Finding the covariance of rv X and Y
Roll a die twice and call $X$ the sum of the outcomes and $Y$ the
difference of the first outcome minus the second one. Compute Cov(X,Y)
Attempt
Let $(i,j)$ denote the first and second outcome. We have $X=i+j$ and $Y=i-j$. WE need to compute $E(XY)-E(X)E(Y)$.
I see the answer will be alot of arithmetic since there is a nice closed for say $p_X(x)$, the pmf of $X$...
Unless I regard $i$ and $j$ as rv and they are uniform discrete on ${1,2,3,4,5,6}$ and so
$$ P(i+j=k) = sum p_i(k)p_j(6-k) = sum_{k=1}^6 frac{1}{(k+1)(7-k)}$$
And from here we compute the E(X) and same for $Y$. But how about $E(XY)$.
Is my approach correct? Or is there a simpler way to solve this problem?
probability
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
add a comment |
Roll a die twice and call $X$ the sum of the outcomes and $Y$ the
difference of the first outcome minus the second one. Compute Cov(X,Y)
Attempt
Let $(i,j)$ denote the first and second outcome. We have $X=i+j$ and $Y=i-j$. WE need to compute $E(XY)-E(X)E(Y)$.
I see the answer will be alot of arithmetic since there is a nice closed for say $p_X(x)$, the pmf of $X$...
Unless I regard $i$ and $j$ as rv and they are uniform discrete on ${1,2,3,4,5,6}$ and so
$$ P(i+j=k) = sum p_i(k)p_j(6-k) = sum_{k=1}^6 frac{1}{(k+1)(7-k)}$$
And from here we compute the E(X) and same for $Y$. But how about $E(XY)$.
Is my approach correct? Or is there a simpler way to solve this problem?
probability
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
add a comment |
Roll a die twice and call $X$ the sum of the outcomes and $Y$ the
difference of the first outcome minus the second one. Compute Cov(X,Y)
Attempt
Let $(i,j)$ denote the first and second outcome. We have $X=i+j$ and $Y=i-j$. WE need to compute $E(XY)-E(X)E(Y)$.
I see the answer will be alot of arithmetic since there is a nice closed for say $p_X(x)$, the pmf of $X$...
Unless I regard $i$ and $j$ as rv and they are uniform discrete on ${1,2,3,4,5,6}$ and so
$$ P(i+j=k) = sum p_i(k)p_j(6-k) = sum_{k=1}^6 frac{1}{(k+1)(7-k)}$$
And from here we compute the E(X) and same for $Y$. But how about $E(XY)$.
Is my approach correct? Or is there a simpler way to solve this problem?
probability
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
Roll a die twice and call $X$ the sum of the outcomes and $Y$ the
difference of the first outcome minus the second one. Compute Cov(X,Y)
Attempt
Let $(i,j)$ denote the first and second outcome. We have $X=i+j$ and $Y=i-j$. WE need to compute $E(XY)-E(X)E(Y)$.
I see the answer will be alot of arithmetic since there is a nice closed for say $p_X(x)$, the pmf of $X$...
Unless I regard $i$ and $j$ as rv and they are uniform discrete on ${1,2,3,4,5,6}$ and so
$$ P(i+j=k) = sum p_i(k)p_j(6-k) = sum_{k=1}^6 frac{1}{(k+1)(7-k)}$$
And from here we compute the E(X) and same for $Y$. But how about $E(XY)$.
Is my approach correct? Or is there a simpler way to solve this problem?
probability
probability
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
asked Nov 30 at 5:42
Jimmy Sabater
1,930219
1,930219
add a comment |
add a comment |
1 Answer
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Note that if we let $A$ be a random variable representing the value of the first roll and $B$ the same for the second roll, we have $X = A + B$ and $Y = A - B$. In particular, $E(Y) = E(A) - E(B)$ by linearity of expectation; since $A$ and $B$ are independent die rolls, $E(Y) = 0$. So then $E(X)E(Y) = 0$ and we only need to worry about $E(XY)$.
Rewriting in terms of $A$ and $B$, we have $E(XY) = E((A+B)(A-B)) = E(A^2 - B^2)$. By linearity of expectation, this is $E(A^2) - E(B^2)$; again, since $A$ and $B$ are identically distributed, this is $0$. Hence, the covariance is just $0$.
answered Nov 30 at 5:50
platty
3,360320
add a comment |
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1 Answer
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1 Answer
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Note that if we let $A$ be a random variable representing the value of the first roll and $B$ the same for the second roll, we have $X = A + B$ and $Y = A - B$. In particular, $E(Y) = E(A) - E(B)$ by linearity of expectation; since $A$ and $B$ are independent die rolls, $E(Y) = 0$. So then $E(X)E(Y) = 0$ and we only need to worry about $E(XY)$.
Rewriting in terms of $A$ and $B$, we have $E(XY) = E((A+B)(A-B)) = E(A^2 - B^2)$. By linearity of expectation, this is $E(A^2) - E(B^2)$; again, since $A$ and $B$ are identically distributed, this is $0$. Hence, the covariance is just $0$.
answered Nov 30 at 5:50
platty
3,360320
add a comment |
Note that if we let $A$ be a random variable representing the value of the first roll and $B$ the same for the second roll, we have $X = A + B$ and $Y = A - B$. In particular, $E(Y) = E(A) - E(B)$ by linearity of expectation; since $A$ and $B$ are independent die rolls, $E(Y) = 0$. So then $E(X)E(Y) = 0$ and we only need to worry about $E(XY)$.
Rewriting in terms of $A$ and $B$, we have $E(XY) = E((A+B)(A-B)) = E(A^2 - B^2)$. By linearity of expectation, this is $E(A^2) - E(B^2)$; again, since $A$ and $B$ are identically distributed, this is $0$. Hence, the covariance is just $0$.
answered Nov 30 at 5:50
platty
3,360320
add a comment |
Note that if we let $A$ be a random variable representing the value of the first roll and $B$ the same for the second roll, we have $X = A + B$ and $Y = A - B$. In particular, $E(Y) = E(A) - E(B)$ by linearity of expectation; since $A$ and $B$ are independent die rolls, $E(Y) = 0$. So then $E(X)E(Y) = 0$ and we only need to worry about $E(XY)$.
Rewriting in terms of $A$ and $B$, we have $E(XY) = E((A+B)(A-B)) = E(A^2 - B^2)$. By linearity of expectation, this is $E(A^2) - E(B^2)$; again, since $A$ and $B$ are identically distributed, this is $0$. Hence, the covariance is just $0$.
answered Nov 30 at 5:50
platty
3,360320
Note that if we let $A$ be a random variable representing the value of the first roll and $B$ the same for the second roll, we have $X = A + B$ and $Y = A - B$. In particular, $E(Y) = E(A) - E(B)$ by linearity of expectation; since $A$ and $B$ are independent die rolls, $E(Y) = 0$. So then $E(X)E(Y) = 0$ and we only need to worry about $E(XY)$.
Rewriting in terms of $A$ and $B$, we have $E(XY) = E((A+B)(A-B)) = E(A^2 - B^2)$. By linearity of expectation, this is $E(A^2) - E(B^2)$; again, since $A$ and $B$ are identically distributed, this is $0$. Hence, the covariance is just $0$.
answered Nov 30 at 5:50
platty
3,360320
answered Nov 30 at 5:50
platty
3,360320
answered Nov 30 at 5:50
platty
3,360320
answered Nov 30 at 5:50
platty
3,360320
3,360320
add a comment |
add a comment |
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