Find combination given expectation and variance
$begingroup$
I have found the expectation of X to be 4 and the Variance to be 3. For Y the expectation is 2 and variance is 2.
Is it possible to find a combination of X and Y which satisfies the expectation to be pi and variance to be sqrt(2)?
probability discrete-mathematics probability-distributions variance expected-value
edited Dec 16 '18 at 8:16
Henry
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
$endgroup$
add a comment |
$begingroup$
I have found the expectation of X to be 4 and the Variance to be 3. For Y the expectation is 2 and variance is 2.
Is it possible to find a combination of X and Y which satisfies the expectation to be pi and variance to be sqrt(2)?
probability discrete-mathematics probability-distributions variance expected-value
edited Dec 16 '18 at 8:16
Henry
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
$endgroup$
add a comment |
$begingroup$
I have found the expectation of X to be 4 and the Variance to be 3. For Y the expectation is 2 and variance is 2.
Is it possible to find a combination of X and Y which satisfies the expectation to be pi and variance to be sqrt(2)?
probability discrete-mathematics probability-distributions variance expected-value
edited Dec 16 '18 at 8:16
Henry
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
$endgroup$
I have found the expectation of X to be 4 and the Variance to be 3. For Y the expectation is 2 and variance is 2.
Is it possible to find a combination of X and Y which satisfies the expectation to be pi and variance to be sqrt(2)?
probability discrete-mathematics probability-distributions variance expected-value
probability discrete-mathematics probability-distributions variance expected-value
edited Dec 16 '18 at 8:16
Henry
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
edited Dec 16 '18 at 8:16
Henry
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
edited Dec 16 '18 at 8:16
Henry
100k480165
edited Dec 16 '18 at 8:16
Henry
100k480165
edited Dec 16 '18 at 8:16
Henry
100k480165
100k480165
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
asked Dec 16 '18 at 5:36
Umade IkloeUmade Ikloe
72
72
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
If by a combination you mean a variable like $aX+bY$ the answer is yes. Write the equations for the mean and variance of $aX+bY$ being equal to your values. You have two equations in two unknowns, $a,b$. Solve them.
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
Write down expressions for
$E[aX + bY]$ and $text{Var}(aX+bY)$ in terms of the expectation and variance of $X$ and $Y$. Then solve for $a$ and $b$.
$E[aX+bY] = aE[X] + bE[Y] = 4a+2b$ and $text{Var}(aX+bY) = a^2 text{Var}(X) + b^2 text{Var}(Y) = 3a^2 + 2b^2$. If $4a+2b=pi$ then $b=pi/2-2a$. Then $3a^2+2b^2 = 3a^2 + 2(pi/2-2a)^2$. Set this equal to $sqrt{2}$ and solve for $a$.
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
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add a comment |
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2 Answers
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2 Answers
2
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active
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$begingroup$
If by a combination you mean a variable like $aX+bY$ the answer is yes. Write the equations for the mean and variance of $aX+bY$ being equal to your values. You have two equations in two unknowns, $a,b$. Solve them.
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
If by a combination you mean a variable like $aX+bY$ the answer is yes. Write the equations for the mean and variance of $aX+bY$ being equal to your values. You have two equations in two unknowns, $a,b$. Solve them.
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
$endgroup$
add a comment |
$begingroup$
If by a combination you mean a variable like $aX+bY$ the answer is yes. Write the equations for the mean and variance of $aX+bY$ being equal to your values. You have two equations in two unknowns, $a,b$. Solve them.
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
$endgroup$
If by a combination you mean a variable like $aX+bY$ the answer is yes. Write the equations for the mean and variance of $aX+bY$ being equal to your values. You have two equations in two unknowns, $a,b$. Solve them.
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
answered Dec 16 '18 at 5:40
Ross MillikanRoss Millikan
296k23198371
296k23198371
add a comment |
add a comment |
$begingroup$
Write down expressions for
$E[aX + bY]$ and $text{Var}(aX+bY)$ in terms of the expectation and variance of $X$ and $Y$. Then solve for $a$ and $b$.
$E[aX+bY] = aE[X] + bE[Y] = 4a+2b$ and $text{Var}(aX+bY) = a^2 text{Var}(X) + b^2 text{Var}(Y) = 3a^2 + 2b^2$. If $4a+2b=pi$ then $b=pi/2-2a$. Then $3a^2+2b^2 = 3a^2 + 2(pi/2-2a)^2$. Set this equal to $sqrt{2}$ and solve for $a$.
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
$endgroup$
add a comment |
$begingroup$
Write down expressions for
$E[aX + bY]$ and $text{Var}(aX+bY)$ in terms of the expectation and variance of $X$ and $Y$. Then solve for $a$ and $b$.
$E[aX+bY] = aE[X] + bE[Y] = 4a+2b$ and $text{Var}(aX+bY) = a^2 text{Var}(X) + b^2 text{Var}(Y) = 3a^2 + 2b^2$. If $4a+2b=pi$ then $b=pi/2-2a$. Then $3a^2+2b^2 = 3a^2 + 2(pi/2-2a)^2$. Set this equal to $sqrt{2}$ and solve for $a$.
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
$endgroup$
add a comment |
$begingroup$
Write down expressions for
$E[aX + bY]$ and $text{Var}(aX+bY)$ in terms of the expectation and variance of $X$ and $Y$. Then solve for $a$ and $b$.
$E[aX+bY] = aE[X] + bE[Y] = 4a+2b$ and $text{Var}(aX+bY) = a^2 text{Var}(X) + b^2 text{Var}(Y) = 3a^2 + 2b^2$. If $4a+2b=pi$ then $b=pi/2-2a$. Then $3a^2+2b^2 = 3a^2 + 2(pi/2-2a)^2$. Set this equal to $sqrt{2}$ and solve for $a$.
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
$endgroup$
Write down expressions for
$E[aX + bY]$ and $text{Var}(aX+bY)$ in terms of the expectation and variance of $X$ and $Y$. Then solve for $a$ and $b$.
$E[aX+bY] = aE[X] + bE[Y] = 4a+2b$ and $text{Var}(aX+bY) = a^2 text{Var}(X) + b^2 text{Var}(Y) = 3a^2 + 2b^2$. If $4a+2b=pi$ then $b=pi/2-2a$. Then $3a^2+2b^2 = 3a^2 + 2(pi/2-2a)^2$. Set this equal to $sqrt{2}$ and solve for $a$.
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
answered Dec 16 '18 at 5:42
angryavianangryavian
40.8k23380
40.8k23380
add a comment |
add a comment |
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