Mixing
Determine the distribution function of a random variable
$begingroup$
I am a bit lost with the following exercise, I would really appreciate your help:
$$
f_X,_Y(x,y)
begin{cases}
c & x^2 + y^2leq 4\
0 & otherwise
end{cases}
$$
What is the distribution function of :
$$
theta=
begin{cases}
arctan frac{X}{Y} & X ne 0\
0 & X=0
end{cases}
$$
Well, I don't even have a clue of how to start with this one.
Thanks
probability
asked Jan 15 at 14:54
superuser123superuser123
47628
$endgroup$
add a comment |
$begingroup$
I am a bit lost with the following exercise, I would really appreciate your help:
$$
f_X,_Y(x,y)
begin{cases}
c & x^2 + y^2leq 4\
0 & otherwise
end{cases}
$$
What is the distribution function of :
$$
theta=
begin{cases}
arctan frac{X}{Y} & X ne 0\
0 & X=0
end{cases}
$$
Well, I don't even have a clue of how to start with this one.
Thanks
probability
asked Jan 15 at 14:54
superuser123superuser123
47628
$endgroup$
add a comment |
$begingroup$
I am a bit lost with the following exercise, I would really appreciate your help:
$$
f_X,_Y(x,y)
begin{cases}
c & x^2 + y^2leq 4\
0 & otherwise
end{cases}
$$
What is the distribution function of :
$$
theta=
begin{cases}
arctan frac{X}{Y} & X ne 0\
0 & X=0
end{cases}
$$
Well, I don't even have a clue of how to start with this one.
Thanks
probability
asked Jan 15 at 14:54
superuser123superuser123
47628
$endgroup$
I am a bit lost with the following exercise, I would really appreciate your help:
$$
f_X,_Y(x,y)
begin{cases}
c & x^2 + y^2leq 4\
0 & otherwise
end{cases}
$$
What is the distribution function of :
$$
theta=
begin{cases}
arctan frac{X}{Y} & X ne 0\
0 & X=0
end{cases}
$$
Well, I don't even have a clue of how to start with this one.
Thanks
probability
probability
asked Jan 15 at 14:54
superuser123superuser123
47628
asked Jan 15 at 14:54
superuser123superuser123
47628
asked Jan 15 at 14:54
superuser123superuser123
47628
asked Jan 15 at 14:54
superuser123superuser123
47628
asked Jan 15 at 14:54
superuser123superuser123
47628
47628
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
add a comment |
$begingroup$
First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
add a comment |
$begingroup$
First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
$endgroup$
First note that since $$int f_{XY}(x,y)dxdy=1$$we obtain $$c={1over 4pi}$$and since the PDF is a circle (more precisely a cylinder) and is maintained under rotation then the distribution of $theta$ is uniform, i.e.$$thetasim Uleft(-{piover 2},{piover 2}right)$$according to standard definition of $arctan$.
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
answered Jan 15 at 15:14
Mostafa AyazMostafa Ayaz
15.6k3939
15.6k3939
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
add a comment |
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
could you please clarify the rotation thing? I am not sure how you could decide that $arctan frac{X}{Y}$ is distributed uniformly
$endgroup$
– superuser123
Jan 15 at 15:21
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
$begingroup$
Surely $theta$ is radial. Since no radius of circle is distinguishable from the others and the PDF of $X$ and $Y$ is uniform we argue that so must be that of $theta$
$endgroup$
– Mostafa Ayaz
Jan 16 at 12:02
add a comment |
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