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Calculate distributions from random vector
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Let $(X,Y)$ be a random vector with $X,Y$ Normal Standard, independent
and $f_x,y(x,y) = frac12pie^-fracx^2+y^22$. Calculate
the distribution of $X^2+Y^2$ and $arctan(fracYX)$
I don't know how to approach this exercise. I've calculated the marginals of $X$ and $Y$ but i don't know where to go from here. I would appreciate any help.
probability-distributions random-variables normal-distribution
asked Jul 20 at 13:55
jscherman
1767
add a comment |Â
up vote
2
down vote
favorite
Let $(X,Y)$ be a random vector with $X,Y$ Normal Standard, independent
and $f_x,y(x,y) = frac12pie^-fracx^2+y^22$. Calculate
the distribution of $X^2+Y^2$ and $arctan(fracYX)$
I don't know how to approach this exercise. I've calculated the marginals of $X$ and $Y$ but i don't know where to go from here. I would appreciate any help.
probability-distributions random-variables normal-distribution
asked Jul 20 at 13:55
jscherman
1767
1
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Let $(X,Y)$ be a random vector with $X,Y$ Normal Standard, independent
and $f_x,y(x,y) = frac12pie^-fracx^2+y^22$. Calculate
the distribution of $X^2+Y^2$ and $arctan(fracYX)$
I don't know how to approach this exercise. I've calculated the marginals of $X$ and $Y$ but i don't know where to go from here. I would appreciate any help.
probability-distributions random-variables normal-distribution
asked Jul 20 at 13:55
jscherman
1767
Let $(X,Y)$ be a random vector with $X,Y$ Normal Standard, independent
and $f_x,y(x,y) = frac12pie^-fracx^2+y^22$. Calculate
the distribution of $X^2+Y^2$ and $arctan(fracYX)$
I don't know how to approach this exercise. I've calculated the marginals of $X$ and $Y$ but i don't know where to go from here. I would appreciate any help.
probability-distributions random-variables normal-distribution
asked Jul 20 at 13:55
jscherman
1767
asked Jul 20 at 13:55
jscherman
1767
asked Jul 20 at 13:55
jscherman
1767
asked Jul 20 at 13:55
jscherman
1767
1767
1
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22
add a comment |Â
1
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22
1
1
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
accepted
The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates ($x = r cos theta$ and $y = r sin theta$) change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < sqrtz$ and no restriction on $theta$)
$$F_Z(z) = int_0^2pifrac12pileft(int_0^sqrtzre^-r^2/2,dr right),dtheta = frac12e^-fracz2$$
The above is the CDF of an exponential distribution with parameter $lambda = frac12$.
The Phase
Let $alpha= arctan fracYX$
Let's compute $$F_alpha(alpha)=iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $arctan fracyx < alpha$ will give $theta < alpha$ and no restriction on $r$ this time)
$$F_alpha(alpha) = int_0^alphafrac12pileft(int_0^inftyre^-r^2/2,dr right),dtheta = frac12e^-fracz2 = fracalpha2 pi$$
which is a CDF of a uniform distribution over $[0,2pi]$.
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
accepted
The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates ($x = r cos theta$ and $y = r sin theta$) change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < sqrtz$ and no restriction on $theta$)
$$F_Z(z) = int_0^2pifrac12pileft(int_0^sqrtzre^-r^2/2,dr right),dtheta = frac12e^-fracz2$$
The above is the CDF of an exponential distribution with parameter $lambda = frac12$.
The Phase
Let $alpha= arctan fracYX$
Let's compute $$F_alpha(alpha)=iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $arctan fracyx < alpha$ will give $theta < alpha$ and no restriction on $r$ this time)
$$F_alpha(alpha) = int_0^alphafrac12pileft(int_0^inftyre^-r^2/2,dr right),dtheta = frac12e^-fracz2 = fracalpha2 pi$$
which is a CDF of a uniform distribution over $[0,2pi]$.
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
add a comment |Â
up vote
3
down vote
accepted
The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates ($x = r cos theta$ and $y = r sin theta$) change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < sqrtz$ and no restriction on $theta$)
$$F_Z(z) = int_0^2pifrac12pileft(int_0^sqrtzre^-r^2/2,dr right),dtheta = frac12e^-fracz2$$
The above is the CDF of an exponential distribution with parameter $lambda = frac12$.
The Phase
Let $alpha= arctan fracYX$
Let's compute $$F_alpha(alpha)=iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $arctan fracyx < alpha$ will give $theta < alpha$ and no restriction on $r$ this time)
$$F_alpha(alpha) = int_0^alphafrac12pileft(int_0^inftyre^-r^2/2,dr right),dtheta = frac12e^-fracz2 = fracalpha2 pi$$
which is a CDF of a uniform distribution over $[0,2pi]$.
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
add a comment |Â
up vote
3
down vote
accepted
up vote
3
down vote
accepted
The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates ($x = r cos theta$ and $y = r sin theta$) change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < sqrtz$ and no restriction on $theta$)
$$F_Z(z) = int_0^2pifrac12pileft(int_0^sqrtzre^-r^2/2,dr right),dtheta = frac12e^-fracz2$$
The above is the CDF of an exponential distribution with parameter $lambda = frac12$.
The Phase
Let $alpha= arctan fracYX$
Let's compute $$F_alpha(alpha)=iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $arctan fracyx < alpha$ will give $theta < alpha$ and no restriction on $r$ this time)
$$F_alpha(alpha) = int_0^alphafrac12pileft(int_0^inftyre^-r^2/2,dr right),dtheta = frac12e^-fracz2 = fracalpha2 pi$$
which is a CDF of a uniform distribution over $[0,2pi]$.
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
The magnitude
Let $Z = X^2 + Y^2$
$$F_Z(z)= Pr(Z < z) = iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates ($x = r cos theta$ and $y = r sin theta$) change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $X^2 + Y^2 < z$ will give $r < sqrtz$ and no restriction on $theta$)
$$F_Z(z) = int_0^2pifrac12pileft(int_0^sqrtzre^-r^2/2,dr right),dtheta = frac12e^-fracz2$$
The above is the CDF of an exponential distribution with parameter $lambda = frac12$.
The Phase
Let $alpha= arctan fracYX$
Let's compute $$F_alpha(alpha)=iint_D frac12pie^-(x^2+y^2)/2,dx,dy.$$
Doing the usual polar coordinates change of variable gives:
$$dx,dy=r,dr,dtheta$$
which yields (keeping in mind that $arctan fracyx < alpha$ will give $theta < alpha$ and no restriction on $r$ this time)
$$F_alpha(alpha) = int_0^alphafrac12pileft(int_0^inftyre^-r^2/2,dr right),dtheta = frac12e^-fracz2 = fracalpha2 pi$$
which is a CDF of a uniform distribution over $[0,2pi]$.
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
answered Jul 20 at 15:17
Ahmad Bazzi
2,6271417
2,6271417
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
add a comment |Â
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
1
1
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
It's written in your question $f_x,y(x,y) = frac12pie^-fracx^2+y^22$
– Ahmad Bazzi
Jul 20 at 18:37
add a comment |Â
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1
You didn't need to calculate the marginals, there were given from the get-go...Anyway, for the sum you could consider that the sum of the square of independent standard Normals follows a chi-square distribution. For the second task you could try to visualize the problem and use a geometric approach...
– user190080
Jul 20 at 14:31
@user190080 For the sum, how one ends up figuring that out? I mean, knowing that fact, i would calculate the momentum generator function and see that corresponds to the Chi Square's one, but what would you do to come up to that idea from the get-go? On the other hand, for the arctan i am not being able to visualize it any more tips for this?
– jscherman
Jul 20 at 14:55
@jscherman sorry my bad i removed the comment.
– Ahmad Bazzi
Jul 20 at 15:22