show that E(Y ) = rE(X_1)












0












$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10


















0












$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10
















0












0








0





$begingroup$


For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes







share|cite|improve this question









$endgroup$




For $Y= X_1 + X_2 + ... + X_r$, with the X's being independent and identically distributed random variables, I'm asked to prove that $E(Y) = rE(X_1)$



All I could do is show $(M_X(t))^r = (E( e^{tx}))^r = E(e^{tx_1}) E(e^{tx_2}) ... E(e^{tx_r}) = E( e^{tx_1} e^{tx_2} ... e^{tx_r}) = E(e^{ty}) = M_Y(t) $



Can anyone help/guide me? Thanks






probability statistics stochastic-processes




probability statistics stochastic-processes






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 16 at 0:16









bluemusebluemuse

976




976












  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10




















  • $begingroup$
    This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
    $endgroup$
    – Makina
    Jan 16 at 0:23












  • $begingroup$
    @makina $r$ isn't random.
    $endgroup$
    – user608030
    Jan 16 at 0:27










  • $begingroup$
    hence the "similar"
    $endgroup$
    – Makina
    Jan 16 at 0:41










  • $begingroup$
    @bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
    $endgroup$
    – Kavi Rama Murthy
    Jan 16 at 6:10


















$begingroup$
This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
$endgroup$
– Makina
Jan 16 at 0:23






$begingroup$
This is something similar to Wald's equation if I'm not wrong. Wiki has the proof
$endgroup$
– Makina
Jan 16 at 0:23














$begingroup$
@makina $r$ isn't random.
$endgroup$
– user608030
Jan 16 at 0:27




$begingroup$
@makina $r$ isn't random.
$endgroup$
– user608030
Jan 16 at 0:27












$begingroup$
hence the "similar"
$endgroup$
– Makina
Jan 16 at 0:41




$begingroup$
hence the "similar"
$endgroup$
– Makina
Jan 16 at 0:41












$begingroup$
@bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 6:10






$begingroup$
@bluemuse Before studying Moment Generating Functions you must have learned that expectation is linear. This question hardly requires a proof.
$endgroup$
– Kavi Rama Murthy
Jan 16 at 6:10












1 Answer
1






active

oldest

votes


















0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09












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
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075154%2fshow-that-ey-rex-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









0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09
















0












$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09














0












0








0





$begingroup$

If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$






share|cite|improve this answer









$endgroup$



If you are going to use moment generating functions (not all distributions have them, but assuming $X$ does) then you can say



$$M'_Y(t) = frac{d}{dt} (M_X(t))^r = r (M_X(t))^{r-1} M'_X(t)$$



and since $M_X(0)=1$ and $M'_X(0)= E[X]$ you can say $$E[Y] = M'_Y(0) = r M'_X(0)=rE[X]$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 16 at 0:47









HenryHenry

101k482169




101k482169












  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09


















  • $begingroup$
    Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
    $endgroup$
    – bluemuse
    Jan 16 at 0:52












  • $begingroup$
    they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
    $endgroup$
    – k.dkhk
    Jan 18 at 13:08










  • $begingroup$
    @k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
    $endgroup$
    – Henry
    Jan 18 at 16:09
















$begingroup$
Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
$endgroup$
– bluemuse
Jan 16 at 0:52






$begingroup$
Thank you, we got this homework after studying moment generating functions so I figured I had to use them. I forgot about $ M'_X(0) = E[X]$ ! Thanks
$endgroup$
– bluemuse
Jan 16 at 0:52














$begingroup$
they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
$endgroup$
– k.dkhk
Jan 18 at 13:08




$begingroup$
they are iid. Isn't sufficient to say that because they are iid then $E[X_i]=E[X_1]$ hence E[Y]=E[X_1]+E[X_2]+...+E[X_r] = E[X_1]+E[X_1]+...+E[X_r]=rE[X_1]?
$endgroup$
– k.dkhk
Jan 18 at 13:08












$begingroup$
@k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
$endgroup$
– Henry
Jan 18 at 16:09




$begingroup$
@k.dkhk - what you say is true in general, independent or not. Can you prove it? With independent random variables a simple alternative approach comes from characteristic functions or (if they exist) moment generating functions.
$endgroup$
– Henry
Jan 18 at 16:09


















draft saved

draft discarded




















































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.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3075154%2fshow-that-ey-rex-1%23new-answer', 'question_page');
}
);

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







-

Popular posts from this blog

Human spaceflight

Image
Human spaceflight From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Space traveler" redirects here. For other uses, see Space traveler (disambiguation). @media all and (max-width:720px){.mw-parser-output .tmulti>.thumbinner{width:100%!important;max-width:none!important}.mw-parser-output .tmulti .tsingle{float:none!important;max-width:none!important;width:100%!important;text-align:center}} Apollo 11 crewmember Buzz Aldrin on the Moon, 1969 International Space Station crewmember Tracy Caldwell Dyson views the Earth, 2010 Space Shuttle Discovery heads into space with a crew aboard, STS-121 in 2006 Inside a space suit on the Canadarm, 1993 Human spaceflight (also referred to as crewed spaceflight or manned spaceflight ) is space travel with a crew or passengers aboard the spacecraft. Spacecraft carrying people may be operated directly, by human crew, or it may be e...
Read more

Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

Image
0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
Read more

張江高科駅

Image
張江高科駅 ちょうこうこうか Zhangjiang Hi-Tech Park 所在地 中国上海市浦東新区 所属事業者 上海地鉄運営有限公司（上海軌道交通・駅詳細） 上海浦東現代有軌交通有限公司（駅詳細） テンプレートを表示 張江高科駅 各種表記 繁体字： 張江高科站 簡体字： 张江高科站 拼音 ： Zhàngjiāng Gāokē Zhàn 発音： ヂャンジャン ガオクー ヂャン 英文： Zhangjiang Hi-Tech Park Station テンプレートを表示 張江高科駅 （ちょうこうこうか-えき）は中華人民共和国上海市浦東新区に位置する、上海軌道交通2号線の駅である。 なお、張江高科駅前にある張江軌道電車の 張江地鉄駅駅 についてもここで述べる。 目次 1 利用可能な鉄道路線 2 沿革 3 駅構造 3.1 上海軌道交通 3.2 張江軌道電車 4 駅周辺 5 隣の駅 6 注釈 7 関連項目 利用可能な鉄道路線 上海軌道交通 ■ 2号線 上海浦東現代有軌交通有限公司 ■ 張江軌道電車 沿革 高架駅時代の姿 2000年12月27日 - 2号線の駅が高架駅として開業。 2010年 1月1日 - 張江軌道電車の開業に伴い、張江地鉄駅駅（停留所）を設置。 2月13日 - 駅地下化に伴い高架駅の運用停止。 2月24日 - 10日間の閉鎖を経て地下新駅の開業。 2017年2月9日 - リニューアル工事を経て再開業。乗降客の増加に対応し、ホームへのエレベーター、エスカレーターを増設、券売機エリアを拡張した [1] 。 駅構造 上海軌道交通 上海軌道交通 張江高科駅 5号口 ちょうこうこうか Zhangjiang Hi-Tech Park ◄ 竜陽路 (2.60km) (1.61km) 金科路 ► 所在地 中国上海市浦東新区松涛路 祖冲之路 （中国語版） 北緯31度12分11秒 東経121度34分50秒  /  北緯31.2...
Read more