Why the expectation of distance to center of disk is $r/3$ and not $r/2$?
A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
probability expected-value
edited Nov 30 at 9:14
David G. Stork
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
|
show 1 more comment
A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
probability expected-value
edited Nov 30 at 9:14
David G. Stork
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
2
Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
1
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30
|
show 1 more comment
A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
probability expected-value
edited Nov 30 at 9:14
David G. Stork
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
A person throw an arrow on a target of radius $r$. The position of the arrow on the target is uniformly distributed. Let $X$ the distance between the arrow and the center of the circle. The score obtained by a person is $r-X$. What is the average score ? The answer is $frac{r}{3}$, whereas I found $frac{r}{2}$ as follow
We have that $X$ is uniform on $[0,r]$. If $Y=r-X$, then $$mathbb E[Y]=int_0^r (r-x)f_X(x)dx=frac{1}{r}int_0^r (r-x)dx=frac{r}{2}.$$
Maybe there is a subtlety than I don't see ?
probability expected-value
probability expected-value
edited Nov 30 at 9:14
David G. Stork
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
edited Nov 30 at 9:14
David G. Stork
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
edited Nov 30 at 9:14
David G. Stork
9,54721232
edited Nov 30 at 9:14
David G. Stork
9,54721232
edited Nov 30 at 9:14
David G. Stork
9,54721232
9,54721232
asked Nov 30 at 9:11
idm
8,57121345
asked Nov 30 at 9:11
idm
8,57121345
asked Nov 30 at 9:11
idm
8,57121345
8,57121345
2
Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
1
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30
|
show 1 more comment
2
Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
1
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30
2
2
Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
1
1
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30
|
show 1 more comment
3 Answers
3
active
oldest
votes
$$F_X(x) = frac{pi x^2}{pi r^2}$$
$$f_X(x)=frac{2x}{r^2}$$
begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
add a comment |
You may first construct the probability density as follows:
- At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$
So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$
answered Nov 30 at 9:43
trancelocation
9,0901521
add a comment |
$$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
answered Nov 30 at 9:23
AmbretteOrrisey
56710
add a comment |
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3 Answers
3
active
oldest
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3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$$F_X(x) = frac{pi x^2}{pi r^2}$$
$$f_X(x)=frac{2x}{r^2}$$
begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
add a comment |
$$F_X(x) = frac{pi x^2}{pi r^2}$$
$$f_X(x)=frac{2x}{r^2}$$
begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
add a comment |
$$F_X(x) = frac{pi x^2}{pi r^2}$$
$$f_X(x)=frac{2x}{r^2}$$
begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
$$F_X(x) = frac{pi x^2}{pi r^2}$$
$$f_X(x)=frac{2x}{r^2}$$
begin{align}
E[r-X]&=r-E[X] \
&=r - frac1{r^2}int_0^r 2x^2, dx\
&= r - frac1{r^2}frac{2r^3}3\
&= frac{r}{3}
end{align}
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
answered Nov 30 at 9:25
Siong Thye Goh
98.4k1463116
98.4k1463116
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
add a comment |
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
I don't get why $F_X(x)=frac{x^2}{r^2}$
– idm
Nov 30 at 9:29
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
That is the meaning of uniform over an area, if you draw a circle of the same size on the target, it is equally likely to hit either of them.
– Siong Thye Goh
Nov 30 at 9:30
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
Ok I see, thank you. But if $Z=(RcosTheta,RsinTheta)$ with $R$ uniform in $[0,r]$ and $Theta$ uniform on $[0,2pi]$, why $$mathbb P{|Z|leq x}=mathbb P{Rleq x,Thetain [0,2pi]}=mathbb P{Rleq x}=frac{x}{r}$$ is not true ? I really don't get this point
– idm
Nov 30 at 9:41
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
the assumption that $R$ is uniform in $[0,r]$ is not valid.
– Siong Thye Goh
Nov 30 at 9:45
ok, stange... thank you :)
– idm
Nov 30 at 9:46
ok, stange... thank you :)
– idm
Nov 30 at 9:46
add a comment |
You may first construct the probability density as follows:
- At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$
So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$
answered Nov 30 at 9:43
trancelocation
9,0901521
add a comment |
You may first construct the probability density as follows:
- At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$
So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$
answered Nov 30 at 9:43
trancelocation
9,0901521
add a comment |
You may first construct the probability density as follows:
- At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$
So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$
answered Nov 30 at 9:43
trancelocation
9,0901521
You may first construct the probability density as follows:
- At distance $x$ from the center of the circle a corresponding annulus of "thickness" $dx$ has a probability weight of
$$frac{1}{pi r^2}cdot 2pi cdot x cdot dx$$
So, you get
$$E(Y) = frac{1}{pi r^2} int_0^r (r-x)2pi cdot x; dx = cdots = frac{r}{3}$$
answered Nov 30 at 9:43
trancelocation
9,0901521
answered Nov 30 at 9:43
trancelocation
9,0901521
answered Nov 30 at 9:43
trancelocation
9,0901521
answered Nov 30 at 9:43
trancelocation
9,0901521
9,0901521
add a comment |
add a comment |
$$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
answered Nov 30 at 9:23
AmbretteOrrisey
56710
add a comment |
$$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
answered Nov 30 at 9:23
AmbretteOrrisey
56710
add a comment |
$$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
answered Nov 30 at 9:23
AmbretteOrrisey
56710
$$frac{int_0^r(r-x)x.dx}{int_0^r x.dx}$$
This gives $r/3$.
The problem with the integral you gave at first, is that the $(r-x)$ needs to be weighted by an $x.dx$, as this is the elemental area of the disc presented by a strip of thickness $dx$ at radius $x$ (or rather $2pi x.dx$ ... but then the $2pi$ appears on the top & bottom & obviously cancels ... or to put it another way, we could do the calculation for any sector of the disc & get the same result) - the catchment, if you like, at radius $x$.
Basically you're calculating the mean value of $r-x$ over a sector of a disc, or over a whole disc - it makes no difference.
answered Nov 30 at 9:23
AmbretteOrrisey
56710
edited Nov 30 at 9:37
answered Nov 30 at 9:23
AmbretteOrrisey
56710
answered Nov 30 at 9:23
AmbretteOrrisey
56710
answered Nov 30 at 9:23
AmbretteOrrisey
56710
56710
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Why do you think $X$ is uniform on $[0,r]$?
– 5xum
Nov 30 at 9:13
@5xum : I set $Z=(Rcos Theta, Rsin Theta)$ with $R$ uniform on $[0,r]$ and $Theta$ uniform on $[0,2pi)$. Then $mathbb P{Xleq x}=mathbb P{Rleq x, Theta in [0,2pi]}=frac{x}{r}.$ It doesn't work ?
– idm
Nov 30 at 9:14
Uniform over the disc does not mean the distribution of the distance from the center is uniform. The probability density of points in a circle sharing a centre with the disc will be inversely proportional to the radius of the circle (if less than $r$), not a constant.$$dfrac{mathsf d ~~}{mathsf d~x}mathsf P(Xleqslant x)~propto~dfrac{1}{x}mathbf 1_{0< xleqslant r}$$
– Graham Kemp
Nov 30 at 9:23
We just had a question like this half a day ago: Average distance from center of circle
– Rahul
Nov 30 at 9:25
1
$mathbb P{Xleq x} = frac{pi x^2}{pi r^2}$ as you are equally likely to land at any point on the area of the board, not equally likely to land at any radial distance from the centre.
– Paul
Nov 30 at 9:30