Expected Value linearity property (simple derivation)
$begingroup$
There is a simple derivation that I am not sure about.
There is a proof in my book for:
E(bX+a) = b × E(X) + a
So, it starts with
This is based on definition of expected value:
But what confuses me is that in my case the random variable is bX(e)+a, but the probability is for P(e). I mean two ingredients of the sigma do not match. Maybe it can be written as P(bX(e)+a), but then I do not know why P(bX(e)+a) will be equal to P(e). How this should be explained intuitively?
Thanks!
probability probability-theory expected-value
asked Dec 21 '18 at 15:37
JohnJohn
1236
$endgroup$
add a comment |
$begingroup$
There is a simple derivation that I am not sure about.
There is a proof in my book for:
E(bX+a) = b × E(X) + a
So, it starts with
This is based on definition of expected value:
But what confuses me is that in my case the random variable is bX(e)+a, but the probability is for P(e). I mean two ingredients of the sigma do not match. Maybe it can be written as P(bX(e)+a), but then I do not know why P(bX(e)+a) will be equal to P(e). How this should be explained intuitively?
Thanks!
probability probability-theory expected-value
asked Dec 21 '18 at 15:37
JohnJohn
1236
$endgroup$
add a comment |
$begingroup$
There is a simple derivation that I am not sure about.
There is a proof in my book for:
E(bX+a) = b × E(X) + a
So, it starts with
This is based on definition of expected value:
But what confuses me is that in my case the random variable is bX(e)+a, but the probability is for P(e). I mean two ingredients of the sigma do not match. Maybe it can be written as P(bX(e)+a), but then I do not know why P(bX(e)+a) will be equal to P(e). How this should be explained intuitively?
Thanks!
probability probability-theory expected-value
asked Dec 21 '18 at 15:37
JohnJohn
1236
$endgroup$
There is a simple derivation that I am not sure about.
There is a proof in my book for:
E(bX+a) = b × E(X) + a
So, it starts with
This is based on definition of expected value:
But what confuses me is that in my case the random variable is bX(e)+a, but the probability is for P(e). I mean two ingredients of the sigma do not match. Maybe it can be written as P(bX(e)+a), but then I do not know why P(bX(e)+a) will be equal to P(e). How this should be explained intuitively?
Thanks!
probability probability-theory expected-value
probability probability-theory expected-value
asked Dec 21 '18 at 15:37
JohnJohn
1236
asked Dec 21 '18 at 15:37
JohnJohn
1236
asked Dec 21 '18 at 15:37
JohnJohn
1236
asked Dec 21 '18 at 15:37
JohnJohn
1236
asked Dec 21 '18 at 15:37
JohnJohn
1236
1236
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Keep in mind that your random variable $X$ depends on the value of $einOmega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $einOmega$. In particular, the expected value of any function of $X$, say $f(X)$, is
$$mathbb{E}[f(X)]=sum_{einOmega}P(e)f(e)$$
In your case, $f(X)=aX+b$
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
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%2f3048617%2fexpected-value-linearity-property-simple-derivation%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
$begingroup$
Keep in mind that your random variable $X$ depends on the value of $einOmega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $einOmega$. In particular, the expected value of any function of $X$, say $f(X)$, is
$$mathbb{E}[f(X)]=sum_{einOmega}P(e)f(e)$$
In your case, $f(X)=aX+b$
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
add a comment |
$begingroup$
Keep in mind that your random variable $X$ depends on the value of $einOmega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $einOmega$. In particular, the expected value of any function of $X$, say $f(X)$, is
$$mathbb{E}[f(X)]=sum_{einOmega}P(e)f(e)$$
In your case, $f(X)=aX+b$
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
$endgroup$
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
add a comment |
$begingroup$
Keep in mind that your random variable $X$ depends on the value of $einOmega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $einOmega$. In particular, the expected value of any function of $X$, say $f(X)$, is
$$mathbb{E}[f(X)]=sum_{einOmega}P(e)f(e)$$
In your case, $f(X)=aX+b$
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
$endgroup$
Keep in mind that your random variable $X$ depends on the value of $einOmega$. So if you want to compute the expected value of $X$, or $2X$, or $5^{x}-sin{X}$, or any function $f(X)$, you need to take a weighted average of the values of $f(X)$, but the weights are still determined by the events $einOmega$. In particular, the expected value of any function of $X$, say $f(X)$, is
$$mathbb{E}[f(X)]=sum_{einOmega}P(e)f(e)$$
In your case, $f(X)=aX+b$
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
answered Dec 21 '18 at 15:51
pwerthpwerth
3,243417
3,243417
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
add a comment |
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
$begingroup$
Ahhh.. I see, thanks a lot!
$endgroup$
– John
Dec 21 '18 at 15:56
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%2f3048617%2fexpected-value-linearity-property-simple-derivation%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
