Why are negative constants removed from variance?
$begingroup$
Let $X_1$ and $X_2$ be random variables such that $X_i sim N(1, 1) $. Why is the constant removed in the case of the variance
$$
mathrm{V}(X_1 + X_2 - 2) = 1 + 1 = 2
$$
but not in the case of the expectation
$$
mathrm{E}(X_1 + X_2 - 2) = 1 + 1 - 2 = 0 ;?
$$
random-variables statistical-inference variance expected-value
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
$endgroup$
add a comment |
$begingroup$
Let $X_1$ and $X_2$ be random variables such that $X_i sim N(1, 1) $. Why is the constant removed in the case of the variance
$$
mathrm{V}(X_1 + X_2 - 2) = 1 + 1 = 2
$$
but not in the case of the expectation
$$
mathrm{E}(X_1 + X_2 - 2) = 1 + 1 - 2 = 0 ;?
$$
random-variables statistical-inference variance expected-value
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
$endgroup$
add a comment |
$begingroup$
Let $X_1$ and $X_2$ be random variables such that $X_i sim N(1, 1) $. Why is the constant removed in the case of the variance
$$
mathrm{V}(X_1 + X_2 - 2) = 1 + 1 = 2
$$
but not in the case of the expectation
$$
mathrm{E}(X_1 + X_2 - 2) = 1 + 1 - 2 = 0 ;?
$$
random-variables statistical-inference variance expected-value
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
$endgroup$
Let $X_1$ and $X_2$ be random variables such that $X_i sim N(1, 1) $. Why is the constant removed in the case of the variance
$$
mathrm{V}(X_1 + X_2 - 2) = 1 + 1 = 2
$$
but not in the case of the expectation
$$
mathrm{E}(X_1 + X_2 - 2) = 1 + 1 - 2 = 0 ;?
$$
random-variables statistical-inference variance expected-value
random-variables statistical-inference variance expected-value
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
edited Dec 13 '18 at 9:20
Björn Friedrich
2,67961831
2,67961831
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
asked Feb 20 '18 at 20:10
Ian SpitzIan Spitz
51
51
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.
answered Feb 20 '18 at 20:17
JBLJBL
463210
$endgroup$
add a comment |
$begingroup$
The variance of a constant is equal to zero.
Assuming $X_1$ and $X_2$ are independent:
$$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
$endgroup$
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%2f2659199%2fwhy-are-negative-constants-removed-from-variance%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
$begingroup$
In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.
answered Feb 20 '18 at 20:17
JBLJBL
463210
$endgroup$
add a comment |
$begingroup$
In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.
answered Feb 20 '18 at 20:17
JBLJBL
463210
$endgroup$
add a comment |
$begingroup$
In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.
answered Feb 20 '18 at 20:17
JBLJBL
463210
$endgroup$
In very non math/statistical terms, variance measures how spread out the data is. Therefore, shifting the data by a constant term does not change how spread out the data is. However, the shift will change the the expected value of the data, because the expected value is the weighted average of the data values.
answered Feb 20 '18 at 20:17
JBLJBL
463210
answered Feb 20 '18 at 20:17
JBLJBL
463210
answered Feb 20 '18 at 20:17
JBLJBL
463210
answered Feb 20 '18 at 20:17
JBLJBL
463210
463210
add a comment |
add a comment |
$begingroup$
The variance of a constant is equal to zero.
Assuming $X_1$ and $X_2$ are independent:
$$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
$endgroup$
add a comment |
$begingroup$
The variance of a constant is equal to zero.
Assuming $X_1$ and $X_2$ are independent:
$$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
$endgroup$
add a comment |
$begingroup$
The variance of a constant is equal to zero.
Assuming $X_1$ and $X_2$ are independent:
$$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
$endgroup$
The variance of a constant is equal to zero.
Assuming $X_1$ and $X_2$ are independent:
$$V(X_1 + X_2 - 2) = V(X_1) + V(X_2) + V(2) = 1+1+0 = 2 $$
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
answered Feb 20 '18 at 20:13
user144410user144410
1,0432719
1,0432719
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%2f2659199%2fwhy-are-negative-constants-removed-from-variance%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
