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







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  • 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






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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






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oldest

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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











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1 Answer
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1 Answer
1






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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


















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