Looking for a proof of : variance of sum is the sum of variances.
$begingroup$
For independent random variables X and Y, the variance of their sum or
difference is the sum of their variances:
I can see why above should be true : if $x_1<X<x_1$
and $y_1 <Y < y_2$, then clearly $x_1+y_1 <X+Y<x_2+y_2$. But proving this seems a bit hard. Here is my attempt :
$mathbb {var(X) = sum[x_i - mean(X)]^2p_i}$
$mathbb {var(Y) = sum[y_i - mean(Y)]^2p_i}$,
then I guess the variance of sum should be :
$mathbb {var(X+Y) = sum[(x_i+y_i) - mean(X+Y)]^2color{red}{p_{??}}}$
There is no way something like (a+b+m)^2 simplifies to (a+m)^2 + (b+m)^2. I'm kinda stuck here, any help ?
statistics random-variables
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
$endgroup$
add a comment |
$begingroup$
For independent random variables X and Y, the variance of their sum or
difference is the sum of their variances:
I can see why above should be true : if $x_1<X<x_1$
and $y_1 <Y < y_2$, then clearly $x_1+y_1 <X+Y<x_2+y_2$. But proving this seems a bit hard. Here is my attempt :
$mathbb {var(X) = sum[x_i - mean(X)]^2p_i}$
$mathbb {var(Y) = sum[y_i - mean(Y)]^2p_i}$,
then I guess the variance of sum should be :
$mathbb {var(X+Y) = sum[(x_i+y_i) - mean(X+Y)]^2color{red}{p_{??}}}$
There is no way something like (a+b+m)^2 simplifies to (a+m)^2 + (b+m)^2. I'm kinda stuck here, any help ?
statistics random-variables
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
$endgroup$
add a comment |
$begingroup$
For independent random variables X and Y, the variance of their sum or
difference is the sum of their variances:
I can see why above should be true : if $x_1<X<x_1$
and $y_1 <Y < y_2$, then clearly $x_1+y_1 <X+Y<x_2+y_2$. But proving this seems a bit hard. Here is my attempt :
$mathbb {var(X) = sum[x_i - mean(X)]^2p_i}$
$mathbb {var(Y) = sum[y_i - mean(Y)]^2p_i}$,
then I guess the variance of sum should be :
$mathbb {var(X+Y) = sum[(x_i+y_i) - mean(X+Y)]^2color{red}{p_{??}}}$
There is no way something like (a+b+m)^2 simplifies to (a+m)^2 + (b+m)^2. I'm kinda stuck here, any help ?
statistics random-variables
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
$endgroup$
For independent random variables X and Y, the variance of their sum or
difference is the sum of their variances:
I can see why above should be true : if $x_1<X<x_1$
and $y_1 <Y < y_2$, then clearly $x_1+y_1 <X+Y<x_2+y_2$. But proving this seems a bit hard. Here is my attempt :
$mathbb {var(X) = sum[x_i - mean(X)]^2p_i}$
$mathbb {var(Y) = sum[y_i - mean(Y)]^2p_i}$,
then I guess the variance of sum should be :
$mathbb {var(X+Y) = sum[(x_i+y_i) - mean(X+Y)]^2color{red}{p_{??}}}$
There is no way something like (a+b+m)^2 simplifies to (a+m)^2 + (b+m)^2. I'm kinda stuck here, any help ?
statistics random-variables
statistics random-variables
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
asked Dec 10 '18 at 11:06
rsadhvikarsadhvika
1,7101228
1,7101228
add a comment |
add a comment |
3 Answers
3
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oldest
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$begingroup$
Using the fact that $V(A) = E(A^2) - E(A)^2$ we have:
begin{align*}
V(X+Y) &= E(X+Y)^2 - [E(X+Y)]^2\
&=[EX^2 + EY^2 + 2E(XY) ] - [(EX)^2 + (EY)^2 + 2(EX)(EY)]\
&=EX^2 - (EX)^2 + EY^2 - (EY)^2 + 2E(XY) - 2(EX)(EY)\
&= V(X) +V(Y) + 2 cov(X,Y)
end{align*}
When is cov$(X,Y)=0$?
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
$endgroup$
add a comment |
$begingroup$
By subtracting off the means, it is sufficient to consider the case when $X$ and $Y$ are centered (i.e., $mathbb EX = mathbb EY=0$). Then
$$
text{Var}(Xpm Y)=mathbb E(Xpm Y)^2=mathbb E X^2pm 2mathbb E(XY)+mathbb EY^2.
$$
Now since $X$ and $Y$ are independent and centered, $mathbb E(XY)=(mathbb EX)(mathbb EY)=0$ and therefore
$$
text{Var}(Xpm Y)=text{Var}( X)+text{Var} (Y).
$$
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
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add a comment |
$begingroup$
Use the definition to expand and simplify
$Var (X+Y)=\
mathbb{E}((X+Y)^2)-(mu_X+mu_Y)^2$.
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Using the fact that $V(A) = E(A^2) - E(A)^2$ we have:
begin{align*}
V(X+Y) &= E(X+Y)^2 - [E(X+Y)]^2\
&=[EX^2 + EY^2 + 2E(XY) ] - [(EX)^2 + (EY)^2 + 2(EX)(EY)]\
&=EX^2 - (EX)^2 + EY^2 - (EY)^2 + 2E(XY) - 2(EX)(EY)\
&= V(X) +V(Y) + 2 cov(X,Y)
end{align*}
When is cov$(X,Y)=0$?
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
$endgroup$
add a comment |
$begingroup$
Using the fact that $V(A) = E(A^2) - E(A)^2$ we have:
begin{align*}
V(X+Y) &= E(X+Y)^2 - [E(X+Y)]^2\
&=[EX^2 + EY^2 + 2E(XY) ] - [(EX)^2 + (EY)^2 + 2(EX)(EY)]\
&=EX^2 - (EX)^2 + EY^2 - (EY)^2 + 2E(XY) - 2(EX)(EY)\
&= V(X) +V(Y) + 2 cov(X,Y)
end{align*}
When is cov$(X,Y)=0$?
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
$endgroup$
add a comment |
$begingroup$
Using the fact that $V(A) = E(A^2) - E(A)^2$ we have:
begin{align*}
V(X+Y) &= E(X+Y)^2 - [E(X+Y)]^2\
&=[EX^2 + EY^2 + 2E(XY) ] - [(EX)^2 + (EY)^2 + 2(EX)(EY)]\
&=EX^2 - (EX)^2 + EY^2 - (EY)^2 + 2E(XY) - 2(EX)(EY)\
&= V(X) +V(Y) + 2 cov(X,Y)
end{align*}
When is cov$(X,Y)=0$?
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
$endgroup$
Using the fact that $V(A) = E(A^2) - E(A)^2$ we have:
begin{align*}
V(X+Y) &= E(X+Y)^2 - [E(X+Y)]^2\
&=[EX^2 + EY^2 + 2E(XY) ] - [(EX)^2 + (EY)^2 + 2(EX)(EY)]\
&=EX^2 - (EX)^2 + EY^2 - (EY)^2 + 2E(XY) - 2(EX)(EY)\
&= V(X) +V(Y) + 2 cov(X,Y)
end{align*}
When is cov$(X,Y)=0$?
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
answered Dec 10 '18 at 11:17
dimebuckerdimebucker
1,67011129
1,67011129
add a comment |
add a comment |
$begingroup$
By subtracting off the means, it is sufficient to consider the case when $X$ and $Y$ are centered (i.e., $mathbb EX = mathbb EY=0$). Then
$$
text{Var}(Xpm Y)=mathbb E(Xpm Y)^2=mathbb E X^2pm 2mathbb E(XY)+mathbb EY^2.
$$
Now since $X$ and $Y$ are independent and centered, $mathbb E(XY)=(mathbb EX)(mathbb EY)=0$ and therefore
$$
text{Var}(Xpm Y)=text{Var}( X)+text{Var} (Y).
$$
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
$endgroup$
add a comment |
$begingroup$
By subtracting off the means, it is sufficient to consider the case when $X$ and $Y$ are centered (i.e., $mathbb EX = mathbb EY=0$). Then
$$
text{Var}(Xpm Y)=mathbb E(Xpm Y)^2=mathbb E X^2pm 2mathbb E(XY)+mathbb EY^2.
$$
Now since $X$ and $Y$ are independent and centered, $mathbb E(XY)=(mathbb EX)(mathbb EY)=0$ and therefore
$$
text{Var}(Xpm Y)=text{Var}( X)+text{Var} (Y).
$$
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
$endgroup$
add a comment |
$begingroup$
By subtracting off the means, it is sufficient to consider the case when $X$ and $Y$ are centered (i.e., $mathbb EX = mathbb EY=0$). Then
$$
text{Var}(Xpm Y)=mathbb E(Xpm Y)^2=mathbb E X^2pm 2mathbb E(XY)+mathbb EY^2.
$$
Now since $X$ and $Y$ are independent and centered, $mathbb E(XY)=(mathbb EX)(mathbb EY)=0$ and therefore
$$
text{Var}(Xpm Y)=text{Var}( X)+text{Var} (Y).
$$
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
$endgroup$
By subtracting off the means, it is sufficient to consider the case when $X$ and $Y$ are centered (i.e., $mathbb EX = mathbb EY=0$). Then
$$
text{Var}(Xpm Y)=mathbb E(Xpm Y)^2=mathbb E X^2pm 2mathbb E(XY)+mathbb EY^2.
$$
Now since $X$ and $Y$ are independent and centered, $mathbb E(XY)=(mathbb EX)(mathbb EY)=0$ and therefore
$$
text{Var}(Xpm Y)=text{Var}( X)+text{Var} (Y).
$$
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
answered Dec 10 '18 at 11:25
pre-kidneypre-kidney
12.9k1849
12.9k1849
add a comment |
add a comment |
$begingroup$
Use the definition to expand and simplify
$Var (X+Y)=\
mathbb{E}((X+Y)^2)-(mu_X+mu_Y)^2$.
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
$endgroup$
add a comment |
$begingroup$
Use the definition to expand and simplify
$Var (X+Y)=\
mathbb{E}((X+Y)^2)-(mu_X+mu_Y)^2$.
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
$endgroup$
add a comment |
$begingroup$
Use the definition to expand and simplify
$Var (X+Y)=\
mathbb{E}((X+Y)^2)-(mu_X+mu_Y)^2$.
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
$endgroup$
Use the definition to expand and simplify
$Var (X+Y)=\
mathbb{E}((X+Y)^2)-(mu_X+mu_Y)^2$.
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
answered Dec 10 '18 at 11:16
AnyADAnyAD
2,113812
2,113812
add a comment |
add a comment |
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