Thread: 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]

Originally Posted by Yan

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:

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

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Originally Posted by Constatine11

Its just arithmetic:

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

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I did the same way, but the answer was different from my teacher. Can you tell me what is you answer?

Originally Posted by Yan

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 )

Originally Posted by mr fantastic

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

Originally Posted by Yan

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.

Originally Posted by mr fantastic

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,

Originally Posted by Yan

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,

$\displaystyle E(X) = (0) \cdot \left(\frac{3}{30} \right) + (1) \cdot \left(\frac{6}{30} \right) + (2) \cdot \left(\frac{9}{30} \right) + (3) \cdot \left(\frac{12}{30} \right) = 2$

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

So it looks OK.

Originally Posted by Yan

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:

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

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 $\displaystyle 10/3$.

So there is an error in your table.

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Originally Posted by Constatine11

Lets compute the maginal distributions, which are the last row and column of the following table:

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

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 $\displaystyle 10/3$.

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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