Mutual independence in probability
$begingroup$
Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.
Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.
probability
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
$endgroup$
add a comment |
$begingroup$
Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.
Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.
probability
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
$endgroup$
add a comment |
$begingroup$
Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.
Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.
probability
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
$endgroup$
Let $P(X_i=1)=P(X_i=-1)=frac{1}{2}$ for $i = 1,2,3$.
Prove that $A={X_1=X_2}, B={ X_1=X_3}, C={X_2=X_3}$ are not MUTUALLY INDEPENDENT. So I know I have to work out $P(Acap{B}cap{C})$ and show it's not equal to $P(A)P(B)P(C)$ but I don't know how to go about it since one variable is in terms of the other.
probability
probability
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
edited Jan 15 at 18:52
Mee Seong Im
2,8151617
2,8151617
asked Jan 15 at 18:04
NoteBookNoteBook
1167
asked Jan 15 at 18:04
NoteBookNoteBook
1167
asked Jan 15 at 18:04
NoteBookNoteBook
1167
1167
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."
Do you know how to compute $P(A) = P(X_1 = X_2)$?
$P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.
Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.
Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?
$A cap B cap C = {X_1 = X_2 = X_3}$
Then can you compute the probability of $A cap B cap C$?
$P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
add a comment |
$begingroup$
Assume $X_1$, $X_2$, and $X_3$ are mutually independent.
Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
$$
A = {X_1=X_2}, qquad
B = { X_1=X_3}, qquad
C = {X_2=X_3 }.
$$
Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.
Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
begin{align*}
P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
&qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
&= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
&= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
&= frac{1}{2},
end{align*}
where the second equality holds since $X_1$ and $X_2$ are independent.
Similarly, $P(B)=P(C)=frac{1}{2}$.
So
begin{align*}
P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
\
&qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
&= P(X_1=1)P(X_2=1)P(X_3=1) \
&qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
&= frac{1}{2^3} + frac{1}{2^3} \
&= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
end{align*}
On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.
Thus the events $A$, $B$, and $C$ are not mutually independent.
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
add a comment |
Your Answer
StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");
StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);
StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});
function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074732%2fmutual-independence-in-probability%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."
Do you know how to compute $P(A) = P(X_1 = X_2)$?
$P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.
Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.
Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?
$A cap B cap C = {X_1 = X_2 = X_3}$
Then can you compute the probability of $A cap B cap C$?
$P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
add a comment |
$begingroup$
Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."
Do you know how to compute $P(A) = P(X_1 = X_2)$?
$P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.
Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.
Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?
$A cap B cap C = {X_1 = X_2 = X_3}$
Then can you compute the probability of $A cap B cap C$?
$P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
$endgroup$
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
add a comment |
$begingroup$
Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."
Do you know how to compute $P(A) = P(X_1 = X_2)$?
$P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.
Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.
Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?
$A cap B cap C = {X_1 = X_2 = X_3}$
Then can you compute the probability of $A cap B cap C$?
$P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
$endgroup$
Nitpicking: you should write "$A = {X_1 = X_2}$" instead of "$A = X_1 = X_2$."
Do you know how to compute $P(A) = P(X_1 = X_2)$?
$P(A) = P(X_1 = X_2) = P(X_1=1, X_2=1) + P(X_1=-1, X_2 = -1) = frac{1}{2} cdot frac{1}{2} + frac{1}{2} cdot frac{1}{2}$.
Computing $P(B)$ and $P(C)$ can be done similarly. Then you can compute $P(A) P(B) P(C)$.
Can you give a simple description (in terms of $X_1, X_2, X_3$) of the event $A cap B cap C$?
$A cap B cap C = {X_1 = X_2 = X_3}$
Then can you compute the probability of $A cap B cap C$?
$P(X_1=X_2=X_3) = P(X_1=1, X_2=1, X_3=1) + P(X_1=-1, X_2=-1, X_3=-1) = cdots$
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
edited Jan 15 at 18:13
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
answered Jan 15 at 18:08
angryavianangryavian
42.5k23481
42.5k23481
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
add a comment |
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
$begingroup$
spotted my mistake, thank you
$endgroup$
– NoteBook
Jan 15 at 18:23
add a comment |
$begingroup$
Assume $X_1$, $X_2$, and $X_3$ are mutually independent.
Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
$$
A = {X_1=X_2}, qquad
B = { X_1=X_3}, qquad
C = {X_2=X_3 }.
$$
Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.
Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
begin{align*}
P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
&qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
&= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
&= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
&= frac{1}{2},
end{align*}
where the second equality holds since $X_1$ and $X_2$ are independent.
Similarly, $P(B)=P(C)=frac{1}{2}$.
So
begin{align*}
P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
\
&qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
&= P(X_1=1)P(X_2=1)P(X_3=1) \
&qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
&= frac{1}{2^3} + frac{1}{2^3} \
&= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
end{align*}
On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.
Thus the events $A$, $B$, and $C$ are not mutually independent.
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
add a comment |
$begingroup$
Assume $X_1$, $X_2$, and $X_3$ are mutually independent.
Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
$$
A = {X_1=X_2}, qquad
B = { X_1=X_3}, qquad
C = {X_2=X_3 }.
$$
Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.
Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
begin{align*}
P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
&qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
&= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
&= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
&= frac{1}{2},
end{align*}
where the second equality holds since $X_1$ and $X_2$ are independent.
Similarly, $P(B)=P(C)=frac{1}{2}$.
So
begin{align*}
P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
\
&qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
&= P(X_1=1)P(X_2=1)P(X_3=1) \
&qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
&= frac{1}{2^3} + frac{1}{2^3} \
&= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
end{align*}
On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.
Thus the events $A$, $B$, and $C$ are not mutually independent.
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
add a comment |
$begingroup$
Assume $X_1$, $X_2$, and $X_3$ are mutually independent.
Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
$$
A = {X_1=X_2}, qquad
B = { X_1=X_3}, qquad
C = {X_2=X_3 }.
$$
Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.
Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
begin{align*}
P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
&qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
&= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
&= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
&= frac{1}{2},
end{align*}
where the second equality holds since $X_1$ and $X_2$ are independent.
Similarly, $P(B)=P(C)=frac{1}{2}$.
So
begin{align*}
P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
\
&qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
&= P(X_1=1)P(X_2=1)P(X_3=1) \
&qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
&= frac{1}{2^3} + frac{1}{2^3} \
&= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
end{align*}
On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.
Thus the events $A$, $B$, and $C$ are not mutually independent.
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
$endgroup$
Assume $X_1$, $X_2$, and $X_3$ are mutually independent.
Let $A$ be the event when $X_1=X_2$, $B$ be the event when $X_1=X_3$, and $C$ be the event when $X_2=X_3$. That is,
$$
A = {X_1=X_2}, qquad
B = { X_1=X_3}, qquad
C = {X_2=X_3 }.
$$
Note that $A$ has the following outcomes: when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$. Events $B$ and $C$ have two similar outcomes.
Since the outcomes for $A$ (when $X_1=1$ and $X_2=1$, and when $X_1=-1$ and $X_2=-1$) cannot occur simultaneously, we have
begin{align*}
P(A) &= P(X_1=1 mbox{ and }X_2=1) + P(X_1=-1 mbox{ and }X_2=-1) \
&qquad -P(X_1=1 mbox{ and }X_2=1 mbox{ and }X_1=-1 mbox{ and }X_2=-1) \
&= P(X_1=1)P(X_2=1) + P(X_1=-1)P(X_2=-1) -0 \
&= frac{1}{2}cdot frac{1}{2} + frac{1}{2}cdot frac{1}{2} \
&= frac{1}{2},
end{align*}
where the second equality holds since $X_1$ and $X_2$ are independent.
Similarly, $P(B)=P(C)=frac{1}{2}$.
So
begin{align*}
P(Acap Bcap C) &= P(X_1=1 mbox{ and } X_2=1 mbox{ and } X_3=1)
\
&qquad + P(X_1=-1 mbox{ and } X_2=-1 mbox{ and } X_3=-1) \
&= P(X_1=1)P(X_2=1)P(X_3=1) \
&qquad + P(X_1=-1)P(X_2=-1)P(X_3=-1) \
&= frac{1}{2^3} + frac{1}{2^3} \
&= require{enclose} enclose{circle}{~~frac{1}{4}~~}.
end{align*}
On the other hand, $P(A)P(B)P(C)=require{enclose}enclose{circle}{~frac{1}{8}~}$.
Thus the events $A$, $B$, and $C$ are not mutually independent.
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
edited Jan 15 at 19:04
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
answered Jan 15 at 18:50
Mee Seong ImMee Seong Im
2,8151617
2,8151617
add a comment |
add a comment |
Thanks for contributing an answer to Mathematics Stack Exchange!
- Please be sure to answer the question. Provide details and share your research!
But avoid …
- Asking for help, clarification, or responding to other answers.
- Making statements based on opinion; back them up with references or personal experience.
Use MathJax to format equations. MathJax reference.
To learn more, see our tips on writing great answers.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3074732%2fmutual-independence-in-probability%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
Required, but never shown
