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CDF for distance between two points on line segment
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I am trying to solve a paper and have a basic question which I am unable to understand.
Given a line segment $[0,a]$, two points $x,y$ are randomly selected such that $mathcalD = |x-y| leq l$.
So the CDF would be
$$P(L leq l) = iint_mathcal DmathcalDf_XY(x,y)dxdy,$$
where $f_XY(x,y) = frac1a^2$ when $0 leq x,y leq a$, else $f_XY(x,y) = 0$
Now I dont understand how the author says, taking into account the bounds of both $mathcalD$ and $f_XY(x,y)$, the CDF for distance distribution is as follows:
$$P(Lleq l) = frac1a^2Big(int_0^lint_0^x+ldydx + int_l^a-lint_x-l^x+ldydx + int_a-l^aint_x-l^adydxBig)$$
Can someone explain 'geometrically' how this expression appears?
The actual problem is attached as the figure below.
probability-distributions geometric-probability
asked Jul 27 at 0:41
Kashan
341110
add a comment |Â
up vote
0
down vote
favorite
I am trying to solve a paper and have a basic question which I am unable to understand.
Given a line segment $[0,a]$, two points $x,y$ are randomly selected such that $mathcalD = |x-y| leq l$.
So the CDF would be
$$P(L leq l) = iint_mathcal DmathcalDf_XY(x,y)dxdy,$$
where $f_XY(x,y) = frac1a^2$ when $0 leq x,y leq a$, else $f_XY(x,y) = 0$
Now I dont understand how the author says, taking into account the bounds of both $mathcalD$ and $f_XY(x,y)$, the CDF for distance distribution is as follows:
$$P(Lleq l) = frac1a^2Big(int_0^lint_0^x+ldydx + int_l^a-lint_x-l^x+ldydx + int_a-l^aint_x-l^adydxBig)$$
Can someone explain 'geometrically' how this expression appears?
The actual problem is attached as the figure below.
probability-distributions geometric-probability
asked Jul 27 at 0:41
Kashan
341110
1
You can get proper spacing in $$ iint_mathcal D $$ by usingiint_mathcal Dinstead ofintint_mathcal D.
– joriki
Jul 27 at 4:52
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I am trying to solve a paper and have a basic question which I am unable to understand.
Given a line segment $[0,a]$, two points $x,y$ are randomly selected such that $mathcalD = |x-y| leq l$.
So the CDF would be
$$P(L leq l) = iint_mathcal DmathcalDf_XY(x,y)dxdy,$$
where $f_XY(x,y) = frac1a^2$ when $0 leq x,y leq a$, else $f_XY(x,y) = 0$
Now I dont understand how the author says, taking into account the bounds of both $mathcalD$ and $f_XY(x,y)$, the CDF for distance distribution is as follows:
$$P(Lleq l) = frac1a^2Big(int_0^lint_0^x+ldydx + int_l^a-lint_x-l^x+ldydx + int_a-l^aint_x-l^adydxBig)$$
Can someone explain 'geometrically' how this expression appears?
The actual problem is attached as the figure below.
probability-distributions geometric-probability
asked Jul 27 at 0:41
Kashan
341110
I am trying to solve a paper and have a basic question which I am unable to understand.
Given a line segment $[0,a]$, two points $x,y$ are randomly selected such that $mathcalD = |x-y| leq l$.
So the CDF would be
$$P(L leq l) = iint_mathcal DmathcalDf_XY(x,y)dxdy,$$
where $f_XY(x,y) = frac1a^2$ when $0 leq x,y leq a$, else $f_XY(x,y) = 0$
Now I dont understand how the author says, taking into account the bounds of both $mathcalD$ and $f_XY(x,y)$, the CDF for distance distribution is as follows:
$$P(Lleq l) = frac1a^2Big(int_0^lint_0^x+ldydx + int_l^a-lint_x-l^x+ldydx + int_a-l^aint_x-l^adydxBig)$$
Can someone explain 'geometrically' how this expression appears?
The actual problem is attached as the figure below.
probability-distributions geometric-probability
asked Jul 27 at 0:41
Kashan
341110
edited Jul 27 at 5:44
asked Jul 27 at 0:41
Kashan
341110
asked Jul 27 at 0:41
Kashan
341110
asked Jul 27 at 0:41
Kashan
341110
341110
1
You can get proper spacing in $$ iint_mathcal D $$ by usingiint_mathcal Dinstead ofintint_mathcal D.
– joriki
Jul 27 at 4:52
add a comment |Â
1
You can get proper spacing in $$ iint_mathcal D $$ by usingiint_mathcal Dinstead ofintint_mathcal D.
– joriki
Jul 27 at 4:52
1
1
You can get proper spacing in $$ iint_mathcal D $$ by using
iint_mathcal D instead of intint_mathcal D.– joriki
Jul 27 at 4:52
You can get proper spacing in $$ iint_mathcal D $$ by using
iint_mathcal D instead of intint_mathcal D.– joriki
Jul 27 at 4:52
add a comment |Â
1 Answer
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The text is cutting a corner here by not saying why this works for $lgtfrac a2$. I'll first explain it for $llefrac a2$.
The outer integration is over $x$, the inner integration is over $y$. We need to distinguish three cases for the outer integration because the bounds for $y$ are different in each case.
- If $xle l$, then the lower limit for $y$ is $0$, because any $y$ below $x$ is within $l$ of $x$. The upper limit is $x+l$. This is the first double integral.
- If $llt xlt a-l$, then any $y$ within $l$ of $x$ lies in $[0,a]$, so the lower and upper limit for $y$ are $x-l$ and $x+l$, respectively. This is the second double integral.
- If $xge a-l$, then the upper limit for $y$ is $a$, because any $y$ above $x$ is within $l$ of $x$. The lower limit is $x-l$. This is the third double integral.
Now comes the tricky part: If $lgtfrac a2$, the second double integral has the limit for $x$ reversed, so it's negative. This is exactly what we need to cancel the overcounting due to the fact that for $xin[a-l,l]$, in the first and third double integrals the limits $x+l$ and $x-l$, respectively, went beyond $[0,a]$.
answered Jul 27 at 5:08
joriki
164k10179328
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
The text is cutting a corner here by not saying why this works for $lgtfrac a2$. I'll first explain it for $llefrac a2$.
The outer integration is over $x$, the inner integration is over $y$. We need to distinguish three cases for the outer integration because the bounds for $y$ are different in each case.
- If $xle l$, then the lower limit for $y$ is $0$, because any $y$ below $x$ is within $l$ of $x$. The upper limit is $x+l$. This is the first double integral.
- If $llt xlt a-l$, then any $y$ within $l$ of $x$ lies in $[0,a]$, so the lower and upper limit for $y$ are $x-l$ and $x+l$, respectively. This is the second double integral.
- If $xge a-l$, then the upper limit for $y$ is $a$, because any $y$ above $x$ is within $l$ of $x$. The lower limit is $x-l$. This is the third double integral.
Now comes the tricky part: If $lgtfrac a2$, the second double integral has the limit for $x$ reversed, so it's negative. This is exactly what we need to cancel the overcounting due to the fact that for $xin[a-l,l]$, in the first and third double integrals the limits $x+l$ and $x-l$, respectively, went beyond $[0,a]$.
answered Jul 27 at 5:08
joriki
164k10179328
add a comment |Â
up vote
1
down vote
accepted
The text is cutting a corner here by not saying why this works for $lgtfrac a2$. I'll first explain it for $llefrac a2$.
The outer integration is over $x$, the inner integration is over $y$. We need to distinguish three cases for the outer integration because the bounds for $y$ are different in each case.
- If $xle l$, then the lower limit for $y$ is $0$, because any $y$ below $x$ is within $l$ of $x$. The upper limit is $x+l$. This is the first double integral.
- If $llt xlt a-l$, then any $y$ within $l$ of $x$ lies in $[0,a]$, so the lower and upper limit for $y$ are $x-l$ and $x+l$, respectively. This is the second double integral.
- If $xge a-l$, then the upper limit for $y$ is $a$, because any $y$ above $x$ is within $l$ of $x$. The lower limit is $x-l$. This is the third double integral.
Now comes the tricky part: If $lgtfrac a2$, the second double integral has the limit for $x$ reversed, so it's negative. This is exactly what we need to cancel the overcounting due to the fact that for $xin[a-l,l]$, in the first and third double integrals the limits $x+l$ and $x-l$, respectively, went beyond $[0,a]$.
answered Jul 27 at 5:08
joriki
164k10179328
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
The text is cutting a corner here by not saying why this works for $lgtfrac a2$. I'll first explain it for $llefrac a2$.
The outer integration is over $x$, the inner integration is over $y$. We need to distinguish three cases for the outer integration because the bounds for $y$ are different in each case.
- If $xle l$, then the lower limit for $y$ is $0$, because any $y$ below $x$ is within $l$ of $x$. The upper limit is $x+l$. This is the first double integral.
- If $llt xlt a-l$, then any $y$ within $l$ of $x$ lies in $[0,a]$, so the lower and upper limit for $y$ are $x-l$ and $x+l$, respectively. This is the second double integral.
- If $xge a-l$, then the upper limit for $y$ is $a$, because any $y$ above $x$ is within $l$ of $x$. The lower limit is $x-l$. This is the third double integral.
Now comes the tricky part: If $lgtfrac a2$, the second double integral has the limit for $x$ reversed, so it's negative. This is exactly what we need to cancel the overcounting due to the fact that for $xin[a-l,l]$, in the first and third double integrals the limits $x+l$ and $x-l$, respectively, went beyond $[0,a]$.
answered Jul 27 at 5:08
joriki
164k10179328
The text is cutting a corner here by not saying why this works for $lgtfrac a2$. I'll first explain it for $llefrac a2$.
The outer integration is over $x$, the inner integration is over $y$. We need to distinguish three cases for the outer integration because the bounds for $y$ are different in each case.
- If $xle l$, then the lower limit for $y$ is $0$, because any $y$ below $x$ is within $l$ of $x$. The upper limit is $x+l$. This is the first double integral.
- If $llt xlt a-l$, then any $y$ within $l$ of $x$ lies in $[0,a]$, so the lower and upper limit for $y$ are $x-l$ and $x+l$, respectively. This is the second double integral.
- If $xge a-l$, then the upper limit for $y$ is $a$, because any $y$ above $x$ is within $l$ of $x$. The lower limit is $x-l$. This is the third double integral.
Now comes the tricky part: If $lgtfrac a2$, the second double integral has the limit for $x$ reversed, so it's negative. This is exactly what we need to cancel the overcounting due to the fact that for $xin[a-l,l]$, in the first and third double integrals the limits $x+l$ and $x-l$, respectively, went beyond $[0,a]$.
answered Jul 27 at 5:08
joriki
164k10179328
answered Jul 27 at 5:08
joriki
164k10179328
answered Jul 27 at 5:08
joriki
164k10179328
answered Jul 27 at 5:08
joriki
164k10179328
164k10179328
add a comment |Â
add a comment |Â
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1
You can get proper spacing in $$ iint_mathcal D $$ by using
iint_mathcal Dinstead ofintint_mathcal D.– joriki
Jul 27 at 4:52