Mixing
Probablity of hitting a square touching the inside of the bull's eye
Clash Royale CLAN TAG#URR8PPP
up vote
1
down vote
favorite
Here, X and Y are independent N(0,1).
Can someone explains to me how do we get (sqrt(2))/2 using rotational symmetry?
I do understand why r = sqrt(2ln(2)), but I do not understand the rotational symmetry part.
Thanks in advance.
probability
asked Jul 25 at 3:39
ibuntu
566
add a comment |Â
up vote
1
down vote
favorite
Here, X and Y are independent N(0,1).
Can someone explains to me how do we get (sqrt(2))/2 using rotational symmetry?
I do understand why r = sqrt(2ln(2)), but I do not understand the rotational symmetry part.
Thanks in advance.
probability
asked Jul 25 at 3:39
ibuntu
566
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Here, X and Y are independent N(0,1).
Can someone explains to me how do we get (sqrt(2))/2 using rotational symmetry?
I do understand why r = sqrt(2ln(2)), but I do not understand the rotational symmetry part.
Thanks in advance.
probability
asked Jul 25 at 3:39
ibuntu
566
Here, X and Y are independent N(0,1).
Can someone explains to me how do we get (sqrt(2))/2 using rotational symmetry?
I do understand why r = sqrt(2ln(2)), but I do not understand the rotational symmetry part.
Thanks in advance.
probability
asked Jul 25 at 3:39
ibuntu
566
asked Jul 25 at 3:39
ibuntu
566
asked Jul 25 at 3:39
ibuntu
566
asked Jul 25 at 3:39
ibuntu
566
566
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07
add a comment |Â
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
2
down vote
accepted
If we just look at the figure as shown, it is bounded by the lines
$lvert X + Yrvert = r$ and $lvert X - Yrvert = r,$
and not by the lines
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
But we can rotate the square by $45$ degrees and then it is bounded by
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
Because of the rotational symmetry of the distribution, we know that the rotation did not affect the probability.
answered Jul 25 at 4:36
David K
48.1k340107
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
2
down vote
accepted
If we just look at the figure as shown, it is bounded by the lines
$lvert X + Yrvert = r$ and $lvert X - Yrvert = r,$
and not by the lines
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
But we can rotate the square by $45$ degrees and then it is bounded by
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
Because of the rotational symmetry of the distribution, we know that the rotation did not affect the probability.
answered Jul 25 at 4:36
David K
48.1k340107
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
add a comment |Â
up vote
2
down vote
accepted
If we just look at the figure as shown, it is bounded by the lines
$lvert X + Yrvert = r$ and $lvert X - Yrvert = r,$
and not by the lines
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
But we can rotate the square by $45$ degrees and then it is bounded by
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
Because of the rotational symmetry of the distribution, we know that the rotation did not affect the probability.
answered Jul 25 at 4:36
David K
48.1k340107
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
add a comment |Â
up vote
2
down vote
accepted
up vote
2
down vote
accepted
If we just look at the figure as shown, it is bounded by the lines
$lvert X + Yrvert = r$ and $lvert X - Yrvert = r,$
and not by the lines
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
But we can rotate the square by $45$ degrees and then it is bounded by
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
Because of the rotational symmetry of the distribution, we know that the rotation did not affect the probability.
answered Jul 25 at 4:36
David K
48.1k340107
If we just look at the figure as shown, it is bounded by the lines
$lvert X + Yrvert = r$ and $lvert X - Yrvert = r,$
and not by the lines
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
But we can rotate the square by $45$ degrees and then it is bounded by
$lvert X rvert = fracsqrt22r$
and $lvert Yrvert = fracsqrt22r.$
Because of the rotational symmetry of the distribution, we know that the rotation did not affect the probability.
answered Jul 25 at 4:36
David K
48.1k340107
answered Jul 25 at 4:36
David K
48.1k340107
answered Jul 25 at 4:36
David K
48.1k340107
answered Jul 25 at 4:36
David K
48.1k340107
48.1k340107
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
add a comment |Â
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
How do we get the (sqrt(2))/2? Does it have to do with the r?
– ibuntu
Jul 25 at 6:00
1
1
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
$r$ is half the diagonal of the square, but when we put the square so its sides are parallel to the axes, the maximum $x$ coordinate is only half the side of the square. The side is $fracsqrt22$ times as long as the diagonal. Another way to see it: draw the square with sides parallel to the axes, centered at the origin. Draw a diagonal segment from the origin to a vertex of the square. Label the length of that segment: it is $r.$ Find the coordinates of the vertex.
– David K
Jul 25 at 12:03
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2861989%2fprobablity-of-hitting-a-square-touching-the-inside-of-the-bulls-eye%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
How did the logarithms get in there?
– joriki
Jul 25 at 9:29
Because the P(shots inside the bull's eye) = 0.5 and then by using F(r)=1-e^(r^2 /2)=1/2, the r obtained is sqrt(2ln(2))
– ibuntu
Jul 25 at 12:50
Aha. Am I missing where it says that the probability for hitting the bull's eye is $frac12$? Or did you forget to state this in the question?
– joriki
Jul 25 at 13:07