出一个题目

[（ヅ）]

大家知道两个随机变量X,Y独立的定义是P(x,y) = P(x)P(y)

请证明或证伪，如果联合分布P(x,y)可以写成P(x,y) = f(x)g(y)，则X,Y独立(反之亦然，但是trivial)

[FoxMe]

很容易证明吧。积分求出P(x), P(y)。

[（ヅ）]

很容易证明吧。积分求出P(x), P(y)。

peng，差评！

[TheMatrix]

大家知道两个随机变量X,Y独立的定义是P(x,y) = P(x)P(y)

请证明或证伪，如果联合分布P(x,y)可以写成P(x,y) = f(x)g(y)，则X,Y独立(反之亦然，但是trivial)

这个我想了一下。大概清楚了里面的问题。关键是定义的用法。


(1) 分布，这里应该是指distribution function，也就是cdf - cumulative distribution function：
f(t)=P(x<=t)
g(t)=P(y<=t)
联合分布应该换一个符号，比如F：
F(s,t)=P(x<=s,y<=t)

(2) 而两随机变量独立的定义，可以写为对于任意interval A 和 B，
P(X in A, Y in B) = P(X in A)P(Y in B)

所以存在一个从(1)推出(2)的问题。这里面有一些细节。但是我觉得大概差不多就行了。

[wind]

刚讲过，你这个结论是错的，得要保证定义区间（support）是长方形的才行，否则pdf里头x，y可以分开没法保证CDF里头也可以。找一个区间是三角形的即可。

大家知道两个随机变量X,Y独立的定义是P(x,y) = P(x)P(y)

请证明或证伪，如果联合分布P(x,y)可以写成P(x,y) = f(x)g(y)，则X,Y独立(反之亦然，但是trivial)

[（ヅ）]

这个我想了一下。大概清楚了里面的问题。关键是定义的用法。


(1) 分布，这里应该是指distribution function，也就是cdf - cumulative distribution function：
f(t)=P(x<=t)
g(t)=P(y<=t)
联合分布应该换一个符号，比如F：
F(s,t)=P(x<=s,y<=t)

(2) 而两随机变量独立的定义，可以写为对于任意interval A 和 B，
P(X in A, Y in B) = P(X in A)P(Y in B)

所以存在一个从(1)推出(2)的问题。这里面有一些细节。但是我觉得大概差不多就行了。

无所谓的，离散pdf或者连续pdf只有细节上有点区别

[TheMatrix]

无所谓的，离散pdf或者连续pdf只有细节上有点区别

属实。

[YWY]

大家知道两个随机变量X,Y独立的定义是P(x,y) = P(x)P(y)

请证明或证伪，如果联合分布P(x,y)可以写成P(x,y) = f(x)g(y)，则X,Y独立(反之亦然，但是trivial)

As P(x,y) is non-negative, we get P(x,y) = |f(x)||g(y)|. So, after renaming f and g, we can assume that f and g are all non-negative functions. Since P(infty, infty) = 1, we see f(infty) * g(infty) = 1. Thus, by multiplying f and g with proper coefficients, we can assume f(infty) = g(infty) = 1. Let P₁(x) and P₂(y) denote the distribution functions of the two random variables X and Y respectively. For any values x and y, we see P₁(x) = P(x, infty) = f(x)g(infty) = f(x) and P₂(y) = P(infty, y) = f(infty)g(y) = g(y). Thus, P(x,y) = f(x)g(y) = P₁(x)P₂(y). In more rigorous term, we see P(X \leq a, Y \leq b) = P(X \leq a)P(Y \leq b) for all a and b, showing that the random variables X and Y are independent.

[（ヅ）]

As P(x,y) is non-negative, we get P(x,y) = |f(x)||g(y)|. So, after renaming f and g, we can assume that f and g are all non-negative functions. Since P(infty, infty) = 1, we see f(infty) * g(infty) = 1. Thus, by multiplying f and g with proper coefficients, we can assume f(infty) = g(infty) = 1. Let P₁(x) and P₂(y) denote the distribution functions of the two random variables X and Y respectively. For any values x and y, we see P₁(x) = P(x, infty) = f(x)g(infty) = f(x) and P₂(y) = P(infty, y) = f(infty)g(y) = g(y). Thus, P(x,y) = f(x)g(y) = P₁(x)P₂(y). In more rigorous term, we see P(X \leq a, Y \leq b) = P(X \leq a)P(Y \leq b) for all a and b, showing that the random variables X and Y are independent.

差不多，我开始想给这个题目难度写成1/5的，因为确实不是很难

离散分布的话，考虑这三个公式

P(x) = \sum_y f(x)g(y) = f(x) \sum_y g(y)
P(y) = g(y) \sum_x f(x)
\sum_x \sum_y P(x,y) = 1

那么就有P(x,y) = P(x)P(y),\forall x,y。连续的pdf求和换成积分也差不多

[TheMatrix]

As P(x,y) is non-negative, we get P(x,y) = |f(x)||g(y)|. So, after renaming f and g, we can assume that f and g are all non-negative functions. Since P(infty, infty) = 1, we see f(infty) * g(infty) = 1. Thus, by multiplying f and g with proper coefficients, we can assume f(infty) = g(infty) = 1. Let P₁(x) and P₂(y) denote the distribution functions of the two random variables X and Y respectively. For any values x and y, we see P₁(x) = P(x, infty) = f(x)g(infty) = f(x) and P₂(y) = P(infty, y) = f(infty)g(y) = g(y). Thus, P(x,y) = f(x)g(y) = P₁(x)P₂(y). In more rigorous term, we see P(X \leq a, Y \leq b) = P(X \leq a)P(Y \leq b) for all a and b, showing that the random variables X and Y are independent.

哦。才发现这个问题只要求联合分布可以分离变量。我隐含假设了f(x)=p(X<=x)，g(y)=P(Y<=y)了。

你把这一步证了一下，你是对的。

[nk]

As P(x,y) is non-negative, we get P(x,y) = |f(x)||g(y)|. So, after renaming f and g, we can assume that f and g are all non-negative functions. Since P(infty, infty) = 1, we see f(infty) * g(infty) = 1. Thus, by multiplying f and g with proper coefficients, we can assume f(infty) = g(infty) = 1. Let P₁(x) and P₂(y) denote the distribution functions of the two random variables X and Y respectively. For any values x and y, we see P₁(x) = P(x, infty) = f(x)g(infty) = f(x) and P₂(y) = P(infty, y) = f(infty)g(y) = g(y). Thus, P(x,y) = f(x)g(y) = P₁(x)P₂(y). In more rigorous term, we see P(X \leq a, Y \leq b) = P(X \leq a)P(Y \leq b) for all a and b, showing that the random variables X and Y are independent.

这个证明是对的。

[pseudo]

刚讲过，你这个结论是错的，得要保证定义区间（support）是长方形的才行，否则pdf里头x，y可以分开没法保证CDF里头也可以。找一个区间是三角形的即可。

如果support不是长方形，P(x,y)也不能写成P(x)P(y)的。

[nk]

差不多，我开始想给这个题目难度写成1/5的，因为确实不是很难

离散分布的话，考虑这三个公式

P(x) = \sum_y f(x)g(y) = f(x) \sum_y g(y)
P(y) = g(y) \sum_x f(x)
\sum_x \sum_y P(x,y) = 1

那么就有P(x,y) = P(x)P(y),\forall x,y。连续的pdf求和换成积分也差不多

这个不就是 FoxMe 的求积分方法吗？连续分布用积分，离散分布用求和。为什么去碰他呢？

当然还是YWY的证明最正确，最严谨。把两种情况都包括了，还包括了离散和连续的混合分布的那种的通用情况。

还有原题出得不严谨，应该把 P(x,y) 理解为CDF 最严谨，而不是可以理解成 pdf (连续分布）, 或者理解成 pmf （离散分布）