Conditional expectation of legitimate emails












-1














While Fred is sleeping one night, $X$ legitimate emails and $Y$ spam emails are sent tohim. Suppose that X and Y are independent, with $X$ ~$ Pois(10)$ and $Y $ ~ $Pois(40)$.When he wakes up, he observes that he has $30$ new emails in his inbox. Given this information, what is the expected value of how many new legitimate emails he has?






probability conditional-probability







share|cite|improve this question






















  • this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
    – D...
    Nov 23 '18 at 10:14






  • 1




    Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
    – user587126
    Nov 23 '18 at 11:40
















-1














While Fred is sleeping one night, $X$ legitimate emails and $Y$ spam emails are sent tohim. Suppose that X and Y are independent, with $X$ ~$ Pois(10)$ and $Y $ ~ $Pois(40)$.When he wakes up, he observes that he has $30$ new emails in his inbox. Given this information, what is the expected value of how many new legitimate emails he has?






probability conditional-probability







share|cite|improve this question






















  • this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
    – D...
    Nov 23 '18 at 10:14






  • 1




    Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
    – user587126
    Nov 23 '18 at 11:40














-1












-1








-1







While Fred is sleeping one night, $X$ legitimate emails and $Y$ spam emails are sent tohim. Suppose that X and Y are independent, with $X$ ~$ Pois(10)$ and $Y $ ~ $Pois(40)$.When he wakes up, he observes that he has $30$ new emails in his inbox. Given this information, what is the expected value of how many new legitimate emails he has?






probability conditional-probability







share|cite|improve this question













While Fred is sleeping one night, $X$ legitimate emails and $Y$ spam emails are sent tohim. Suppose that X and Y are independent, with $X$ ~$ Pois(10)$ and $Y $ ~ $Pois(40)$.When he wakes up, he observes that he has $30$ new emails in his inbox. Given this information, what is the expected value of how many new legitimate emails he has?






probability conditional-probability




probability conditional-probability






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 23 '18 at 7:43









user587126user587126

155




155












  • this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
    – D...
    Nov 23 '18 at 10:14






  • 1




    Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
    – user587126
    Nov 23 '18 at 11:40


















  • this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
    – D...
    Nov 23 '18 at 10:14






  • 1




    Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
    – user587126
    Nov 23 '18 at 11:40
















this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
– D...
Nov 23 '18 at 10:14




this seems a question of some homework or assignment. you should show what you have tried and that you have made an effort to solve the question.
– D...
Nov 23 '18 at 10:14




1




1




Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
– user587126
Nov 23 '18 at 11:40




Well , X+Y follows pois(50)...so we have to find E(X|X+Y=30). I tried to evaluate the sum and somehow i got an absurd answer.
– user587126
Nov 23 '18 at 11:40










1 Answer
1






active

oldest

votes


















0














Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.



Due to independence, $p(X,Y)=p(X)p(Y)$.



As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.



We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:



$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.



The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.



Finally, your result is given by:



$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.



Now, it's just a matter of computing $k$, $f(X)$ and the sum above.






share|cite|improve this answer

















  • 1




    Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
    – user587126
    Nov 23 '18 at 13:13










  • That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
    – D...
    Nov 23 '18 at 14:01






  • 1




    Okay...i will keep that in mind from now on.
    – user587126
    Nov 23 '18 at 15:59











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%2f3010086%2fconditional-expectation-of-legitimate-emails%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














Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.



Due to independence, $p(X,Y)=p(X)p(Y)$.



As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.



We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:



$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.



The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.



Finally, your result is given by:



$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.



Now, it's just a matter of computing $k$, $f(X)$ and the sum above.






share|cite|improve this answer

















  • 1




    Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
    – user587126
    Nov 23 '18 at 13:13










  • That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
    – D...
    Nov 23 '18 at 14:01






  • 1




    Okay...i will keep that in mind from now on.
    – user587126
    Nov 23 '18 at 15:59
















0














Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.



Due to independence, $p(X,Y)=p(X)p(Y)$.



As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.



We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:



$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.



The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.



Finally, your result is given by:



$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.



Now, it's just a matter of computing $k$, $f(X)$ and the sum above.






share|cite|improve this answer

















  • 1




    Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
    – user587126
    Nov 23 '18 at 13:13










  • That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
    – D...
    Nov 23 '18 at 14:01






  • 1




    Okay...i will keep that in mind from now on.
    – user587126
    Nov 23 '18 at 15:59














0












0








0






Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.



Due to independence, $p(X,Y)=p(X)p(Y)$.



As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.



We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:



$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.



The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.



Finally, your result is given by:



$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.



Now, it's just a matter of computing $k$, $f(X)$ and the sum above.






share|cite|improve this answer












Let $p(X,Y)$ be the joint distribution of $X$ and $Y$ and $p(X)=text{Pois}(X|lambda=10)$ and $p(Y)=text{Pois}(Y|lambda=40)$ the marginals for $X$ and $Y$ respectively.



Due to independence, $p(X,Y)=p(X)p(Y)$.



As you said, the random variable $X+Y$ follows a distribution $p(X+Y)=text{Pois}(X+Y|lambda=50)$.



We also need the conditional distribution $p(X|X+Y=30)$, which, by definition, is given by:



$p(X|X+Y=30) = frac{p(X,X+Y=30)}{p(X+Y=30)}=frac{p(X,Y=30-X)}{p(X+Y=30)}=frac{p(X)p(Y=30-X)}{p(X+Y=30)}$.



The denominator ${p(X+Y=30)}$ is constant, let's call it $k$. The numerator is some function of $X$ which you can easily compute by taking the product of the two Poisson distributions $p(X)$ and $p(Y=30-X)$. Let's call it $f(X)$. Then, $p(X|X+Y=30) = frac{1}{k}f(X)$.



Finally, your result is given by:



$mathbb{E}_{p(X|X+Y)}[X]=frac{1}{k}sum_{x=0}^{30}xf(x)$.



Now, it's just a matter of computing $k$, $f(X)$ and the sum above.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Nov 23 '18 at 12:27









D...D...

213113




213113








  • 1




    Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
    – user587126
    Nov 23 '18 at 13:13










  • That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
    – D...
    Nov 23 '18 at 14:01






  • 1




    Okay...i will keep that in mind from now on.
    – user587126
    Nov 23 '18 at 15:59














  • 1




    Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
    – user587126
    Nov 23 '18 at 13:13










  • That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
    – D...
    Nov 23 '18 at 14:01






  • 1




    Okay...i will keep that in mind from now on.
    – user587126
    Nov 23 '18 at 15:59








1




1




Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
– user587126
Nov 23 '18 at 13:13




Thanks a lot! Actually i thought $p(X=x,Y=30-x)$=$p(Y=30-x) $ and things got messed up! ☺
– user587126
Nov 23 '18 at 13:13












That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
– D...
Nov 23 '18 at 14:01




That's ok ;) when you have a question, it is useful to show what you have tried and add the steps that you have followed (even if there's some mistake on them). This shows that you have actually made an effort to solve the question and helps people who are willing to help you. :)
– D...
Nov 23 '18 at 14:01




1




1




Okay...i will keep that in mind from now on.
– user587126
Nov 23 '18 at 15:59




Okay...i will keep that in mind from now on.
– user587126
Nov 23 '18 at 15:59


















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.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • 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.


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%2f3010086%2fconditional-expectation-of-legitimate-emails%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

How to change which sound is reproduced for terminal bell?

Image
3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
Read more

Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

Image
up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx
Read more

Can I use Tabulator js library in my java Spring + Thymeleaf project?

Image
0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
Read more