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Math Help - Expectation Problem-need help!!!

  1. #1
    Yan
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    Expectation Problem-need help!!!

    f(x,y)=(x+y)/30 for x=0,1,2,3; y=0,1,2

    here is the chart for the probability of f(x,y), i.e., P(0,0)=0, P(1,0)=1/30, P(1,1)=2/15 ...

    0 1/30 1/10 1/5
    1/30 2/15 3/10 8/15
    1/10 3/10 3/5 1

    Find E[2X-Y]
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  2. #2
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    Quote Originally Posted by Yan View Post
    f(x,y)=(x+y)/30 for x=0,1,2,3; y=0,1,2

    here is the chart for the probability of f(x,y), i.e., P(0,0)=0, P(1,0)=1/30, P(1,1)=2/15 ...

    0 1/30 1/10 1/5
    1/30 2/15 3/10 8/15
    1/10 3/10 3/5 1

    Find E[2X-Y]
    Its just arithmetic:

    E[2X-Y]=\sum_{x=0}^3\sum_{y=0}^2 (2x+y)\ p(x,y)

    .
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  3. #3
    Yan
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    Quote Originally Posted by Constatine11 View Post
    Its just arithmetic:

    E[2X-Y]=\sum_{x=0}^3\sum_{y=0}^2 (2x+y)\ p(x,y)

    .
    I did the same way, but the answer was different from my teacher. Can you tell me what is you answer?
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  4. #4
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    Quote Originally Posted by Yan View Post
    I did the same way, but the answer was different from my teacher. Can you tell me what is you answer?
    It would be better if you showed your working so that it can be reviewed for mistakes (maybe your teacher should post his/her working too )
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  5. #5
    Yan
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    Quote Originally Posted by mr fantastic View Post
    It would be better if you showed your working so that it can be reviewed for mistakes (maybe your teacher should post his/her working too )
    E[2X-Y]=2E[X]-E[Y]

    g(x)=1/30*[(x+0)+(x+1)+(x+2)]=(x+1)/10
    h(y)=1/30*[(y+0)+(y+1)+(y+2)+(y+3)]=(2y+3)/15

    2E[X]=4
    E[Y]=19/15

    So E[2X-Y]=41/15,

    But when I use the chart, the answer is 31/10. So I don't know why, can you tell me how will you gonna do to solve this problem? Thanks
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    Quote Originally Posted by Yan View Post
    E[2X-Y]=2E[X]-E[Y]

    g(x)=1/30*[(x+0)+(x+1)+(x+2)]=(x+1)/10
    h(y)=1/30*[(y+0)+(y+1)+(y+2)+(y+3)]=(2y+3)/15

    2E[X]=4
    E[Y]=19/15

    So E[2X-Y]=41/15,

    But when I use the chart, the answer is 31/10. So I don't know why, can you tell me how will you gonna do to solve this problem? Thanks
    Many of the probabilities in your chart are wrong. The probabilities don't add up to 1. Do it again.

    eg. Pr(X = 1, Y = 1) = (1 + 1)/30 = 2/30, NOT 2/15 (how did you get this??) etc.
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  7. #7
    Yan
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    Quote Originally Posted by mr fantastic View Post
    Many of the probabilities in your chart are wrong. The probabilities don't add up to 1. Do it again.

    eg. Pr(X = 1, Y = 1) = (1 + 1)/30 = 2/30, NOT 2/15 (how did you get this??) etc.
    so did I did right, if I did it in this way?
    E[2X-Y]=2E[X]-E[Y]

    g(x)=1/30*[(x+0)+(x+1)+(x+2)]=(x+1)/10
    h(y)=1/30*[(y+0)+(y+1)+(y+2)+(y+3)]=(2y+3)/15

    2E[X]=4
    E[Y]=19/15

    So E[2X-Y]=41/15,
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  8. #8
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    Quote Originally Posted by Yan View Post
    so did I did right, if I did it in this way?
    E[2X-Y]=2E[X]-E[Y]

    g(x)=1/30*[(x+0)+(x+1)+(x+2)]=(x+1)/10
    h(y)=1/30*[(y+0)+(y+1)+(y+2)+(y+3)]=(2y+3)/15

    2E[X]=4
    E[Y]=19/15

    So E[2X-Y]=41/15,
    E(X) = (0) \cdot \left(\frac{3}{30} \right) + (1) \cdot \left(\frac{6}{30} \right) + (2) \cdot \left(\frac{9}{30} \right) + (3) \cdot \left(\frac{12}{30} \right) = 2

    E(Y) = (0) \cdot \left(\frac{6}{30} \right) + (1) \cdot \left(\frac{10}{30} \right) + (2) \cdot \left(\frac{14}{30} \right) = \frac{38}{30}.

    So it looks OK.
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    Quote Originally Posted by Yan View Post
    f(x,y)=(x+y)/30 for x=0,1,2,3; y=0,1,2

    here is the chart for the probability of f(x,y), i.e., P(0,0)=0, P(1,0)=1/30, P(1,1)=2/15 ...

    0 1/30 1/10 1/5
    1/30 2/15 3/10 8/15
    1/10 3/10 3/5 1

    Find E[2X-Y]
    Lets compute the maginal distributions, which are the last row and column of the following table:

     <br />
\begin{array}{cccc|c}<br />
0 & 1/30 & 1/10 & 1/5 & 1/3 \\<br />
1/30 & 2/15 & 3/10 & 8/15 & 1 \\<br />
1/10 & 3/10 & 3/5 & 1 & 2 \\<br />
\hline<br />
2/15 & 7/15 & 1 & 26/15 &10/3<br />
\end{array}<br />

    The sums of the last row and column should both be 1 which should be the sum of all the elements in the table if this truly represents the joint distribution of (x,y), but they do not, they are 10/3.

    So there is an error in your table.

    .
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  10. #10
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    Quote Originally Posted by Constatine11 View Post
    Lets compute the maginal distributions, which are the last row and column of the following table:

     <br />
\begin{array}{cccc|c}<br />
0 & 1/30 & 1/10 & 1/5 & 1/3 \\<br />
1/30 & 2/15 & 3/10 & 8/15 & 1 \\<br />
1/10 & 3/10 & 3/5 & 1 & 2 \\<br />
\hline<br />
2/15 & 7/15 & 1 & 26/15 &10/3<br />
\end{array}<br />

    The sums of the last row and column should both be 1 which should be the sum of all the elements in the table if this truly represents the joint distribution of (x,y), but they do not, they are 10/3.

    So there is an error in your table.

    .
    The fact that there is a 1 in the table and at least one other non-zero element is a sure give away that this is not a joint distribution.

    .
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