Why does it need to calculate E(X/ Y=1) first before E(X)? What is the purpose of doing this [closed]

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Given the conditional probabilities in the table below and the unconditional probabilities $P(Y = 1) = 0.3$ and $P(Y = 2) = 0.7$, what is the expected value of X.



$$boxedbeginarrayc:cx_i & P(X=x_imid Y=1) & P(X=x_imid Y=2) \hline 0 & 0.2 & 0.1\hdashline 5 & 0.4 & 0.8 \hdashline 10 & 0.4 & 0.1endarray$$




probability




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closed as off-topic by saulspatz, Arnaud D., amWhy, Isaac Browne, Xander Henderson Aug 1 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud D., amWhy, Isaac Browne, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What table are you referring to?
    – David G. Stork
    Jul 31 at 1:17










  • XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
    – jAX WJ
    Jul 31 at 1:18







  • 3




    Huh? What kind of a "table" is that?!
    – David G. Stork
    Jul 31 at 1:20










  • @DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
    – Graham Kemp
    Jul 31 at 1:54














up vote
-6
down vote

favorite












Given the conditional probabilities in the table below and the unconditional probabilities $P(Y = 1) = 0.3$ and $P(Y = 2) = 0.7$, what is the expected value of X.



$$boxedbeginarrayc:cx_i & P(X=x_imid Y=1) & P(X=x_imid Y=2) \hline 0 & 0.2 & 0.1\hdashline 5 & 0.4 & 0.8 \hdashline 10 & 0.4 & 0.1endarray$$




probability




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closed as off-topic by saulspatz, Arnaud D., amWhy, Isaac Browne, Xander Henderson Aug 1 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud D., amWhy, Isaac Browne, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.








  • 1




    What table are you referring to?
    – David G. Stork
    Jul 31 at 1:17










  • XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
    – jAX WJ
    Jul 31 at 1:18







  • 3




    Huh? What kind of a "table" is that?!
    – David G. Stork
    Jul 31 at 1:20










  • @DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
    – Graham Kemp
    Jul 31 at 1:54












up vote
-6
down vote

favorite









up vote
-6
down vote

favorite











Given the conditional probabilities in the table below and the unconditional probabilities $P(Y = 1) = 0.3$ and $P(Y = 2) = 0.7$, what is the expected value of X.



$$boxedbeginarrayc:cx_i & P(X=x_imid Y=1) & P(X=x_imid Y=2) \hline 0 & 0.2 & 0.1\hdashline 5 & 0.4 & 0.8 \hdashline 10 & 0.4 & 0.1endarray$$




probability




share|cite|improve this question













Given the conditional probabilities in the table below and the unconditional probabilities $P(Y = 1) = 0.3$ and $P(Y = 2) = 0.7$, what is the expected value of X.



$$boxedbeginarrayc:cx_i & P(X=x_imid Y=1) & P(X=x_imid Y=2) \hline 0 & 0.2 & 0.1\hdashline 5 & 0.4 & 0.8 \hdashline 10 & 0.4 & 0.1endarray$$





probability





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edited Jul 31 at 1:53









Graham Kemp

80k43275




80k43275









asked Jul 31 at 1:15









jAX WJ

1




1




closed as off-topic by saulspatz, Arnaud D., amWhy, Isaac Browne, Xander Henderson Aug 1 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud D., amWhy, Isaac Browne, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by saulspatz, Arnaud D., amWhy, Isaac Browne, Xander Henderson Aug 1 at 2:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Arnaud D., amWhy, Isaac Browne, Xander Henderson
If this question can be reworded to fit the rules in the help center, please edit the question.







  • 1




    What table are you referring to?
    – David G. Stork
    Jul 31 at 1:17










  • XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
    – jAX WJ
    Jul 31 at 1:18







  • 3




    Huh? What kind of a "table" is that?!
    – David G. Stork
    Jul 31 at 1:20










  • @DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
    – Graham Kemp
    Jul 31 at 1:54












  • 1




    What table are you referring to?
    – David G. Stork
    Jul 31 at 1:17










  • XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
    – jAX WJ
    Jul 31 at 1:18







  • 3




    Huh? What kind of a "table" is that?!
    – David G. Stork
    Jul 31 at 1:20










  • @DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
    – Graham Kemp
    Jul 31 at 1:54







1




1




What table are you referring to?
– David G. Stork
Jul 31 at 1:17




What table are you referring to?
– David G. Stork
Jul 31 at 1:17












XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
– jAX WJ
Jul 31 at 1:18





XI P(X| Y=1) P(X|Y=2) 0 0.2 0.1 5 0.4 0.8 10 0.4 0.1
– jAX WJ
Jul 31 at 1:18





3




3




Huh? What kind of a "table" is that?!
– David G. Stork
Jul 31 at 1:20




Huh? What kind of a "table" is that?!
– David G. Stork
Jul 31 at 1:20












@DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
– Graham Kemp
Jul 31 at 1:54




@DavidG.Stork One stripped of formatting by the look. Someone with the moniker jAX WJ really should take the effort to learn MathJax.
– Graham Kemp
Jul 31 at 1:54










1 Answer
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By the Law of Total Expectation (LoTE): $mathsf E(X) =mathsf E(mathsf E(Xmid Y))$



Then by the definition of Expectation for Discrete Random Variables, you have:$$mathsf E(X)=mathsf E(Xmid Y=1)~mathsf P(Y=1)+mathsf E(Xmid Y=2)~mathsf P(Y=2)$$



There are other ways to find $mathsf E(X)$, but since you know those two probabilities, and can easily find those two conditional expectations from the table, you may as well use the easiest method available.






share|cite|improve this answer




























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    1
    down vote













    By the Law of Total Expectation (LoTE): $mathsf E(X) =mathsf E(mathsf E(Xmid Y))$



    Then by the definition of Expectation for Discrete Random Variables, you have:$$mathsf E(X)=mathsf E(Xmid Y=1)~mathsf P(Y=1)+mathsf E(Xmid Y=2)~mathsf P(Y=2)$$



    There are other ways to find $mathsf E(X)$, but since you know those two probabilities, and can easily find those two conditional expectations from the table, you may as well use the easiest method available.






    share|cite|improve this answer

























      up vote
      1
      down vote













      By the Law of Total Expectation (LoTE): $mathsf E(X) =mathsf E(mathsf E(Xmid Y))$



      Then by the definition of Expectation for Discrete Random Variables, you have:$$mathsf E(X)=mathsf E(Xmid Y=1)~mathsf P(Y=1)+mathsf E(Xmid Y=2)~mathsf P(Y=2)$$



      There are other ways to find $mathsf E(X)$, but since you know those two probabilities, and can easily find those two conditional expectations from the table, you may as well use the easiest method available.






      share|cite|improve this answer























        up vote
        1
        down vote










        up vote
        1
        down vote









        By the Law of Total Expectation (LoTE): $mathsf E(X) =mathsf E(mathsf E(Xmid Y))$



        Then by the definition of Expectation for Discrete Random Variables, you have:$$mathsf E(X)=mathsf E(Xmid Y=1)~mathsf P(Y=1)+mathsf E(Xmid Y=2)~mathsf P(Y=2)$$



        There are other ways to find $mathsf E(X)$, but since you know those two probabilities, and can easily find those two conditional expectations from the table, you may as well use the easiest method available.






        share|cite|improve this answer













        By the Law of Total Expectation (LoTE): $mathsf E(X) =mathsf E(mathsf E(Xmid Y))$



        Then by the definition of Expectation for Discrete Random Variables, you have:$$mathsf E(X)=mathsf E(Xmid Y=1)~mathsf P(Y=1)+mathsf E(Xmid Y=2)~mathsf P(Y=2)$$



        There are other ways to find $mathsf E(X)$, but since you know those two probabilities, and can easily find those two conditional expectations from the table, you may as well use the easiest method available.







        share|cite|improve this answer













        share|cite|improve this answer



        share|cite|improve this answer











        answered Jul 31 at 1:57









        Graham Kemp

        80k43275




        80k43275












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