Probability Distribution on unit circle
$begingroup$
I currently stuck at following task:
So I have a probability distribution:
$$
f^{X,Y}(x,y) = frac{1}{pi}; forall(x,y)in A
$$
where $$
A = { (x,y) : x^2+y^2 leq 1 }
$$
-I now want to calculate the Variance(X), Variance(Y) and the Covariance(X,Y). I know that Cov(X,Y) = 0, but how do I show this ?
-Also I want to show that X and Y are not indipendent. I mean intuitively it is pretty clear that X and Y are dependent of each other, but how do i show it mathematically? So obviously i need to show that
$$
mathbb{P}(Xcap Y) =mathbb{P}(X)mathbb{P}(Y)
$$
but how ?
probability
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
$endgroup$
add a comment |
$begingroup$
I currently stuck at following task:
So I have a probability distribution:
$$
f^{X,Y}(x,y) = frac{1}{pi}; forall(x,y)in A
$$
where $$
A = { (x,y) : x^2+y^2 leq 1 }
$$
-I now want to calculate the Variance(X), Variance(Y) and the Covariance(X,Y). I know that Cov(X,Y) = 0, but how do I show this ?
-Also I want to show that X and Y are not indipendent. I mean intuitively it is pretty clear that X and Y are dependent of each other, but how do i show it mathematically? So obviously i need to show that
$$
mathbb{P}(Xcap Y) =mathbb{P}(X)mathbb{P}(Y)
$$
but how ?
probability
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
$endgroup$
add a comment |
$begingroup$
I currently stuck at following task:
So I have a probability distribution:
$$
f^{X,Y}(x,y) = frac{1}{pi}; forall(x,y)in A
$$
where $$
A = { (x,y) : x^2+y^2 leq 1 }
$$
-I now want to calculate the Variance(X), Variance(Y) and the Covariance(X,Y). I know that Cov(X,Y) = 0, but how do I show this ?
-Also I want to show that X and Y are not indipendent. I mean intuitively it is pretty clear that X and Y are dependent of each other, but how do i show it mathematically? So obviously i need to show that
$$
mathbb{P}(Xcap Y) =mathbb{P}(X)mathbb{P}(Y)
$$
but how ?
probability
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
$endgroup$
I currently stuck at following task:
So I have a probability distribution:
$$
f^{X,Y}(x,y) = frac{1}{pi}; forall(x,y)in A
$$
where $$
A = { (x,y) : x^2+y^2 leq 1 }
$$
-I now want to calculate the Variance(X), Variance(Y) and the Covariance(X,Y). I know that Cov(X,Y) = 0, but how do I show this ?
-Also I want to show that X and Y are not indipendent. I mean intuitively it is pretty clear that X and Y are dependent of each other, but how do i show it mathematically? So obviously i need to show that
$$
mathbb{P}(Xcap Y) =mathbb{P}(X)mathbb{P}(Y)
$$
but how ?
probability
probability
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
asked Dec 18 '18 at 23:22
RedCrayonRedCrayon
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Hints: covariance can easily be calculated using polar coordinates: $cov(X,Y)=int_0^{1} int_0^{2pi} (rcos , theta)(rsin , theta) rdrdtheta$. To show that $X$ and $Y$ are not independent consider ${Y>frac 1 2}$ and ${|X| leqfrac {sqrt 3} 2}$. Note that $X^{2}+Y^{2} leq 1$ so $Y>frac 1 2$ implies $|X| leqfrac {sqrt 3} 2$. If $X$ and $Y$ were independent we would have $P{Y>frac 1 2}P{|X| leqfrac {sqrt 3} 2}=P{ Y>frac 1 2,|X| leqfrac {sqrt 3} 2}$. But the right side is simply $P{Y>frac 1 2}$. So the equation becomes $P{|X| leqfrac {sqrt 3} 2}=1$ (because $P{Y>frac 1 2} neq 0$). Can you check that this is a contardiction? (Use polar coordinates for all probabilities).
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
add a comment |
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1 Answer
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active
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1 Answer
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oldest
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$begingroup$
Hints: covariance can easily be calculated using polar coordinates: $cov(X,Y)=int_0^{1} int_0^{2pi} (rcos , theta)(rsin , theta) rdrdtheta$. To show that $X$ and $Y$ are not independent consider ${Y>frac 1 2}$ and ${|X| leqfrac {sqrt 3} 2}$. Note that $X^{2}+Y^{2} leq 1$ so $Y>frac 1 2$ implies $|X| leqfrac {sqrt 3} 2$. If $X$ and $Y$ were independent we would have $P{Y>frac 1 2}P{|X| leqfrac {sqrt 3} 2}=P{ Y>frac 1 2,|X| leqfrac {sqrt 3} 2}$. But the right side is simply $P{Y>frac 1 2}$. So the equation becomes $P{|X| leqfrac {sqrt 3} 2}=1$ (because $P{Y>frac 1 2} neq 0$). Can you check that this is a contardiction? (Use polar coordinates for all probabilities).
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
add a comment |
$begingroup$
Hints: covariance can easily be calculated using polar coordinates: $cov(X,Y)=int_0^{1} int_0^{2pi} (rcos , theta)(rsin , theta) rdrdtheta$. To show that $X$ and $Y$ are not independent consider ${Y>frac 1 2}$ and ${|X| leqfrac {sqrt 3} 2}$. Note that $X^{2}+Y^{2} leq 1$ so $Y>frac 1 2$ implies $|X| leqfrac {sqrt 3} 2$. If $X$ and $Y$ were independent we would have $P{Y>frac 1 2}P{|X| leqfrac {sqrt 3} 2}=P{ Y>frac 1 2,|X| leqfrac {sqrt 3} 2}$. But the right side is simply $P{Y>frac 1 2}$. So the equation becomes $P{|X| leqfrac {sqrt 3} 2}=1$ (because $P{Y>frac 1 2} neq 0$). Can you check that this is a contardiction? (Use polar coordinates for all probabilities).
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
add a comment |
$begingroup$
Hints: covariance can easily be calculated using polar coordinates: $cov(X,Y)=int_0^{1} int_0^{2pi} (rcos , theta)(rsin , theta) rdrdtheta$. To show that $X$ and $Y$ are not independent consider ${Y>frac 1 2}$ and ${|X| leqfrac {sqrt 3} 2}$. Note that $X^{2}+Y^{2} leq 1$ so $Y>frac 1 2$ implies $|X| leqfrac {sqrt 3} 2$. If $X$ and $Y$ were independent we would have $P{Y>frac 1 2}P{|X| leqfrac {sqrt 3} 2}=P{ Y>frac 1 2,|X| leqfrac {sqrt 3} 2}$. But the right side is simply $P{Y>frac 1 2}$. So the equation becomes $P{|X| leqfrac {sqrt 3} 2}=1$ (because $P{Y>frac 1 2} neq 0$). Can you check that this is a contardiction? (Use polar coordinates for all probabilities).
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
$endgroup$
Hints: covariance can easily be calculated using polar coordinates: $cov(X,Y)=int_0^{1} int_0^{2pi} (rcos , theta)(rsin , theta) rdrdtheta$. To show that $X$ and $Y$ are not independent consider ${Y>frac 1 2}$ and ${|X| leqfrac {sqrt 3} 2}$. Note that $X^{2}+Y^{2} leq 1$ so $Y>frac 1 2$ implies $|X| leqfrac {sqrt 3} 2$. If $X$ and $Y$ were independent we would have $P{Y>frac 1 2}P{|X| leqfrac {sqrt 3} 2}=P{ Y>frac 1 2,|X| leqfrac {sqrt 3} 2}$. But the right side is simply $P{Y>frac 1 2}$. So the equation becomes $P{|X| leqfrac {sqrt 3} 2}=1$ (because $P{Y>frac 1 2} neq 0$). Can you check that this is a contardiction? (Use polar coordinates for all probabilities).
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
answered Dec 18 '18 at 23:43
Kavi Rama MurthyKavi Rama Murthy
60.3k42161
60.3k42161
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
add a comment |
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
But in order to calculate Var(x) and Var(y) i first have to calculate the marginal distributions right ? So that I have: $$ mathbb{E}(X) = int x f^X(x) dx $$ and then I substitute $ x = r cdot cos(theta)$ right ?
$endgroup$
– RedCrayon
Dec 18 '18 at 23:55
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
$Var(X)=int_A x^{2} frac 1 {pi} dxdy$ which can be calculated easily using polar coordinates. (Note that $EX=EY=0$ by symmetry).
$endgroup$
– Kavi Rama Murthy
Dec 18 '18 at 23:58
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
$begingroup$
@RedCrayon The formula $Eg(X,Y) =int int g(x,y) f_{X,Y}dxdy$ can be used when $g(x,y)$ depends only on one variable (like $g(x,y)=x^{2})$. So you can find $var (X)$ without calculating the density of $X$.
$endgroup$
– Kavi Rama Murthy
Dec 19 '18 at 5:13
add a comment |
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