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Math Help - Simple statistics

  1. #1
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    Simple statistics

    if E(X) = 1.6 and E(Y) = 2

    does E (Y/X) = E(Y)/E(X)

    2/1.6 ??
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  2. #2
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    Quote Originally Posted by djmccabie View Post
    if E(X) = 1.6 and E(Y) = 2

    does E (Y/X) = E(Y)/E(X)

    2/1.6 ??
    No. It might be best if you post the entire question.
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  3. #3
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    The random variables X and Y are independent and have the following probability distributions.



    x 1 2
    P(X=x) 0.4 0.6


    Y 1 2 3
    P(Y=y) 0.3 0.4 0.3


    i) Evaluate E(Y/X)
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  4. #4
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    Quote Originally Posted by djmccabie View Post
    The random variables X and Y are independent and have the following probability distributions.



    x 1 2
    P(X=x) 0.4 0.6


    Y 1 2 3
    P(Y=y) 0.3 0.4 0.3


    i) Evaluate E(Y/X)
     <br />
E(Y/X)= 1 \times 0.4 \times 0.3 + 2 \times 0.4 \times 0.4 + 3\times 0.4 \times 0.3 +

    ............... \ <br />
(1/2) \times 0.6 \times 0.3+(2/2) \times 0.6 \times 0.4+(3/2) \times 0.6 \times 0.3<br />

    CB
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  5. #5
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    i dont understand any of that
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  6. #6
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    Quote Originally Posted by djmccabie View Post
    The random variables X and Y are independent and have the following probability distributions.



    x 1 2
    P(X=x) 0.4 0.6


    Y 1 2 3
    P(Y=y) 0.3 0.4 0.3


    i) Evaluate E(Y/X)
    Quote Originally Posted by CaptainBlack View Post
     <br />
E(Y/X)= 1 \times 0.4 \times 0.3 + 2 \times 0.4 \times 0.4 + 3\times 0.4 \times 0.3 +

    ............... \ <br />
(1/2) \times 0.6 \times 0.3+(2/2) \times 0.6 \times 0.4+(3/2) \times 0.6 \times 0.3<br />

    CB
    Quote Originally Posted by djmccabie View Post
    i dont understand any of that
    That doesn't bode well.

    What are the possible values of Y/X ......? 1/1 = 1, 1/2, 2/1 = 2, 2/2 = 1 (already considered), 3/1 = 3, 3/2.

    Pr(Y/X = 1) = Pr(Y = 1 and X = 1) + Pr(Y = 2 and X = 2) = (0.3)(0.4) + (0.4)(0.6) = 0.12 + 0.24 = 0.36.

    Pr(Y/X = 1/2) = Pr(Y = 1 and X = 2) = (0.3)(0.6) = 0.18.

    etc.

    Then E(Y/X) = (1)(0.36) + (1/2)(0.18) + ......

    This is the same calculation as CaptainB did - all I've done is simplified the arithmetic.
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  7. #7
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    Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?
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  8. #8
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    Quote Originally Posted by HallsofIvy View Post
    Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?
    I'm not sure who your post is addressed is to. Nevertheless, X and Y are not events, they are random variables and have a probability of assuming particular values.
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  9. #9
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    Quote Originally Posted by djmccabie View Post
    i dont understand any of that
    It uses the definition of expectation for a discrete random variable:

    E(U)=\sum_i u_i p(u_i)

    where the u_i 's are the possible values of U, and p(u_i) are the probabilities of those values.

    Also it uses the definition of independent RV that p(a,b)=p(a)p(b).

    So in this case we have:

    E(Y/X)=\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i,y=j) =\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i)p(y=j)

    CB
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  10. #10
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    Quote Originally Posted by HallsofIvy View Post
    Do you uderstand what P(Y/X) means? Remember that Y and X are events not numbers. What is meant by Y/X?
    They are random variables, not events. We are finding expectations, which are properties of RV.

    Also no one has used P(Y/X) as a notation (which if they had might be mistaken for the conditional probability P(Y|X) but neither I nor MrF would do that -would we?)

    Also ^2, please quote what you are replying to so we know what you are talking about (and if you don't know what you are talking about don't talk about it).

    CB
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  11. #11
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    Quote Originally Posted by CaptainBlack View Post
    It uses the definition of expectation for a discrete random variable:

    E(U)=\sum_i u_i p(u_i)

    where the u_i 's are the possible values of U, and p(u_i) are the probabilities of those values.

    Also it uses the definition of independent RV that p(a,b)=p(a)p(b).

    So in this case we have:

    E(Y/X)=\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i,y=j) =\sum_{i \in \{1,2\}, j \in \{1,2,3\}} \frac{j}{i}\ p(x=i)p(y=j)

    CB

    Not 1 for simplicity are you? lol

    Its ok tho, i know how to work it out now.

    Ill ask my tutor why when college starts again
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