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Circle-shaped integration of a bivariate normal distribution
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I have a bivariate Normal Distribution of positions in a 2D space given by the marginal distributions of x and y components: $N_x(mu_x,sigma_x)$, $N_y(mu_y,sigma_y)$.
I want to determine the probability of a coordinate $(x,y)$ to be located within a certain region of the space. In a first step, this region is a rectangle $R$ defined by a tuple of coordinates $R = (x_a, y_a, x_b, y_b)$, which represents two corners.
The probability of $(x,y)$ to be located within this rectangular is
$$ beginalign
P((x,y) in R) & = P(x in [x_a,x_b]) cdot P(y in [y_a, y_b]) \
& = int_x_a^x_bp_x(u) du cdot int_y_a^y_bp_y(v)dv
endalign
$$
The next step is to have a circular region $C$ given by its center $(x_c,y_c)$ and radius $r$.
I know that I have to somehow incorporate the condition $(x-x_c)^2 + (y-y_c)^2 = r^2$ into the integration, which introduces a non-trivial dependency between the two dimensions. And at this point I am lost.
So I want to ask you, how to best calculate $P((x,y) in C)$?
EDIT
Thanks to Henry for his inspiring comment! It's now obvious to me, how to solve this. However, integration boundaries are dependent and non-constant. This makes it difficult for me, to use this solution in my special case.
Is the following transformation into polar coordinates equivalent?
$$
int_0^r int_0^2pi p_x(x_c + tcdot cos(phi)) cdot p_y(y_c + tcdot sin(phi)) dphi dt
$$
multivariable-calculus normal-distribution bivariate-distributions
asked Jul 17 at 7:32
Stanley F.
1064
add a comment |Â
up vote
1
down vote
favorite
I have a bivariate Normal Distribution of positions in a 2D space given by the marginal distributions of x and y components: $N_x(mu_x,sigma_x)$, $N_y(mu_y,sigma_y)$.
I want to determine the probability of a coordinate $(x,y)$ to be located within a certain region of the space. In a first step, this region is a rectangle $R$ defined by a tuple of coordinates $R = (x_a, y_a, x_b, y_b)$, which represents two corners.
The probability of $(x,y)$ to be located within this rectangular is
$$ beginalign
P((x,y) in R) & = P(x in [x_a,x_b]) cdot P(y in [y_a, y_b]) \
& = int_x_a^x_bp_x(u) du cdot int_y_a^y_bp_y(v)dv
endalign
$$
The next step is to have a circular region $C$ given by its center $(x_c,y_c)$ and radius $r$.
I know that I have to somehow incorporate the condition $(x-x_c)^2 + (y-y_c)^2 = r^2$ into the integration, which introduces a non-trivial dependency between the two dimensions. And at this point I am lost.
So I want to ask you, how to best calculate $P((x,y) in C)$?
EDIT
Thanks to Henry for his inspiring comment! It's now obvious to me, how to solve this. However, integration boundaries are dependent and non-constant. This makes it difficult for me, to use this solution in my special case.
Is the following transformation into polar coordinates equivalent?
$$
int_0^r int_0^2pi p_x(x_c + tcdot cos(phi)) cdot p_y(y_c + tcdot sin(phi)) dphi dt
$$
multivariable-calculus normal-distribution bivariate-distributions
asked Jul 17 at 7:32
Stanley F.
1064
1
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I have a bivariate Normal Distribution of positions in a 2D space given by the marginal distributions of x and y components: $N_x(mu_x,sigma_x)$, $N_y(mu_y,sigma_y)$.
I want to determine the probability of a coordinate $(x,y)$ to be located within a certain region of the space. In a first step, this region is a rectangle $R$ defined by a tuple of coordinates $R = (x_a, y_a, x_b, y_b)$, which represents two corners.
The probability of $(x,y)$ to be located within this rectangular is
$$ beginalign
P((x,y) in R) & = P(x in [x_a,x_b]) cdot P(y in [y_a, y_b]) \
& = int_x_a^x_bp_x(u) du cdot int_y_a^y_bp_y(v)dv
endalign
$$
The next step is to have a circular region $C$ given by its center $(x_c,y_c)$ and radius $r$.
I know that I have to somehow incorporate the condition $(x-x_c)^2 + (y-y_c)^2 = r^2$ into the integration, which introduces a non-trivial dependency between the two dimensions. And at this point I am lost.
So I want to ask you, how to best calculate $P((x,y) in C)$?
EDIT
Thanks to Henry for his inspiring comment! It's now obvious to me, how to solve this. However, integration boundaries are dependent and non-constant. This makes it difficult for me, to use this solution in my special case.
Is the following transformation into polar coordinates equivalent?
$$
int_0^r int_0^2pi p_x(x_c + tcdot cos(phi)) cdot p_y(y_c + tcdot sin(phi)) dphi dt
$$
multivariable-calculus normal-distribution bivariate-distributions
asked Jul 17 at 7:32
Stanley F.
1064
I have a bivariate Normal Distribution of positions in a 2D space given by the marginal distributions of x and y components: $N_x(mu_x,sigma_x)$, $N_y(mu_y,sigma_y)$.
I want to determine the probability of a coordinate $(x,y)$ to be located within a certain region of the space. In a first step, this region is a rectangle $R$ defined by a tuple of coordinates $R = (x_a, y_a, x_b, y_b)$, which represents two corners.
The probability of $(x,y)$ to be located within this rectangular is
$$ beginalign
P((x,y) in R) & = P(x in [x_a,x_b]) cdot P(y in [y_a, y_b]) \
& = int_x_a^x_bp_x(u) du cdot int_y_a^y_bp_y(v)dv
endalign
$$
The next step is to have a circular region $C$ given by its center $(x_c,y_c)$ and radius $r$.
I know that I have to somehow incorporate the condition $(x-x_c)^2 + (y-y_c)^2 = r^2$ into the integration, which introduces a non-trivial dependency between the two dimensions. And at this point I am lost.
So I want to ask you, how to best calculate $P((x,y) in C)$?
EDIT
Thanks to Henry for his inspiring comment! It's now obvious to me, how to solve this. However, integration boundaries are dependent and non-constant. This makes it difficult for me, to use this solution in my special case.
Is the following transformation into polar coordinates equivalent?
$$
int_0^r int_0^2pi p_x(x_c + tcdot cos(phi)) cdot p_y(y_c + tcdot sin(phi)) dphi dt
$$
multivariable-calculus normal-distribution bivariate-distributions
asked Jul 17 at 7:32
Stanley F.
1064
edited Jul 23 at 8:53
asked Jul 17 at 7:32
Stanley F.
1064
asked Jul 17 at 7:32
Stanley F.
1064
asked Jul 17 at 7:32
Stanley F.
1064
1064
1
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00
add a comment |Â
1
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00
1
1
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00
add a comment |Â
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1
Presumably the two random variables are independent
– Henry
Jul 17 at 7:54
It may not be the best way, but I think one way of writing this is $$int_u=-r^r p_x(x_c+u) int_v=-sqrtr^2-u^2^sqrtr^2-u^2 p_y(y_c+v),dv , du$$
– Henry
Jul 17 at 8:00