Finding the probability of two random variables being equal to 1
$begingroup$
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
$endgroup$
add a comment |
$begingroup$
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
$endgroup$
add a comment |
$begingroup$
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
$endgroup$
Question:
A die is thrown $n+2$ times. After each throw a '$+$' is recorded for $4$, $5$, or $6$ and '$-$' for $1$,$2$, or $3$, the signs forming an ordered sequence. To each, except the first and the last sign, is attached a characteristic random variable which takes the value $1$ if both the neighboring signs differ from the one between them and $0$ otherwise. If $X_1, X_2, ldots , X_n$ are the characteristic random variables, find the mean and variance of $X = sum_{i=1}^n X_i$
Problematic part:
$$ V(X) = V(X_1+X_2+cdots+X_n) = sum_{i=1}^n V(X_i) +2sum_{i=1}^nsum_{j=1, j>i}^n Cov(X_i,X_j) $$
Calculating the variance of $X$ requires calculating the covariance of two arbitrarily chosen random variables $X_i$ and $X_j$.
The formula then used is $Cov(X_i,X_j) = E(X_iX_j) - E(X_i)E(X_j)$
Which brings us to the essence of my problem -- How to find $E(X_iX_j)$?
It is certain that $X_iX_j$ can take only two values, namely, $0$ and $1$. Therefore $E(X_iX_j) = 1.P(X_iX_j=1) + 0.P(X_iX_j=0)= P(X_iX_j=1)$
But at this point I'm not sure how to compute the probability. The book I'm using says the probability is $1/8$, but I can't seem to wrap my head around the reasoning. An intuitive explanation would be highly appreciated!
probability statistics random-variables covariance expected-value
probability statistics random-variables covariance expected-value
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
asked Nov 26 '18 at 8:28
s0ulr3aper07s0ulr3aper07
1088
1088
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
$endgroup$
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
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%2f3014039%2ffinding-the-probability-of-two-random-variables-being-equal-to-1%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
$endgroup$
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
add a comment |
$begingroup$
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
$endgroup$
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
add a comment |
$begingroup$
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
$endgroup$
Assuming the die is thrown independently, $X_i$ and $X_j$ are independent if $j geq i + 3$. So in that case the convariance is $0$. So you only have to calculate $E[X_iX_{i+1}] = E[X_1X_2]$ and $E[X_iX_{i+2}]=E[X_1X_3]$ which should be easy by just looking at all the possible outcomes.
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
edited Nov 26 '18 at 9:36
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
answered Nov 26 '18 at 9:30
Tki DenebTki Deneb
32710
32710
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
add a comment |
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
$begingroup$
Thanks! The textbook I'm using did not talk about the case where j = i + 2. This lead to some confusion.
$endgroup$
– s0ulr3aper07
Nov 29 '18 at 9:30
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%2f3014039%2ffinding-the-probability-of-two-random-variables-being-equal-to-1%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
