Expected value E(xy) of dependent bivariate distribution?
$begingroup$
**Two fair dice are thrown and
the number of times a 1 comes up is recorded as X and the number of
times a 6 comes up is Y.
The joint distribution of X and Y is shown in the following
table.
$begin{array}{ccccc}
& & & y\
& & 0 & 1 & 2\
& 0 & frac{16}{36} & frac{8}{36} & frac{1}{36}\
x & 1 & frac{8}{36} & frac{2}{36} & 0\
& 2 & frac{1}{36} & 0 & 0
end{array}$}
Calculate E(XY)
Looking at the problem, the joint distribution does not look to be
independent since f$_{X|Y}(x|y)neq$f$_{X}(x).f_{Y}(y)$ for all
values
x and y, given the fact it is not independent we cannot use the formulae
E(XY)=E(X).E(Y).
How can the expected value be correctly calculated ?
statistics
asked Jun 12 '15 at 10:09
MassinMassin
1641215
$endgroup$
add a comment |
$begingroup$
**Two fair dice are thrown and
the number of times a 1 comes up is recorded as X and the number of
times a 6 comes up is Y.
The joint distribution of X and Y is shown in the following
table.
$begin{array}{ccccc}
& & & y\
& & 0 & 1 & 2\
& 0 & frac{16}{36} & frac{8}{36} & frac{1}{36}\
x & 1 & frac{8}{36} & frac{2}{36} & 0\
& 2 & frac{1}{36} & 0 & 0
end{array}$}
Calculate E(XY)
Looking at the problem, the joint distribution does not look to be
independent since f$_{X|Y}(x|y)neq$f$_{X}(x).f_{Y}(y)$ for all
values
x and y, given the fact it is not independent we cannot use the formulae
E(XY)=E(X).E(Y).
How can the expected value be correctly calculated ?
statistics
asked Jun 12 '15 at 10:09
MassinMassin
1641215
$endgroup$
$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42
add a comment |
$begingroup$
**Two fair dice are thrown and
the number of times a 1 comes up is recorded as X and the number of
times a 6 comes up is Y.
The joint distribution of X and Y is shown in the following
table.
$begin{array}{ccccc}
& & & y\
& & 0 & 1 & 2\
& 0 & frac{16}{36} & frac{8}{36} & frac{1}{36}\
x & 1 & frac{8}{36} & frac{2}{36} & 0\
& 2 & frac{1}{36} & 0 & 0
end{array}$}
Calculate E(XY)
Looking at the problem, the joint distribution does not look to be
independent since f$_{X|Y}(x|y)neq$f$_{X}(x).f_{Y}(y)$ for all
values
x and y, given the fact it is not independent we cannot use the formulae
E(XY)=E(X).E(Y).
How can the expected value be correctly calculated ?
statistics
asked Jun 12 '15 at 10:09
MassinMassin
1641215
$endgroup$
**Two fair dice are thrown and
the number of times a 1 comes up is recorded as X and the number of
times a 6 comes up is Y.
The joint distribution of X and Y is shown in the following
table.
$begin{array}{ccccc}
& & & y\
& & 0 & 1 & 2\
& 0 & frac{16}{36} & frac{8}{36} & frac{1}{36}\
x & 1 & frac{8}{36} & frac{2}{36} & 0\
& 2 & frac{1}{36} & 0 & 0
end{array}$}
Calculate E(XY)
Looking at the problem, the joint distribution does not look to be
independent since f$_{X|Y}(x|y)neq$f$_{X}(x).f_{Y}(y)$ for all
values
x and y, given the fact it is not independent we cannot use the formulae
E(XY)=E(X).E(Y).
How can the expected value be correctly calculated ?
statistics
statistics
asked Jun 12 '15 at 10:09
MassinMassin
1641215
asked Jun 12 '15 at 10:09
MassinMassin
1641215
asked Jun 12 '15 at 10:09
MassinMassin
1641215
asked Jun 12 '15 at 10:09
MassinMassin
1641215
asked Jun 12 '15 at 10:09
MassinMassin
1641215
1641215
$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42
add a comment |
$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42
$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42
$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
We have $XY=0$, unless $X=1$ and $Y=1$. In that case, $XY=1$.
Thus $Pr(XY=1)$, from the table, is $frac{2}{36}$. And therefore $Pr(XY=0)=frac{34}{36}$. Now we know the complete distribution of $XY$, so we can find its expectation, which is $(0)left(frac{34}{36}right)+(1)left(frac{2}{36}right)$.
Remark: Computing $Pr(XY=0)$ was unnecessary, since it would later be multiplied by $0$. We did it to fit this computation into the standard expectation calculation framework.
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
$endgroup$
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
$begingroup$
We have $XY=0$, unless $X=1$ and $Y=1$. In that case, $XY=1$.
Thus $Pr(XY=1)$, from the table, is $frac{2}{36}$. And therefore $Pr(XY=0)=frac{34}{36}$. Now we know the complete distribution of $XY$, so we can find its expectation, which is $(0)left(frac{34}{36}right)+(1)left(frac{2}{36}right)$.
Remark: Computing $Pr(XY=0)$ was unnecessary, since it would later be multiplied by $0$. We did it to fit this computation into the standard expectation calculation framework.
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
$endgroup$
add a comment |
$begingroup$
We have $XY=0$, unless $X=1$ and $Y=1$. In that case, $XY=1$.
Thus $Pr(XY=1)$, from the table, is $frac{2}{36}$. And therefore $Pr(XY=0)=frac{34}{36}$. Now we know the complete distribution of $XY$, so we can find its expectation, which is $(0)left(frac{34}{36}right)+(1)left(frac{2}{36}right)$.
Remark: Computing $Pr(XY=0)$ was unnecessary, since it would later be multiplied by $0$. We did it to fit this computation into the standard expectation calculation framework.
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
$endgroup$
add a comment |
$begingroup$
We have $XY=0$, unless $X=1$ and $Y=1$. In that case, $XY=1$.
Thus $Pr(XY=1)$, from the table, is $frac{2}{36}$. And therefore $Pr(XY=0)=frac{34}{36}$. Now we know the complete distribution of $XY$, so we can find its expectation, which is $(0)left(frac{34}{36}right)+(1)left(frac{2}{36}right)$.
Remark: Computing $Pr(XY=0)$ was unnecessary, since it would later be multiplied by $0$. We did it to fit this computation into the standard expectation calculation framework.
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
$endgroup$
We have $XY=0$, unless $X=1$ and $Y=1$. In that case, $XY=1$.
Thus $Pr(XY=1)$, from the table, is $frac{2}{36}$. And therefore $Pr(XY=0)=frac{34}{36}$. Now we know the complete distribution of $XY$, so we can find its expectation, which is $(0)left(frac{34}{36}right)+(1)left(frac{2}{36}right)$.
Remark: Computing $Pr(XY=0)$ was unnecessary, since it would later be multiplied by $0$. We did it to fit this computation into the standard expectation calculation framework.
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
answered Jun 12 '15 at 10:59
André NicolasAndré Nicolas
452k36423808
452k36423808
add a comment |
add a comment |
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$begingroup$
Where does the value (34/36) come from in your solution?
$endgroup$
– Massin
Jun 12 '15 at 10:42