공분산이 0과 같은가 이항 랜덤 변수에 대한 독립성을 의미합니까?

만약 $X$ 와 $Y$ 있습니다 나는 것을 보여줄 수있는 방법 만이 개 가능한 상태를 취할 수있는 두 개의 확률 변수, $Cov(X,Y) = 0$ 독립을 의미? $Cov(X,Y) = 0$ 이 독립성을 의미하지 않는 날에 배운 것에 반하는 이런 종류의 ...

힌트는 가능한 상태로 $1$ 과 $0$ 으로 시작하여 거기에서 일반화한다고 말합니다. 그리고 나는 그것을 할 수 있고 보여줄 수 $E(XY) = E(X)E(Y)$있지만 이것은 독립성을 의미하지는 않습니까 ???

수학적 으로이 작업을 수행하는 방법을 혼란스럽게 생각합니다.

이진 변수의 경우 예상 값은 1과 같을 확률과 같습니다. 따라서,

$$E(XY) = P(XY = 1) = P(X=1 \cap Y=1) \\ E(X) = P(X=1) \\ E(Y) = P(Y=1) \\ $$

두 제로 공분산 수단이있는 경우 수단$E(XY) = E(X)E(Y)$

$$P(X=1 \cap Y=1) = P(X=1) \cdot P(Y=1)$$

독립 사건에 대한 기본 규칙을 사용하여 (예 : 와 B 가 독립하고 그 상보가 독립 등) 다른 모든 공동 확률도 곱하는 것을 보는 것은 사소한 일입니다. 두 개의 임의 변수 중 독립적입니다.$A$$B$

Both correlation and covariance measure linear association between two given variables and it has no obligation to detect any other form of association else.

So those two variables might be associated in several other non-linear ways and covariance (and, therefore, correlation) could not distinguish from independent case.

As a very didactic, artificial and non realistic example, one can consider $X$ such that $P(X=x)=1/3$ for $x=−1,0,1$ and also consider $Y=X^2$. Notice that they are not only associated, but one is a function of the other. Nonetheless, their covariance is 0, for their association is orthogonal to the association that covariance can detect.

EDIT

Indeed, as indicated by @whuber, the above original answer was actually a comment on how the assertion is not universally true if both variables were not necessarily dichotomous. My bad!

So let's math up. (The local equivalent of Barney Stinson's "Suit up!")

Particular Case

두 경우 및 Y는 이분법 있었다 그럼 모두 값만한다고 가정, 일반성의 손실없이, 취할 수 0 및 1 , 임의의 확률로 , P , Q를 그리고 r에 의해 제공된 P ( X = 1 ) = P ∈ [ 0 , 1 ] P ( Y = 1 ) = q ∈ [ 0 , 1 ] P ( X = 1 , $X$$Y$$0$$1$$p$$q$$r$ 완전히 공동 분포를 특성화하는

$$\begin{align*} P(X=1) = p \in [0,1] \\ P(Y=1) = q \in [0,1] \\ P(X=1,Y=1) = r \in [0,1], \end{align*}$$ 및 Y를 . @DilipSarwate의 힌트를 취하면 P ( X = 0 , Y = 1 ) 이므로 세 값이 ( X , Y ) 의 공동 분포를 결정하기에 충분하다는 점에 유의하십시오.$X$$Y$$(X,Y)$ (측면에서, 물론r은p-r∈[0,1],q-r∈[0,1]및1-p−q-r∈ $$\begin{align*} P(X=0,Y=1) &= P(Y=1) - P(X=1,Y=1) = q - r\\ P(X=1,Y=0) &= P(X=1) - P(X=1,Y=1) = p - r\\ P(X=0,Y=0) &= 1 - P(X=0,Y=1) - P(X=1,Y=0) - P(X=1,Y=1) \\ &= 1 - (q - r) - (p - r) - r = 1 - p - q - r. \end{align*}$$ $r$$p-r\in[0,1]$$q-r\in[0,1]$ 이후 R ∈ [ 0 , 1 ] 말하자면, R을 ∈ [ 0 , 분 ( P , Q , 1 - P - Q ) ] ).$1-p-q-r\in[0,1]$$r\in[0,1]$$r\in[0,\min(p,q,1-p-q)]$

Notice that $r = P(X=1,Y=1)$ might be equal to the product $p\cdot q = P(X=1) P(Y=1)$, which would render $X$ and $Y$ independent, since

$$\begin{align*} P(X=0,Y=0) &= 1 - p - q - pq = (1-p)(1-q) = P(X=0)P(Y=0)\\ P(X=1,Y=0) &= p - pq = p(1-q) = P(X=1)P(Y=0)\\ P(X=0,Y=1) &= q - pq = (1-p)q = P(X=0)P(Y=1). \end{align*}$$

Yes, $r$ might be equal to $pq$, BUT it can be different, as long as it respects the boundaries above.

Well, from the above joint distribution, we would have

$$\begin{align*} E(X) &= 0\cdot P(X=0) + 1\cdot P(X=1) = P(X=1) = p \\ E(Y) &= 0\cdot P(Y=0) + 1\cdot P(Y=1) = P(Y=1) = q \\ E(XY) &= 0\cdot P(XY=0) + 1\cdot P(XY=1) \\ &= P(XY=1) = P(X=1,Y=1) = r\\ Cov(X,Y) &= E(XY) - E(X)E(Y) = r - pq \end{align*}$$

Now, notice then that $X$ and $Y$ are independent if and only if $Cov(X,Y)=0$. Indeed, if $X$ and $Y$ are independent, then $P(X=1,Y=1)=P(X=1)P(Y=1)$, which is to say $r=pq$. Therefore, $Cov(X,Y)=r-pq=0$; and, on the other hand, if $Cov(X,Y)=0$, then $r-pq=0$, which is to say $r=pq$. Therefore, $X$ and $Y$ are independent.

General Case

About the without loss of generality clause above, if $X$ and $Y$ were distributed otherwise, let's say, for $a<b$ and $c<d$,

$$\begin{align*} P(X=b)=p \\ P(Y=d)=q \\ P(X=b, Y=d)=r \end{align*}$$ then $X'$ and $Y'$ given by $$X'=\frac{X-a}{b-a} \qquad \text{and} \qquad Y'=\frac{Y-c}{d-c}$$ would be distributed just as characterized above, since $$X=a \Leftrightarrow X'=0, \quad X=b \Leftrightarrow X'=1, \quad Y=c \Leftrightarrow Y'=0 \quad \text{and} \quad Y=d \Leftrightarrow Y'=1.$$ So $X$ and $Y$ are independent if and only if $X'$ and $Y'$ are independent.

Also, we would have

$$\begin{align*} E(X') &= E\left(\frac{X-a}{b-a}\right) = \frac{E(X)-a}{b-a} \\ E(Y') &= E\left(\frac{Y-c}{d-c}\right) = \frac{E(Y)-c}{d-c} \\ E(X'Y') &= E\left(\frac{X-a}{b-a} \frac{Y-c}{d-c}\right) = \frac{E[(X-a)(Y-c)]}{(b-a)(d-c)} \\ &= \frac{E(XY-Xc-aY+ac)}{(b-a)(d-c)} = \frac{E(XY)-cE(X)-aE(Y)+ac}{(b-a)(d-c)} \\ Cov(X',Y') &= E(X'Y')-E(X')E(Y') \\ &= \frac{E(XY)-cE(X)-aE(Y)+ac}{(b-a)(d-c)} - \frac{E(X)-a}{b-a} \frac{E(Y)-c}{d-c} \\ &= \frac{[E(XY)-cE(X)-aE(Y)+ac] - [E(X)-a] [E(Y)-c]}{(b-a)(d-c)}\\ &= \frac{[E(XY)-cE(X)-aE(Y)+ac] - [E(X)E(Y)-cE(X)-aE(Y)+ac]}{(b-a)(d-c)}\\ &= \frac{E(XY)-E(X)E(Y)}{(b-a)(d-c)} = \frac{1}{(b-a)(d-c)} Cov(X,Y). \end{align*}$$ So $Cov(X,Y)=0$ if and only $Cov(X',Y')=0$.

=D

IN GENERAL:

The criterion for independence is $F(x,y) = F_X(x)F_Y(y)$. Or

$$f_{X,Y}(x,y)=f_X(x)\,f_Y(y)\tag 1$$

"If two variables are independent, their covariance is $0.$ But, having a covariance of $0$ does not imply the variables are independent."

This is nicely explained by Macro here, and in the Wikipedia entry for independence.

$\text {independence} \Rightarrow \text{zero cov}$, yet

$\text{zero cov}\nRightarrow \text{independence}.$

Great example: $X \sim N(0,1)$, and $Y= X^2.$ Covariance is zero (and $\mathbb E(XY)=0$, which is the criterion for orthogonality), yet they are dependent. Credit goes to this post.

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IN PARTICULAR (OP problem):

These are Bernoulli rv's, $X$ and $Y$ with probability of success $\Pr(X=1)$, and $\Pr(Y=1)$.

$\begin{align}\mathrm{cov}(X,Y)&=\mathrm E[XY] - \mathrm E[X]\,\mathrm E[Y]\\[2ex] &\underset{*}{=} \Pr(X=1 \cap Y=1) - \Pr(X=1)\, \Pr(Y=1)\\[2ex] &\implies \Pr(X=1 , Y=1) = \Pr (X=1)\,\Pr(Y=1). \end{align}$

This is equivalent to the condition for independence in Eq. $(1).$

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$(*)$:

$$\mathrm E[XY]\quad \underset{**}{=} \quad \displaystyle \sum_{\text{domain X, Y}} \Pr(X=x\cap Y=y)\, x\,y \underset{\neq\,0\text{ iff } x \times y\neq 0}= \Pr(X=1 \cap Y=1).$$

$(**)$: by LOTUS.

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As pointed out below, the argument is incomplete without what Dilip Sarwate had pointed out in his comments shortly after the OP appeared. After searching around, I found this proof of the missing part here:

If events $A$ and $B$ are independent, then events $A^c$ and $B$ are independent, and events $A^c$ and $B^c$ are also independent.

Proof By definition,

$A$ and $B$ are independent $\iff P(A\cap B) = P(A)P(B).$

But $B=(A\cap B) + ( A^c \cup B)$, so $P(B)= P(A\cap B) + P(A^c \cup B)$, which yields:

$\small P(A^c \cap B) = P(B) - P(A\cap B) = P(B) - P(A)\,P(B) = P(B) \left[1 - P(A)\right] = P(B)\,P( A^c).$

Repeat the argument for the events $A^c$ and $B^c,$ this time starting from the statement that $A^c$ and $B$ are independent and taking the complement of $B.$

Similarly. $A$ and $B^c$ are independent events.

So, we have shown already that

$$\Pr(X=1 , Y=1) = \Pr (X=1)\,\Pr(Y=1)$$ and the above shows that this implies that $$\Pr(X=i , Y=j) = \Pr (X=i)\,\Pr(Y=j), ~~i, j \in \{0,1\}$$ that is, the joint pmf factors into the product of marginal pmfs everywhere, not just at $(1,1)$. Hence, uncorrelated Bernoulli random variables $X$ and $Y$ are also independent random variables.