15-Distribution_and_Expectation
If we toss a fair coin n times, then there are \(2^{n}\) possible outcomes, each of which is equally likely and has probability \(\frac{1}{2^{n}}\) .
But now we want to know what is the number of heads in n coin tosses; call this number X. For n = 4, the result is shown in the following picture
I Random Variables（随机变量）¶
[!DEFINITION 15.1]
( Random Variable ). A random variable X on a sample space Ω is a function X : Ω → R that assigns to each sample point ω ∈ Ω a real number X(ω).(abbreviated r.v.).
As we see from the example above, a random variable X typically does not have a definitive value, but instead only has a probability distribution over the set of possible values X can take, which is why it is called random.
Until further notice, we will restrict our attention to random variables that are discrete , i.e., they take values in a range that is finite or countably infinite.
[!ATTENTION]
Note that the term “random variable” is really something of a misnomer: it is a function so there is nothing random about it and it is definitely not a variable!
I.1 Fixed Points of Permutations（固定点排列）¶
[!DEFINITION 15.2]
Permutation : In mathematics, a permutation of a set is, loosely speaking, an arrangement of its members into a sequence or linear order, or if the set is already ordered, a rearrangement of its elements. The word "permutation" also refers to the act or process of changing the linear order of an ordered set. [!QUESTION]
Suppose we collect the homeworks of n students, randomly shuffle them, and return them to the students. How many students receive their own homework?
这是一个比较经典的问题，即讨论将一个序列重排后没有改变位置的点的数量，我们将这个数字设为 \(X_{n}\) ，下面是当 n = 3 时的映射情况：
II Probability Distribution（概率分布）¶
Since a random variable is defined on a probability space, we can calculate these probabilities given the probabilities of the sample points. Let a be any number in the range of a random variable X. Then the set \(\{\omega \in\Omega: X(\omega)=a\}\) is an event in the sample space (because it is a subset of Ω). We usually abbreviate this event to simply “X = a” , and the probability of it is P[X = a].
[!DEFINITION 15.3]
(Distribution ). The distribution of a discrete random variable X is the collection of values {(a,P[X = a]) : a ∈ A }, where A is the set of all possible values taken by X.
II.1 Bernoulli Distribution（两点分布 / 伯努利分布）¶
A simple yet very useful probability distribution is the Bernoulli distribution of a random variable which takes value in {0,1}:
We say that X is distributed as a Bernoulli random variable with parameter p, and write
II.2 Binomial Distribution（二项分布）¶
If we conduct Bernoulli Distribution for n times, we find that the
which is named Binomial Distribution and we write
II.3 Hypergeometric Distribution（超几何分布）¶
Consider an urn containing N = B + W balls, where B balls are black and W are white. Suppose you randomly sample n ≤ N balls from the urn , and let X denote the number of black balls in your sample. - if you put it back i.e. get ball with replacement, that is Binomial Distribution - if you don't put it back i.e. without replacement, then we call it Hypergeometric Distribution we can get that
This probability distribution is called the hypergeometric distribution with parameters N,B,n, and write: \(Y ∼ Hypergeometric(N, B, n)\)
[!ATTENTION]
Multiple Random Variables and Independence¶
[!DEFINITION 15.4]
The joint distribution of two discrete random variables X and Y is the collection of values {((a,b),P[X = a,Y = b]) : a ∈ A , b ∈ B}, where A is the set of all possible values taken by X and B is the set of all possible values taken by Y.
Given a joint distribution of X and Y, the distribution P[X = a] of X is called the marginal distribution of X, and can be found by summing over the values of Y.
and if P[X=a, Y=b] = P[X=a]P[Y=b], we say that X = a and Y = b are independent for all values a,b.
Expectation（期望）¶
我们在高中应当都已经学习过 数学期望 ，因此这里只陈列重要结论而省略证明过程。
[!DEFINITION 15.5]
(Expectation). The expectation of a discrete random variable X is defined as \(E(X) = \sum_{a \in A} a×P[X = a]\) [!THEOREM (many)]
(15.1) For any two random variables X and Y on the same probability space, we have \(E[X+Y] = E[X] + E[Y]\quad E[cX] = cE[X]\)
(15.2) \(E[f(x)] = \sum_{x}f(x)P_{X}[X=x]\)## Practice
Q 1 Diversify Your Hand
You are dealt 5 cards from a standard 52 card deck. Let X be the number of distinct values in your hand. For instance, the hand (A, A, A, 2, 3) has 3 distinct values.
- (a) Calculate E[X]. (Hint: Consider indicator variables Xi representing whether i appears in the hand.)
- (b) Calculate Var(X)
[!HELP]
答案的想法很奇特（个人认为
） ，记 P[X_i = 0] 为 i 代表的牌不出现的概率，那么 1-P[X_i = 0] 就是某张牌出现的概率，此时 values += 1，那我们看作两点分布即可，即 E[X_i] = P [X_i != 0] (b) 没看懂 qwq
Q 2 Swaps and cycles
We’ll say that a permutation π = (π(1),...,π(n)) contains a swap if there exist i, j ∈ {1,...,n} so that π(i) = j and π(j) = i, where i ̸= j.
- (a) What is the expected number of swaps in a random permutation?
In the same spirit as above, we’ll say that π contains a k-cycle if there exist i_1,...,i_k ∈ {1,...,n} with π(i_1) = i_2,π(i_2) = i_3,...,π(i_k) = i_1.
- (b) Compute the expectation of the numbe of k-cycles.
We can find that swap is a 2-cycles,
Let's take k numbers (1, 2, ... , k) for example, as they in cycle (with \((k-1)!\) kinds of permutations, the rest of the numbers (k+1, k+2, ... , n) have \((n-k)!\) kinds of permutations.
Since k numbers are not special, we get \(( ^{n}_{k})\) kinds of k numbers.
Ok, then we get that the expectation of k-cycles is:
$$
\frac{(k-1)!(n-k)!}{n!}* (^{n}_{k} ) = \frac{1}{k}
$$