Marginal probability distribution of unit circle random variable
$begingroup$
I am given a task stating that there is a bivariate random vector X such that:
$$begin{equation}
p_x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2+y^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
and I am asked to find margianl distribution with respect to X and Y.
I tried to catch it up by google and found some answers. e.g.
$$f(x)_x=int_sqrt{1^2-x^2}^sqrt{1^2+x^2}frac1pi dy$$
One thing I am interested in. Where the integration bounds came from. Can anyone explain pls?
probability probability-theory probability-distributions
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
$endgroup$
add a comment |
$begingroup$
I am given a task stating that there is a bivariate random vector X such that:
$$begin{equation}
p_x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2+y^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
and I am asked to find margianl distribution with respect to X and Y.
I tried to catch it up by google and found some answers. e.g.
$$f(x)_x=int_sqrt{1^2-x^2}^sqrt{1^2+x^2}frac1pi dy$$
One thing I am interested in. Where the integration bounds came from. Can anyone explain pls?
probability probability-theory probability-distributions
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
$endgroup$
$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42
add a comment |
$begingroup$
I am given a task stating that there is a bivariate random vector X such that:
$$begin{equation}
p_x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2+y^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
and I am asked to find margianl distribution with respect to X and Y.
I tried to catch it up by google and found some answers. e.g.
$$f(x)_x=int_sqrt{1^2-x^2}^sqrt{1^2+x^2}frac1pi dy$$
One thing I am interested in. Where the integration bounds came from. Can anyone explain pls?
probability probability-theory probability-distributions
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
$endgroup$
I am given a task stating that there is a bivariate random vector X such that:
$$begin{equation}
p_x(x)=left{
begin{array}{@{}ll@{}}
frac1pi, & text{if} x^2+y^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation} $$
and I am asked to find margianl distribution with respect to X and Y.
I tried to catch it up by google and found some answers. e.g.
$$f(x)_x=int_sqrt{1^2-x^2}^sqrt{1^2+x^2}frac1pi dy$$
One thing I am interested in. Where the integration bounds came from. Can anyone explain pls?
probability probability-theory probability-distributions
probability probability-theory probability-distributions
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
edited Jan 9 at 19:16
Hillbilly Joe
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
asked Jan 9 at 19:07
Hillbilly JoeHillbilly Joe
164
164
$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42
add a comment |
$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42
$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Define the domain $mathcal{D} = {(x_1,x2) vert x_1^2 + x_2^2 <1}$. Given a fix $x_1$, if $(x_1, x_2)$ belongs to $mathcal{D}$, $x_2$ has to satisfy $ - sqrt{1-x_1^2}< x_2 < sqrt{1-x_1^2}$, where we are taken positive roots:
begin{equation}
f_{X_1}(x_1) = int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = left{
begin{array}{@{}ll@{}}
int_{- sqrt{1-x_1^2}}^{ sqrt{1-x_1^2}} frac{1}{pi} d x_2 = frac{2sqrt{1-x_1^2}}{pi} & text{if} x_1^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation}
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
$endgroup$
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
add a comment |
Your Answer
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Define the domain $mathcal{D} = {(x_1,x2) vert x_1^2 + x_2^2 <1}$. Given a fix $x_1$, if $(x_1, x_2)$ belongs to $mathcal{D}$, $x_2$ has to satisfy $ - sqrt{1-x_1^2}< x_2 < sqrt{1-x_1^2}$, where we are taken positive roots:
begin{equation}
f_{X_1}(x_1) = int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = left{
begin{array}{@{}ll@{}}
int_{- sqrt{1-x_1^2}}^{ sqrt{1-x_1^2}} frac{1}{pi} d x_2 = frac{2sqrt{1-x_1^2}}{pi} & text{if} x_1^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation}
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
$endgroup$
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
add a comment |
$begingroup$
Define the domain $mathcal{D} = {(x_1,x2) vert x_1^2 + x_2^2 <1}$. Given a fix $x_1$, if $(x_1, x_2)$ belongs to $mathcal{D}$, $x_2$ has to satisfy $ - sqrt{1-x_1^2}< x_2 < sqrt{1-x_1^2}$, where we are taken positive roots:
begin{equation}
f_{X_1}(x_1) = int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = left{
begin{array}{@{}ll@{}}
int_{- sqrt{1-x_1^2}}^{ sqrt{1-x_1^2}} frac{1}{pi} d x_2 = frac{2sqrt{1-x_1^2}}{pi} & text{if} x_1^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation}
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
$endgroup$
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
add a comment |
$begingroup$
Define the domain $mathcal{D} = {(x_1,x2) vert x_1^2 + x_2^2 <1}$. Given a fix $x_1$, if $(x_1, x_2)$ belongs to $mathcal{D}$, $x_2$ has to satisfy $ - sqrt{1-x_1^2}< x_2 < sqrt{1-x_1^2}$, where we are taken positive roots:
begin{equation}
f_{X_1}(x_1) = int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = left{
begin{array}{@{}ll@{}}
int_{- sqrt{1-x_1^2}}^{ sqrt{1-x_1^2}} frac{1}{pi} d x_2 = frac{2sqrt{1-x_1^2}}{pi} & text{if} x_1^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation}
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
$endgroup$
Define the domain $mathcal{D} = {(x_1,x2) vert x_1^2 + x_2^2 <1}$. Given a fix $x_1$, if $(x_1, x_2)$ belongs to $mathcal{D}$, $x_2$ has to satisfy $ - sqrt{1-x_1^2}< x_2 < sqrt{1-x_1^2}$, where we are taken positive roots:
begin{equation}
f_{X_1}(x_1) = int_{D} f_{X_1,X_2}(x_1,x_2) d x_2 = left{
begin{array}{@{}ll@{}}
int_{- sqrt{1-x_1^2}}^{ sqrt{1-x_1^2}} frac{1}{pi} d x_2 = frac{2sqrt{1-x_1^2}}{pi} & text{if} x_1^2 < 1 \
0, & text{otherwise}
end{array}right.
end{equation}
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
answered Jan 9 at 19:42
Carlos CamposCarlos Campos
57439
57439
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
add a comment |
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
$begingroup$
Thank you for your explanation.
$endgroup$
– Hillbilly Joe
Jan 9 at 19:46
add a comment |
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$begingroup$
I find different limits: $- sqrt(1-x^2)$ and $sqrt(1-x^2)$
$endgroup$
– Carlos Campos
Jan 9 at 19:35
$begingroup$
Thanks, just realized that I can do in such way
$endgroup$
– Hillbilly Joe
Jan 9 at 19:42