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How to proof variance of the sum $x+y$ is the sum of variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$...
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Show that the variance of the sum $x + y$ of the random variable $x$ and the random variable $y$ is the sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$?
My attempt:
Is my first time doing statistics and there are alot of terms i am quite unfamiliar with.
Random variable: Value of a variable result from a random experiment: e.g. (the score when a die is rolled once)
Variance($sigma^2$) : Average of squared deviations from the mean or square of standard deviation?
I know variance of sum $x + y$ means $var(x+y) = $E(x+y)^2 -$(E(x+y))^2$
But what does sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$ means?
Edited:
My 2nd attempt:
Var(x + y ) = E$(x+y)^2$ - $(E(x+y)^2)$ = E($x^2$ + 2xy +$y^2$) - [($Ex)^2$ + 2ExEy + ($Ey)^2$ ] = [E$x^2$ - (E$x)^2$ ] + [E$y^2$ - (E$y)^2$ ] + 2[Exy - ExEy] = Var x + Var y + 2cov (x,y)
Now i assume(not sure is it correct) sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of
$y$
= var ($sigma_x^2$ ) + var($sigma_y^2$)
= E($sigma_x^2$ ) -[E($sigma_x)]^2$ + E($sigma_y^2$ ) -[E($sigma_y)]^2$
Since var(x) = $sigma_x^2$,
=E(E($x^2)$ - $[E(x)]^2$) - [$E^2$ ((E($x^2)$ - $[E(x)]^2$)) + E(E($y^2)$ - $[E(y)]^2$) - [$E^2$ ((E($y^2)$
I did not further solve it because i realise is wrong as the above does not have an expression that has an xy expression. So where have i done wrong?
statistics variance standard-deviation
edited Jan 31 at 14:01
Peter Taylor
9,15712343
asked Jan 30 at 13:34
johnjohn
749
$endgroup$
add a comment |
$begingroup$
Show that the variance of the sum $x + y$ of the random variable $x$ and the random variable $y$ is the sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$?
My attempt:
Is my first time doing statistics and there are alot of terms i am quite unfamiliar with.
Random variable: Value of a variable result from a random experiment: e.g. (the score when a die is rolled once)
Variance($sigma^2$) : Average of squared deviations from the mean or square of standard deviation?
I know variance of sum $x + y$ means $var(x+y) = $E(x+y)^2 -$(E(x+y))^2$
But what does sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$ means?
Edited:
My 2nd attempt:
Var(x + y ) = E$(x+y)^2$ - $(E(x+y)^2)$ = E($x^2$ + 2xy +$y^2$) - [($Ex)^2$ + 2ExEy + ($Ey)^2$ ] = [E$x^2$ - (E$x)^2$ ] + [E$y^2$ - (E$y)^2$ ] + 2[Exy - ExEy] = Var x + Var y + 2cov (x,y)
Now i assume(not sure is it correct) sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of
$y$
= var ($sigma_x^2$ ) + var($sigma_y^2$)
= E($sigma_x^2$ ) -[E($sigma_x)]^2$ + E($sigma_y^2$ ) -[E($sigma_y)]^2$
Since var(x) = $sigma_x^2$,
=E(E($x^2)$ - $[E(x)]^2$) - [$E^2$ ((E($x^2)$ - $[E(x)]^2$)) + E(E($y^2)$ - $[E(y)]^2$) - [$E^2$ ((E($y^2)$
I did not further solve it because i realise is wrong as the above does not have an expression that has an xy expression. So where have i done wrong?
statistics variance standard-deviation
edited Jan 31 at 14:01
Peter Taylor
9,15712343
asked Jan 30 at 13:34
johnjohn
749
$endgroup$
$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
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@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16
add a comment |
$begingroup$
Show that the variance of the sum $x + y$ of the random variable $x$ and the random variable $y$ is the sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$?
My attempt:
Is my first time doing statistics and there are alot of terms i am quite unfamiliar with.
Random variable: Value of a variable result from a random experiment: e.g. (the score when a die is rolled once)
Variance($sigma^2$) : Average of squared deviations from the mean or square of standard deviation?
I know variance of sum $x + y$ means $var(x+y) = $E(x+y)^2 -$(E(x+y))^2$
But what does sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$ means?
Edited:
My 2nd attempt:
Var(x + y ) = E$(x+y)^2$ - $(E(x+y)^2)$ = E($x^2$ + 2xy +$y^2$) - [($Ex)^2$ + 2ExEy + ($Ey)^2$ ] = [E$x^2$ - (E$x)^2$ ] + [E$y^2$ - (E$y)^2$ ] + 2[Exy - ExEy] = Var x + Var y + 2cov (x,y)
Now i assume(not sure is it correct) sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of
$y$
= var ($sigma_x^2$ ) + var($sigma_y^2$)
= E($sigma_x^2$ ) -[E($sigma_x)]^2$ + E($sigma_y^2$ ) -[E($sigma_y)]^2$
Since var(x) = $sigma_x^2$,
=E(E($x^2)$ - $[E(x)]^2$) - [$E^2$ ((E($x^2)$ - $[E(x)]^2$)) + E(E($y^2)$ - $[E(y)]^2$) - [$E^2$ ((E($y^2)$
I did not further solve it because i realise is wrong as the above does not have an expression that has an xy expression. So where have i done wrong?
statistics variance standard-deviation
edited Jan 31 at 14:01
Peter Taylor
9,15712343
asked Jan 30 at 13:34
johnjohn
749
$endgroup$
Show that the variance of the sum $x + y$ of the random variable $x$ and the random variable $y$ is the sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$?
My attempt:
Is my first time doing statistics and there are alot of terms i am quite unfamiliar with.
Random variable: Value of a variable result from a random experiment: e.g. (the score when a die is rolled once)
Variance($sigma^2$) : Average of squared deviations from the mean or square of standard deviation?
I know variance of sum $x + y$ means $var(x+y) = $E(x+y)^2 -$(E(x+y))^2$
But what does sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of $y$ means?
Edited:
My 2nd attempt:
Var(x + y ) = E$(x+y)^2$ - $(E(x+y)^2)$ = E($x^2$ + 2xy +$y^2$) - [($Ex)^2$ + 2ExEy + ($Ey)^2$ ] = [E$x^2$ - (E$x)^2$ ] + [E$y^2$ - (E$y)^2$ ] + 2[Exy - ExEy] = Var x + Var y + 2cov (x,y)
Now i assume(not sure is it correct) sum of the variance $sigma_x^2$ of $x$ and the variance $sigma_y^2$ of
$y$
= var ($sigma_x^2$ ) + var($sigma_y^2$)
= E($sigma_x^2$ ) -[E($sigma_x)]^2$ + E($sigma_y^2$ ) -[E($sigma_y)]^2$
Since var(x) = $sigma_x^2$,
=E(E($x^2)$ - $[E(x)]^2$) - [$E^2$ ((E($x^2)$ - $[E(x)]^2$)) + E(E($y^2)$ - $[E(y)]^2$) - [$E^2$ ((E($y^2)$
I did not further solve it because i realise is wrong as the above does not have an expression that has an xy expression. So where have i done wrong?
statistics variance standard-deviation
statistics variance standard-deviation
edited Jan 31 at 14:01
Peter Taylor
9,15712343
asked Jan 30 at 13:34
johnjohn
749
edited Jan 31 at 14:01
Peter Taylor
9,15712343
asked Jan 30 at 13:34
johnjohn
749
edited Jan 31 at 14:01
Peter Taylor
9,15712343
edited Jan 31 at 14:01
Peter Taylor
9,15712343
edited Jan 31 at 14:01
Peter Taylor
9,15712343
9,15712343
asked Jan 30 at 13:34
johnjohn
749
asked Jan 30 at 13:34
johnjohn
749
asked Jan 30 at 13:34
johnjohn
749
749
$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
$begingroup$
@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16
add a comment |
$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
$begingroup$
@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16
$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
$begingroup$
@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16
add a comment |
1 Answer
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The variance of $X$ is $Eleft[big(X - E[X]big)^2right]$ so here we are looking at
$$Eleft[big((X+Y) - E[X+Y]big)^2right] \= Eleft[big((X-E[X])+(Y-E[Y])big)^2right] \ = Eleft[big(X - E[X]big)^2right] +Eleft[big(Y - E[Y]big)^2right] +2Eleft[big(X - E[X]big)big(Y - E[Y]big)right] $$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $0$, which will happen when $X$ and $Y$ are independent or in a few other special cases
answered Jan 30 at 16:36
HenryHenry
101k482170
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add a comment |
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1 Answer
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$begingroup$
The variance of $X$ is $Eleft[big(X - E[X]big)^2right]$ so here we are looking at
$$Eleft[big((X+Y) - E[X+Y]big)^2right] \= Eleft[big((X-E[X])+(Y-E[Y])big)^2right] \ = Eleft[big(X - E[X]big)^2right] +Eleft[big(Y - E[Y]big)^2right] +2Eleft[big(X - E[X]big)big(Y - E[Y]big)right] $$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $0$, which will happen when $X$ and $Y$ are independent or in a few other special cases
answered Jan 30 at 16:36
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
The variance of $X$ is $Eleft[big(X - E[X]big)^2right]$ so here we are looking at
$$Eleft[big((X+Y) - E[X+Y]big)^2right] \= Eleft[big((X-E[X])+(Y-E[Y])big)^2right] \ = Eleft[big(X - E[X]big)^2right] +Eleft[big(Y - E[Y]big)^2right] +2Eleft[big(X - E[X]big)big(Y - E[Y]big)right] $$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $0$, which will happen when $X$ and $Y$ are independent or in a few other special cases
answered Jan 30 at 16:36
HenryHenry
101k482170
$endgroup$
add a comment |
$begingroup$
The variance of $X$ is $Eleft[big(X - E[X]big)^2right]$ so here we are looking at
$$Eleft[big((X+Y) - E[X+Y]big)^2right] \= Eleft[big((X-E[X])+(Y-E[Y])big)^2right] \ = Eleft[big(X - E[X]big)^2right] +Eleft[big(Y - E[Y]big)^2right] +2Eleft[big(X - E[X]big)big(Y - E[Y]big)right] $$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $0$, which will happen when $X$ and $Y$ are independent or in a few other special cases
answered Jan 30 at 16:36
HenryHenry
101k482170
$endgroup$
The variance of $X$ is $Eleft[big(X - E[X]big)^2right]$ so here we are looking at
$$Eleft[big((X+Y) - E[X+Y]big)^2right] \= Eleft[big((X-E[X])+(Y-E[Y])big)^2right] \ = Eleft[big(X - E[X]big)^2right] +Eleft[big(Y - E[Y]big)^2right] +2Eleft[big(X - E[X]big)big(Y - E[Y]big)right] $$
You actually want the result to be the two left terms. The right-hand term cancels out if and only if the covariance is $0$, which will happen when $X$ and $Y$ are independent or in a few other special cases
answered Jan 30 at 16:36
HenryHenry
101k482170
answered Jan 30 at 16:36
HenryHenry
101k482170
answered Jan 30 at 16:36
HenryHenry
101k482170
answered Jan 30 at 16:36
HenryHenry
101k482170
101k482170
add a comment |
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$begingroup$
It means $E((X+Y)^2-(E(X+Y))^2$ instead.
$endgroup$
– Michael Hoppe
Jan 30 at 13:49
$begingroup$
@MichaelHoppe sorry, just now i made a mistake about variance of sum x + y. Anyways what you said i believe is referring to var(x+y) ? How about the case for the sum of variance σx^2 of x and the variance σy^2 of y? What is the expression?
$endgroup$
– john
Jan 30 at 13:56
$begingroup$
Possible duplicate of Finding the mean and variance of a linear combination of independent random variables
$endgroup$
– Did
Jan 30 at 17:16