Mixing
Finding E(YX) from a table
$begingroup$
Taking this table as an example of how would one be able to find E(XY)
I know that if X and Y are independent then:
$$E(XY) = E(X)*E(Y)$$
What method should be used if one doesn't know if X and Y are independent?
Thank you very much for your response.
(In this case, the answer is 12)
probability
asked Jan 22 at 16:04
FozoroFozoro
1265
$endgroup$
add a comment |
$begingroup$
Taking this table as an example of how would one be able to find E(XY)
I know that if X and Y are independent then:
$$E(XY) = E(X)*E(Y)$$
What method should be used if one doesn't know if X and Y are independent?
Thank you very much for your response.
(In this case, the answer is 12)
probability
asked Jan 22 at 16:04
FozoroFozoro
1265
$endgroup$
add a comment |
$begingroup$
Taking this table as an example of how would one be able to find E(XY)
I know that if X and Y are independent then:
$$E(XY) = E(X)*E(Y)$$
What method should be used if one doesn't know if X and Y are independent?
Thank you very much for your response.
(In this case, the answer is 12)
probability
asked Jan 22 at 16:04
FozoroFozoro
1265
$endgroup$
Taking this table as an example of how would one be able to find E(XY)
I know that if X and Y are independent then:
$$E(XY) = E(X)*E(Y)$$
What method should be used if one doesn't know if X and Y are independent?
Thank you very much for your response.
(In this case, the answer is 12)
probability
probability
asked Jan 22 at 16:04
FozoroFozoro
1265
asked Jan 22 at 16:04
FozoroFozoro
1265
edited Jan 23 at 22:39
Fozoro
asked Jan 22 at 16:04
FozoroFozoro
1265
asked Jan 22 at 16:04
FozoroFozoro
1265
asked Jan 22 at 16:04
FozoroFozoro
1265
1265
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Since you don't know if the variables are independent, just use the definition of expectation:
$$E[XY]=sum_{x,y}xyP(X=x,Y=y)=2cdot 1 cdot frac{1}{8}+2cdot 3cdot frac{1}{24}+...=12$$
answered Jan 22 at 16:07
pwerthpwerth
3,265417
$endgroup$
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Since you don't know if the variables are independent, just use the definition of expectation:
$$E[XY]=sum_{x,y}xyP(X=x,Y=y)=2cdot 1 cdot frac{1}{8}+2cdot 3cdot frac{1}{24}+...=12$$
answered Jan 22 at 16:07
pwerthpwerth
3,265417
$endgroup$
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
add a comment |
$begingroup$
Since you don't know if the variables are independent, just use the definition of expectation:
$$E[XY]=sum_{x,y}xyP(X=x,Y=y)=2cdot 1 cdot frac{1}{8}+2cdot 3cdot frac{1}{24}+...=12$$
answered Jan 22 at 16:07
pwerthpwerth
3,265417
$endgroup$
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
add a comment |
$begingroup$
Since you don't know if the variables are independent, just use the definition of expectation:
$$E[XY]=sum_{x,y}xyP(X=x,Y=y)=2cdot 1 cdot frac{1}{8}+2cdot 3cdot frac{1}{24}+...=12$$
answered Jan 22 at 16:07
pwerthpwerth
3,265417
$endgroup$
Since you don't know if the variables are independent, just use the definition of expectation:
$$E[XY]=sum_{x,y}xyP(X=x,Y=y)=2cdot 1 cdot frac{1}{8}+2cdot 3cdot frac{1}{24}+...=12$$
answered Jan 22 at 16:07
pwerthpwerth
3,265417
answered Jan 22 at 16:07
pwerthpwerth
3,265417
answered Jan 22 at 16:07
pwerthpwerth
3,265417
answered Jan 22 at 16:07
pwerthpwerth
3,265417
3,265417
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
add a comment |
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
$begingroup$
In fact, we know the variables are dependent, since $mathsf P(X=4, Y=9)=0$.
$endgroup$
– Graham Kemp
Jan 24 at 0:29
add a comment |
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