Prove $operatorname{Var}(X+Y) le 2operatorname{Var}(X) + 2operatorname{Var}(Y)$ where $X$ and $Y$ are not...
up vote
1
down vote
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I know how to show that $operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)$
I was also given the hint that I should use the triangle inequality to get
$|operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)| le |operatorname{Var}(x)| + |operatorname{Var}(Y)| + |2operatorname{Cov}(X,Y)|$
Honestly, I have no idea where to go from here.
statistics covariance variance
edited Oct 11 at 21:38
Bernard
117k637110
asked Oct 11 at 21:32
mattfdz
61
add a comment |
up vote
1
down vote
favorite
I know how to show that $operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)$
I was also given the hint that I should use the triangle inequality to get
$|operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)| le |operatorname{Var}(x)| + |operatorname{Var}(Y)| + |2operatorname{Cov}(X,Y)|$
Honestly, I have no idea where to go from here.
statistics covariance variance
edited Oct 11 at 21:38
Bernard
117k637110
asked Oct 11 at 21:32
mattfdz
61
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I know how to show that $operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)$
I was also given the hint that I should use the triangle inequality to get
$|operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)| le |operatorname{Var}(x)| + |operatorname{Var}(Y)| + |2operatorname{Cov}(X,Y)|$
Honestly, I have no idea where to go from here.
statistics covariance variance
edited Oct 11 at 21:38
Bernard
117k637110
asked Oct 11 at 21:32
mattfdz
61
I know how to show that $operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)$
I was also given the hint that I should use the triangle inequality to get
$|operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)| le |operatorname{Var}(x)| + |operatorname{Var}(Y)| + |2operatorname{Cov}(X,Y)|$
Honestly, I have no idea where to go from here.
statistics covariance variance
statistics covariance variance
edited Oct 11 at 21:38
Bernard
117k637110
asked Oct 11 at 21:32
mattfdz
61
edited Oct 11 at 21:38
Bernard
117k637110
asked Oct 11 at 21:32
mattfdz
61
edited Oct 11 at 21:38
Bernard
117k637110
edited Oct 11 at 21:38
Bernard
117k637110
edited Oct 11 at 21:38
Bernard
117k637110
117k637110
asked Oct 11 at 21:32
mattfdz
61
asked Oct 11 at 21:32
mattfdz
61
asked Oct 11 at 21:32
mattfdz
61
61
add a comment |
add a comment |
5 Answers
5
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oldest
votes
up vote
0
down vote
The difference is
$$2E((X-mu_X)^2)+2E((Y-mu_Y)^2)-E((X+Y-mu_X-mu_Y)^2)
=2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$
where $X'=X-mu_X$, $Y'=Y-mu_Y$. Can you simplify that to something conveniently positive?
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
add a comment |
up vote
0
down vote
The key is to show that $|2 operatorname{Cov}(X,Y)| leq operatorname{Var}(X) + operatorname{Var}(Y)$.
Here's a hint towards that: Write $hat{X} = X - mu_X$ and $hat{Y} = Y - mu_Y$. Then note that $$E[(hat{X} - hat{Y})^2] geq 0,.$$
What happens if you expand the expectation?
edited Oct 11 at 21:39
Bernard
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
add a comment |
up vote
0
down vote
Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
0
down vote
Start as you did with:
$$
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
$$
The Cauchy–Schwarz inequality then gives us:
$$
2Cov(X, Y) leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}.
$$
Finally, Young's inequality for products gives us:
$$
2[Var(X)]^{1/2}[Var(Y)]^{1/2} leq Var(X) + Var(Y),
$$
so we get, as desired:
$$
Var(X + Y) leq 2Var(X) + 2Var(Y).
$$
answered Nov 29 at 5:22
Fredrik Savje
1304
add a comment |
up vote
0
down vote
If you know how to show
$$operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)tag1$$
then you can show
$$
operatorname{Var}(X-Y) = operatorname{Var}(X) + operatorname{Var}(Y) - 2operatorname{Cov}(X,Y).tag2
$$
Now add equations (1) and (2) to get
$$
operatorname{Var}(X+Y)+operatorname{Var}(X-Y) = 2operatorname{Var}(X) + 2operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
operatorname{Var}(X+Y)le operatorname{Var}(X+Y)+operatorname{Var}(X-Y).
$$
answered Nov 29 at 14:56
grand_chat
19.9k11225
add a comment |
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5 Answers
5
active
oldest
votes
5 Answers
5
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
The difference is
$$2E((X-mu_X)^2)+2E((Y-mu_Y)^2)-E((X+Y-mu_X-mu_Y)^2)
=2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$
where $X'=X-mu_X$, $Y'=Y-mu_Y$. Can you simplify that to something conveniently positive?
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
add a comment |
up vote
0
down vote
The difference is
$$2E((X-mu_X)^2)+2E((Y-mu_Y)^2)-E((X+Y-mu_X-mu_Y)^2)
=2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$
where $X'=X-mu_X$, $Y'=Y-mu_Y$. Can you simplify that to something conveniently positive?
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
add a comment |
up vote
0
down vote
up vote
0
down vote
The difference is
$$2E((X-mu_X)^2)+2E((Y-mu_Y)^2)-E((X+Y-mu_X-mu_Y)^2)
=2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$
where $X'=X-mu_X$, $Y'=Y-mu_Y$. Can you simplify that to something conveniently positive?
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
The difference is
$$2E((X-mu_X)^2)+2E((Y-mu_Y)^2)-E((X+Y-mu_X-mu_Y)^2)
=2E(X'^2)+2E(Y'^2)-E((X'+Y')^2)$$
where $X'=X-mu_X$, $Y'=Y-mu_Y$. Can you simplify that to something conveniently positive?
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
answered Oct 11 at 21:36
Lord Shark the Unknown
100k958131
100k958131
add a comment |
add a comment |
up vote
0
down vote
The key is to show that $|2 operatorname{Cov}(X,Y)| leq operatorname{Var}(X) + operatorname{Var}(Y)$.
Here's a hint towards that: Write $hat{X} = X - mu_X$ and $hat{Y} = Y - mu_Y$. Then note that $$E[(hat{X} - hat{Y})^2] geq 0,.$$
What happens if you expand the expectation?
edited Oct 11 at 21:39
Bernard
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
add a comment |
up vote
0
down vote
The key is to show that $|2 operatorname{Cov}(X,Y)| leq operatorname{Var}(X) + operatorname{Var}(Y)$.
Here's a hint towards that: Write $hat{X} = X - mu_X$ and $hat{Y} = Y - mu_Y$. Then note that $$E[(hat{X} - hat{Y})^2] geq 0,.$$
What happens if you expand the expectation?
edited Oct 11 at 21:39
Bernard
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
add a comment |
up vote
0
down vote
up vote
0
down vote
The key is to show that $|2 operatorname{Cov}(X,Y)| leq operatorname{Var}(X) + operatorname{Var}(Y)$.
Here's a hint towards that: Write $hat{X} = X - mu_X$ and $hat{Y} = Y - mu_Y$. Then note that $$E[(hat{X} - hat{Y})^2] geq 0,.$$
What happens if you expand the expectation?
edited Oct 11 at 21:39
Bernard
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
The key is to show that $|2 operatorname{Cov}(X,Y)| leq operatorname{Var}(X) + operatorname{Var}(Y)$.
Here's a hint towards that: Write $hat{X} = X - mu_X$ and $hat{Y} = Y - mu_Y$. Then note that $$E[(hat{X} - hat{Y})^2] geq 0,.$$
What happens if you expand the expectation?
edited Oct 11 at 21:39
Bernard
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
edited Oct 11 at 21:39
Bernard
117k637110
edited Oct 11 at 21:39
Bernard
117k637110
edited Oct 11 at 21:39
Bernard
117k637110
117k637110
answered Oct 11 at 21:38
Marcus M
8,7281947
answered Oct 11 at 21:38
Marcus M
8,7281947
answered Oct 11 at 21:38
Marcus M
8,7281947
8,7281947
add a comment |
add a comment |
up vote
0
down vote
Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
0
down vote
Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
add a comment |
up vote
0
down vote
up vote
0
down vote
Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
Let $X'=X-EX,Y'=Y-EY$. Then the inequality to be proved is $E(X'+Y')^{2} leq 2EX'^{2}+2EY'^{2}$ whic follows from the fact that $(a+b)^{2} leq 2a^{2} +2b^{2}$ for any two real numbers $a$ and $b$.
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
answered Oct 11 at 23:45
Kavi Rama Murthy
48.2k31854
48.2k31854
add a comment |
add a comment |
up vote
0
down vote
Start as you did with:
$$
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
$$
The Cauchy–Schwarz inequality then gives us:
$$
2Cov(X, Y) leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}.
$$
Finally, Young's inequality for products gives us:
$$
2[Var(X)]^{1/2}[Var(Y)]^{1/2} leq Var(X) + Var(Y),
$$
so we get, as desired:
$$
Var(X + Y) leq 2Var(X) + 2Var(Y).
$$
answered Nov 29 at 5:22
Fredrik Savje
1304
add a comment |
up vote
0
down vote
Start as you did with:
$$
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
$$
The Cauchy–Schwarz inequality then gives us:
$$
2Cov(X, Y) leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}.
$$
Finally, Young's inequality for products gives us:
$$
2[Var(X)]^{1/2}[Var(Y)]^{1/2} leq Var(X) + Var(Y),
$$
so we get, as desired:
$$
Var(X + Y) leq 2Var(X) + 2Var(Y).
$$
answered Nov 29 at 5:22
Fredrik Savje
1304
add a comment |
up vote
0
down vote
up vote
0
down vote
Start as you did with:
$$
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
$$
The Cauchy–Schwarz inequality then gives us:
$$
2Cov(X, Y) leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}.
$$
Finally, Young's inequality for products gives us:
$$
2[Var(X)]^{1/2}[Var(Y)]^{1/2} leq Var(X) + Var(Y),
$$
so we get, as desired:
$$
Var(X + Y) leq 2Var(X) + 2Var(Y).
$$
answered Nov 29 at 5:22
Fredrik Savje
1304
Start as you did with:
$$
Var(X + Y) = Var(X) + Var(Y) + 2Cov(X, Y).
$$
The Cauchy–Schwarz inequality then gives us:
$$
2Cov(X, Y) leq 2[Var(X)]^{1/2}[Var(Y)]^{1/2}.
$$
Finally, Young's inequality for products gives us:
$$
2[Var(X)]^{1/2}[Var(Y)]^{1/2} leq Var(X) + Var(Y),
$$
so we get, as desired:
$$
Var(X + Y) leq 2Var(X) + 2Var(Y).
$$
answered Nov 29 at 5:22
Fredrik Savje
1304
answered Nov 29 at 5:22
Fredrik Savje
1304
answered Nov 29 at 5:22
Fredrik Savje
1304
answered Nov 29 at 5:22
Fredrik Savje
1304
1304
add a comment |
add a comment |
up vote
0
down vote
If you know how to show
$$operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)tag1$$
then you can show
$$
operatorname{Var}(X-Y) = operatorname{Var}(X) + operatorname{Var}(Y) - 2operatorname{Cov}(X,Y).tag2
$$
Now add equations (1) and (2) to get
$$
operatorname{Var}(X+Y)+operatorname{Var}(X-Y) = 2operatorname{Var}(X) + 2operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
operatorname{Var}(X+Y)le operatorname{Var}(X+Y)+operatorname{Var}(X-Y).
$$
answered Nov 29 at 14:56
grand_chat
19.9k11225
add a comment |
up vote
0
down vote
If you know how to show
$$operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)tag1$$
then you can show
$$
operatorname{Var}(X-Y) = operatorname{Var}(X) + operatorname{Var}(Y) - 2operatorname{Cov}(X,Y).tag2
$$
Now add equations (1) and (2) to get
$$
operatorname{Var}(X+Y)+operatorname{Var}(X-Y) = 2operatorname{Var}(X) + 2operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
operatorname{Var}(X+Y)le operatorname{Var}(X+Y)+operatorname{Var}(X-Y).
$$
answered Nov 29 at 14:56
grand_chat
19.9k11225
add a comment |
up vote
0
down vote
up vote
0
down vote
If you know how to show
$$operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)tag1$$
then you can show
$$
operatorname{Var}(X-Y) = operatorname{Var}(X) + operatorname{Var}(Y) - 2operatorname{Cov}(X,Y).tag2
$$
Now add equations (1) and (2) to get
$$
operatorname{Var}(X+Y)+operatorname{Var}(X-Y) = 2operatorname{Var}(X) + 2operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
operatorname{Var}(X+Y)le operatorname{Var}(X+Y)+operatorname{Var}(X-Y).
$$
answered Nov 29 at 14:56
grand_chat
19.9k11225
If you know how to show
$$operatorname{Var}(X+Y) = operatorname{Var}(X) + operatorname{Var}(Y) + 2operatorname{Cov}(X,Y)tag1$$
then you can show
$$
operatorname{Var}(X-Y) = operatorname{Var}(X) + operatorname{Var}(Y) - 2operatorname{Cov}(X,Y).tag2
$$
Now add equations (1) and (2) to get
$$
operatorname{Var}(X+Y)+operatorname{Var}(X-Y) = 2operatorname{Var}(X) + 2operatorname{Var}(Y).
$$
Finally the fact that variance of anything is non-negative implies that
$$
operatorname{Var}(X+Y)le operatorname{Var}(X+Y)+operatorname{Var}(X-Y).
$$
answered Nov 29 at 14:56
grand_chat
19.9k11225
answered Nov 29 at 14:56
grand_chat
19.9k11225
answered Nov 29 at 14:56
grand_chat
19.9k11225
answered Nov 29 at 14:56
grand_chat
19.9k11225
19.9k11225
add a comment |
add a comment |
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