Apakah kovarians sama dengan nol menyiratkan independensi untuk variabel acak biner?

Jika dan adalah dua variabel acak yang hanya dapat mengambil dua kemungkinan keadaan, bagaimana saya bisa menunjukkan bahwa menyiratkan independensi? Jenis ini bertentangan dengan apa yang saya pelajari kembali pada hari dimana tidak menyiratkan independensi ... $X$ $Y$ $Cov(X,Y) = 0$ $Cov(X,Y) = 0$

Petunjuk mengatakan mulai dengan dan sebagai status yang memungkinkan dan digeneralisasi dari sana. Dan saya bisa melakukan itu dan menunjukkan , tetapi ini tidak menyiratkan independensi ??? $1$ $0$ $E(XY) = E(X)E(Y)$

Agak bingung bagaimana melakukan ini secara matematis, saya kira.

covariance independence

— pengguna3604869
sumber

Itu tidak benar secara umum seperti judul pertanyaan Anda menyarankan ..

— Michael R. Chernick

Pernyataan yang Anda coba buktikan memang benar. Jika dan adalah variabel acak Bernoulli masing-masing dengan parameter

dan

, maka

dan

. Jadi,

sama dengan

X

$X$

hanya jika

Y

$Y$

p_{1}

$p_1$

p_{2}

$p_2$

E [X] = p_{1}

$E[X]=p_1$

E [Y] = p_{2}

$E[Y]=p_2$

cov (X, Y) = E [X Y] - E [X] E [Y]

$\operatorname{cov}(X,Y)=E[XY]-E[X]E[Y]$

0

$0$

E [X Y] = P {X = 1, Y = 1}

$E[XY]=P\{X=1,Y=1\}$ sama dengan

p_{1} p_{2} = P {X = 1} P {Y = 1}

$p_1p_2=P\{X=1\}P\{Y=1\}$ menunjukkan bahwa

{X = 1}

$\{X=1\}$ dan

{Y = 1}

$\{Y=1\}$ adalahindependen acara . Ini adalah hasil standar bahwa jika

A

$A$ dan

B

$B$ adalah sepasang peristiwa independen, maka begitu juga

A, B^{c}

$A,B^c$ , dan

A^{c}, B

$A^c,B$ , dan

A^{c}, B^{c}

$A^c,B^c$ peristiwa independen, yaitu

X

$X$ dan

Y

$Y$ adalah variabel acak independen. Sekarang menggeneralisasi.

— Dilip Sarwate

Jawaban:

Untuk variabel biner, nilai yang diharapkan sama dengan probabilitas bahwa mereka sama dengan satu. Karena itu,

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

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

Jika keduanya memiliki nol kovarians ini berarti $E(XY) = E(X)E(Y)$ , yang berarti

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

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

Sepele untuk melihat semua probabilitas gabungan lainnya berlipat ganda juga, menggunakan aturan dasar tentang peristiwa independen (yaitu jika $A$ dan $B$ independen, maka pelengkap mereka independen, dll.), Yang berarti fungsi massa sendi difaktorkan, yang merupakan definisi dari dua variabel acak yang independen.

— nyai
sumber

Ringkas dan elegan. Berkelas! +1 = D

— Marcelo Ventura

Baik korelasi dan kovarian mengukur hubungan linier antara dua variabel yang diberikan dan tidak berkewajiban untuk mendeteksi bentuk asosiasi lain apa pun.

Jadi kedua variabel tersebut dapat dikaitkan dengan beberapa cara non-linear dan kovarian (dan, oleh karena itu, korelasi) tidak dapat dibedakan dari kasus independen.

Sebagai sangat didaktik, buatan dan contoh non realistis, seseorang dapat mempertimbangkan $X$ sehingga $P(X=x)=1/3$ untuk $x=−1,0,1$ dan juga mempertimbangkan $Y=X^2$ . Perhatikan bahwa mereka tidak hanya terkait, tetapi satu adalah fungsi dari yang lain. Meskipun demikian, kovarians mereka adalah 0, karena hubungan mereka ortogonal dengan asosiasi yang dapat dideteksi oleh kovarians.

EDIT

Memang, seperti yang ditunjukkan oleh @whuber, jawaban asli di atas sebenarnya adalah komentar tentang bagaimana pernyataan itu tidak benar secara universal jika kedua variabel tidak harus dikotomis. Salahku!

Jadi mari kita berhitung. (Setara dengan Barney Stinson's "Suit up!")

Kasus tertentu

Jika $X$ dan $Y$ keduanya dikotomis, maka Anda dapat mengasumsikan, tanpa kehilangan keumuman, bahwa keduanya hanya mengasumsikan nilai $0$ dan $1$ dengan probabilitas arbitrer $p$ , $q$ dan $r$ diberikan oleh

\begin{aligned} P (X = 1) = p \in [0, 1] \\ P (Y = 1) = q \in [0, 1] \\ P (X = 1, Y = 1) = r \in [0, 1], \end{aligned}

$\begin{align*} P(X=1) = p \in [0,1] \\ P(Y=1) = q \in [0,1] \\ P(X=1,Y=1) = r \in [0,1], \end{align*}$ yang mencirikan sepenuhnya distribusi gabungan dari

X

$X$ dan

Y

$Y$ . Mengambil petunjuk @ DilipSarwate, perhatikan bahwa ketiga nilai tersebut cukup untuk menentukan distribusi gabungan

(X, Y)

$(X,Y)$ , karena

(Di samping catatan, tentu saja

terikat untuk menghormati

dan

\begin{aligned} 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{aligned}

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

r

$r$

p - r \in [0, 1]

$p-r\in[0,1]$

q - r \in [0, 1]

$q-r\in[0,1]$

luar

, yaitu

1 - p - q - r \in [0, 1]

$1-p-q-r\in[0,1]$

r \in [0, 1]

$r\in[0,1]$

r \in [0, min (p, q, 1 - p - q)]

$r\in[0,\min(p,q,1-p-q)]$

Perhatikan bahwa mungkin sama dengan produk , yang akan membuat dan independen, karena $r = P(X=1,Y=1)$ $p\cdot q = P(X=1) P(Y=1)$ $X$ $Y$

\begin{aligned} P (X = 0, Y = 0) & = 1 - p - q - p q = (1 - p) (1 - q) = P (X = 0) P (Y = 0) \\ P (X = 1, Y = 0) & = p - p q = p (1 - q) = P (X = 1) P (Y = 0) \\ P (X = 0, Y = 1) & = q - p q = (1 - p) q = P (X = 0) P (Y = 1) . \end{aligned}

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

Ya, mungkin sama dengan , TAPI itu bisa berbeda, asalkan itu menghormati batas-batas di atas. $r$ $pq$

Nah, dari distribusi bersama di atas, kita akan memiliki

\begin{aligned} 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 (X Y) & = 0 \cdot P (X Y = 0) + 1 \cdot P (X Y = 1) \\ = P (X Y = 1) = P (X = 1, Y = 1) = r \\ C o v (X, Y) & = E (X Y) - E (X) E (Y) = r - p q \end{aligned}

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

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{aligned} P (X = b) = p \\ P (Y = d) = q \\ P (X = b, Y = d) = r \end{aligned}

$\begin{align*} P(X=b)=p \\ P(Y=d)=q \\ P(X=b, Y=d)=r \end{align*}$ then

X^{'}

$X'$ and

Y^{'}

$Y'$ given by

X^{'} = \frac{X - a}{b - a} and Y^{'} = \frac{Y - c}{d - c}

$X'=\frac{X-a}{b-a} \qquad \text{and} \qquad Y'=\frac{Y-c}{d-c}$ would be distributed just as characterized above, since

X = a \Leftrightarrow X^{'} = 0, X = b \Leftrightarrow X^{'} = 1, Y = c \Leftrightarrow Y^{'} = 0 and Y = d \Leftrightarrow Y^{'} = 1.

$X=a \Leftrightarrow X'=0, \quad X=b \Leftrightarrow X'=1, \quad Y=c \Leftrightarrow Y'=0 \quad \text{and} \quad Y=d \Leftrightarrow Y'=1.$ So

X

$X$ and

Y

$Y$ are independent if and only if

X^{'}

$X'$ and

Y^{'}

$Y'$ are independent.

Also, we would have

\begin{aligned} E (X^{'}) & = E (\frac{X - a}{b - a}) = \frac{E (X) - a}{b - a} \\ E (Y^{'}) & = E (\frac{Y - c}{d - c}) = \frac{E (Y) - c}{d - c} \\ E (X^{'} Y^{'}) & = E (\frac{X - a}{b - a} \frac{Y - c}{d - c}) = \frac{E [(X - a) (Y - c)]}{(b - a) (d - c)} \\ = \frac{E (X Y - X c - a Y + a c)}{(b - a) (d - c)} = \frac{E (X Y) - c E (X) - a E (Y) + a c}{(b - a) (d - c)} \\ C o v (X^{'}, Y^{'}) & = E (X^{'} Y^{'}) - E (X^{'}) E (Y^{'}) \\ = \frac{E (X Y) - c E (X) - a E (Y) + a c}{(b - a) (d - c)} - \frac{E (X) - a}{b - a} \frac{E (Y) - c}{d - c} \\ = \frac{[E (X Y) - c E (X) - a E (Y) + a c] - [E (X) - a] [E (Y) - c]}{(b - a) (d - c)} \\ = \frac{[E (X Y) - c E (X) - a E (Y) + a c] - [E (X) E (Y) - c E (X) - a E (Y) + a c]}{(b - a) (d - c)} \\ = \frac{E (X Y) - E (X) E (Y)}{(b - a) (d - c)} = \frac{1}{(b - a) (d - c)} C o v (X, Y) . \end{aligned}

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

C o v (X, Y) = 0

$Cov(X,Y)=0$ if and only

C o v (X^{'}, Y^{'}) = 0

$Cov(X',Y')=0$ .

— Marcelo Ventura
sumber

I recycled that answer from this post.

— Marcelo Ventura

Verbatim cut and paste from your other post. Love it. +1

— gammer

The problem with copy-and-paste is that your answer no longer seems to address the question: it is merely a comment on the question. It would be better, then, to post a comment with a link to your other answer.

— whuber

How is thus an answer to the question asked?

— Dilip Sarwate

Your edits still don't answer the question, at least not at the level the question is asked. You write "Notice that

r \dots

$r~\ldots$ not necessarily equal to the product

p q

$pq$ . That exceptional situation corresponds to the case of independence between

X

$X$ and

Y

$Y$ ." which is a perfectly true statement but only for the cognoscenti because for the hoi polloi, independence requires not just that

\begin{matrix} (1) & P (X = 1, Y = 1) = P (X = 1) P (Y = 1) \end{matrix}

$P(X=1,Y=1)=P(X=1)P(Y=1)\tag 1$ but also

\begin{matrix} (2) & P (X = u, Y = v) = P (X = u) P (Y = v), u . v \in {0, 1} . \end{matrix}

$P(X=u,Y=v)=P(X=u)P(Y=v),~u.v\in\{0,1\}.\tag 2$ Yes,

(1) ⟹ (2)

$(1) \implies(2)$ as the cognoscenti know; for lesser mortals, a proof that

(1) ⟹ (2)

$(1) \implies (2)$ is helpful.

— Dilip Sarwate

IN GENERAL:

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

\begin{matrix} (1) & f_{X, Y} (x, y) = f_{X} (x) f_{Y} (y) \end{matrix}

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

"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.

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. $(1).$

$(*)$ :

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

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

$(**)$ : by LOTUS.

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)

$\Pr(X=1 , Y=1) = \Pr (X=1)\,\Pr(Y=1)$ and the above shows that this implies that

Pr (X = i, Y = j) = Pr (X = i) Pr (Y = j), i, j \in {0, 1}

$\Pr(X=i , Y=j) = \Pr (X=i)\,\Pr(Y=j), ~~i, j \in \{0,1\}$ that is, the joint pmf factors into the product of marginal pmfs everywhere, not just at

(1, 1)

$(1,1)$ . Hence, uncorrelated Bernoulli random variables

X

$X$ and

Y

$Y$ are also independent random variables.

— Antoni Parellada
sumber

Actually that's not an equivalent condition to Eq (1). All you showed was that

f_{X, Y} (1, 1) = f_{X} (1) f_{Y} (1)

$f_{X,Y}(1,1) = f_{X}(1) f_{Y}(1)$

— gammer

Please consider replacing that image with your own equations, preferably ones that don't use overbars to denote complements. The overbars in the image are very hard to see.

— Dilip Sarwate

@DilipSarwate No problem. Is it better, now?

— Antoni Parellada

Thanks. Also, note that strictly speaking, you also need to show that

A

$A$ and

B^{c}

$B^c$ are independent events since the factorization of the joint pdf into the product of the marginal pmts must hold at all four points. Perhaps adding the sentence "Similarly.

A

$A$ and

B^{c}

$B^c$ are independent events" right after the proof that

A^{c}

$A^c$ and

B

$B$ are independent events will work.

— Dilip Sarwate

@DilipSarwate Thank you very much for your help getting it right. The proof as it was before all the editing seemed self-explanatory, because of all the inherent symmetry, but it clearly couldn't be taken for granted. I am very appreciative of your assistance.

— Antoni Parellada