is more insightful.
Story proof: Consider a group of $n$ people, from which a team of $k$ will be chosen, one of whom will be the team captain. To specify a possibility, we could first choose the team captain and then choose the remaining $k-1$ team members; this gives the left-hand side. Equivalently, we could first choose the $k$ team members and then choose one of them to be captain; this gives the right-hand side.
Example 1.5.3 (Vandermonde’s identity). A famous relationship between binomial coefficients, called Vandermonde’s identity, ${ }^{2}$ says that
$$
\left(\begin{array}{c}
m+n \
k
\end{array}\right)=\sum_{j=0}^{k}\left(\begin{array}{c}
m \
j
\end{array}\right)\left(\begin{array}{c}
n \
k-j
\end{array}\right)
$$
This identity will come up several times in this book. Trying to prove it with a brute force expansion of all the binomial coefficients would be a nightmare. But a story proves the result elegantly and makes it clear why the identity holds.
Story proof: Consider a student organization consisting of $m$ juniors and $n$ seniors, from which a committee of size $k$ will be chosen. There are $\left(\begin{array}{c}m+n \ k\end{array}\right)$ possibilities. If there are $j$ juniors in the committee, then there must be $k-j$ seniors in the committee. The right-hand side of the identity sums up the cases for $j$.
Example 1.5.4 (Partnerships). Let’s use a story proof to show that
$$
\frac{(2 n) !}{2^{n} \cdot n !}=(2 n-1)(2 n-3) \cdots 3 \cdot 1
$$
Story proof: We will show that both sides count the number of ways to break $2 n$ people into $n$ partnerships. Take $2 n$ people, and give them ID numbers from 1 to $2 n$. We can form partnerships by lining up the people in some order and then saying the first two are a pair, the next two are a pair, etc. This overcounts by a factor of $n ! \cdot 2^{n}$ since the order of pairs doesn’t matter, nor does the order within each pair. Alternatively, count the number of possibilities by noting that there are $2 n-1$ choices for the partner of person 1 , then $2 n$ – 3 choices for person 2 (or person 3 , if person 2 was already paired to person 1), and so on.
统计代考
故事证明是解释的证明。对于计算问题，这通常意味着用两种不同的方式计算同一事物，而不是做繁琐的代数。故事证明通常避免混乱的计算，并且比代数证明更能解释为什么结果为真。 “故事”一词有多种含义，其中一些含义比其他含义更数学化，但故事证明（在我们使用该术语的意义上）是完全有效的数学证明。这里有一些故事证明的例子，它们也可以作为计数的进一步例子。
示例 1.5.1（选择补码）。对于任何具有 $k \leq n$ 的非负整数 $n$ 和 $k$，我们有
$$
\left(\begin{数组}{l}
n \
ķ
\end{array}\right)=\left(\begin{array}{c}
n \
n-k
\end{数组}\右）
$$
这很容易通过代数检查（通过用阶乘写出二项式系数），但是故事证明使结果更容易直观地理解。
故事证明：考虑在一组 $n$ 人中选择一个规模为 $k$ 的委员会。我们知道有 $\left(\begin{array}{l}n \ k\end{array}\right)$ 种可能性。但另一种选择委员会的方法是指定哪些 $n-k$ 人不在委员会中；指定谁在委员会决定谁不在委员会，反之亦然。所以两边是相等的，因为它们是计算同一事物的两种方式。
示例 1.5.2（队长）。对于任何具有 $k \leq n$ 的正整数 $n$ 和 $k$，
$$
n\left(\begin{数组}{l}
n-1 \
k-1
\end{array}\right)=k\left(\begin{array}{l}
n \
ķ
\end{数组}\右）
$$
这再次很容易通过代数检查（使用 $m !=m(m-1) !$ 对于任何正整数 $m$ 的事实），但故事证明更有洞察力。
故事证明：考虑一组 $n$ 人，从中选择一个 $k$ 的团队，其中一人将担任队长。要指定一种可能性，我们可以先选择队长，然后再选择剩余的$k-1$个团队成员；这给出了左侧。等效地，我们可以先选择$k$的团队成员，然后选择其中一个担任队长；这给出了右手边。
示例 1.5.3（Vandermonde 的身份）。二项式系数之间的著名关系，称为 Vandermonde 恒等式，${ }^{2}$ 表示
$$
\left(\begin{数组}{c}
m+n \
ķ
\end{array}\right)=\sum_{j=0}^{k}\left(\begin{array}{c}
米\
j
\end{array}\right)\left(\begin{array}{c}
n \
k-j
\end{数组}\右）
$$
这个身份将在本书中多次出现。试图用所有二项式系数的强力扩展来证明它将是一场噩梦。但是一个故事优雅地证明了结果，并清楚地说明了身份为何成立。
故事证明：考虑一个由 $m$ 大三学生和 $n$ 大四学生组成的学生组织，从中选出一个规模为 $k$ 的委员会。有 $\left(\begin{array}{c}m+n \ k\end{array}\right)$ 种可能性。如果委员会中有$j$ 的初级，那么委员会中必须有$k-j$ 的高级。恒等式的右侧总结了 $j$ 的情况。
示例 1.5.4（伙伴关系）。让我们用一个故事证明来证明
$$
\frac{(2 n) !}{2^{n} \cdot n !}=(2 n-1)(2 n-3) \cdots 3 \cdot 1
$$
故事证明：我们将展示双方计算将 2 n 美元的人分解为 n 美元伙伴关系的方法的数量。带上 $2 n$ 人，给他们从 1 到 $2 n$ 的 ID 号。我们可以通过将人员按某种顺序排列起来，然后说前两个是一对，接下来的两个是一对，等等来建立伙伴关系。这多计了 $n ！ \cdot 2^{n}$ 因为对的顺序无关紧要，每对中的顺序也不重要。或者，通过注意第 1 个人的合作伙伴有 $2 n-1$ 选择，然后第 2 个人（或第 3 个人，如果第 2 个人已经与第 1 个人配对）有 $2 n$ – 3 个选择来计算可能性的数量， 等等。
R语言代写
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