Mixing
Distribution of a random convergent sequence of nested intervals
$begingroup$
Starting from the interval $[0,1]$, generate two uniform random numbers $x_1,y_1$ and sort them so $x_1<y_1$. This yields an interval $[x_1,y_1]$. Generate two numbers uniformly from this interval, sort them as $x_2<y_2$, and repeat. This produces a nested sequence of intervals whose lengths decrease geometrically (on average), and converges to some $x^*$ almost surely. What is the distribution of $x^*$?
So far, I have the following: Let $f(a)=P(x^*le a)$. If the two random numbers $x_1,y_1$ are both smaller than $a$ then $x^*le a$ for sure. If $x_1<c<y_1$, then we have the same problem again with a new value for $a$, namely $frac{a-x_1}{y_1-x_1}$. Similarly if $y_1<c<x_1$. Thus we obtain the following integral equation for $f$:
$$f(a)=a^2+2int_0^aint_a^1fleft(frac{a-x}{y-x}right),dy,dx$$
At this point I'm stuck. Is this equation elementarily solvable?
real-analysis probability probability-distributions integral-equations
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
Starting from the interval $[0,1]$, generate two uniform random numbers $x_1,y_1$ and sort them so $x_1<y_1$. This yields an interval $[x_1,y_1]$. Generate two numbers uniformly from this interval, sort them as $x_2<y_2$, and repeat. This produces a nested sequence of intervals whose lengths decrease geometrically (on average), and converges to some $x^*$ almost surely. What is the distribution of $x^*$?
So far, I have the following: Let $f(a)=P(x^*le a)$. If the two random numbers $x_1,y_1$ are both smaller than $a$ then $x^*le a$ for sure. If $x_1<c<y_1$, then we have the same problem again with a new value for $a$, namely $frac{a-x_1}{y_1-x_1}$. Similarly if $y_1<c<x_1$. Thus we obtain the following integral equation for $f$:
$$f(a)=a^2+2int_0^aint_a^1fleft(frac{a-x}{y-x}right),dy,dx$$
At this point I'm stuck. Is this equation elementarily solvable?
real-analysis probability probability-distributions integral-equations
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
$begingroup$
I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
$endgroup$
– Paul Sinclair
Jan 27 at 17:55
$begingroup$
@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
$endgroup$
– Mario Carneiro
Jan 29 at 7:01
add a comment |
$begingroup$
Starting from the interval $[0,1]$, generate two uniform random numbers $x_1,y_1$ and sort them so $x_1<y_1$. This yields an interval $[x_1,y_1]$. Generate two numbers uniformly from this interval, sort them as $x_2<y_2$, and repeat. This produces a nested sequence of intervals whose lengths decrease geometrically (on average), and converges to some $x^*$ almost surely. What is the distribution of $x^*$?
So far, I have the following: Let $f(a)=P(x^*le a)$. If the two random numbers $x_1,y_1$ are both smaller than $a$ then $x^*le a$ for sure. If $x_1<c<y_1$, then we have the same problem again with a new value for $a$, namely $frac{a-x_1}{y_1-x_1}$. Similarly if $y_1<c<x_1$. Thus we obtain the following integral equation for $f$:
$$f(a)=a^2+2int_0^aint_a^1fleft(frac{a-x}{y-x}right),dy,dx$$
At this point I'm stuck. Is this equation elementarily solvable?
real-analysis probability probability-distributions integral-equations
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
Starting from the interval $[0,1]$, generate two uniform random numbers $x_1,y_1$ and sort them so $x_1<y_1$. This yields an interval $[x_1,y_1]$. Generate two numbers uniformly from this interval, sort them as $x_2<y_2$, and repeat. This produces a nested sequence of intervals whose lengths decrease geometrically (on average), and converges to some $x^*$ almost surely. What is the distribution of $x^*$?
So far, I have the following: Let $f(a)=P(x^*le a)$. If the two random numbers $x_1,y_1$ are both smaller than $a$ then $x^*le a$ for sure. If $x_1<c<y_1$, then we have the same problem again with a new value for $a$, namely $frac{a-x_1}{y_1-x_1}$. Similarly if $y_1<c<x_1$. Thus we obtain the following integral equation for $f$:
$$f(a)=a^2+2int_0^aint_a^1fleft(frac{a-x}{y-x}right),dy,dx$$
At this point I'm stuck. Is this equation elementarily solvable?
real-analysis probability probability-distributions integral-equations
real-analysis probability probability-distributions integral-equations
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
asked Jan 27 at 9:09
Mario CarneiroMario Carneiro
18.6k34090
18.6k34090
$begingroup$
I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
$endgroup$
– Paul Sinclair
Jan 27 at 17:55
$begingroup$
@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
$endgroup$
– Mario Carneiro
Jan 29 at 7:01
add a comment |
$begingroup$
I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
$endgroup$
– Paul Sinclair
Jan 27 at 17:55
$begingroup$
@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
$endgroup$
– Mario Carneiro
Jan 29 at 7:01
$begingroup$
I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
$endgroup$
– Paul Sinclair
Jan 27 at 17:55
$begingroup$
I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
$endgroup$
– Paul Sinclair
Jan 27 at 17:55
$begingroup$
@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
$endgroup$
– Mario Carneiro
Jan 29 at 7:01
$begingroup$
@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
$endgroup$
– Mario Carneiro
Jan 29 at 7:01
add a comment |
1 Answer
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$begingroup$
We can reparameterize the integral with $x=a-u$, $y=a+v$, and $t=frac u{u+v}$ to eliminate one of the integrals, and use $f(1-a)=1-f(a)$ to make the integral more symmetric:
begin{align*}
f(a)&=a^2+2int_0^aint_0^{1-a}fleft(frac{u}{u+v}right),dv,du\
&=a^2+2int_0^aint_{u/(1-a+u)}^1frac{u}{t^2}f(t),dt,du\
&=a^2+int_0^1frac{f(t)}{t^2}int_0^{min((1-a)t/(1-t),a)}2u,du,dt\
&=a^2+int_0^1frac{f(t)}{t^2}minleft(frac{(1-a)t}{1-t},aright)^2,dt\
&=a^2+int_a^1f(t)frac{a^2}{t^2},dt+int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt\
end{align*}
Let $g(a)=int_a^1f(t)frac{a^2}{t^2},dt$, $h(a)=int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt$, and $K=g(1)=int_0^1f(t)/t^2;dt$. Then we have the ODE:
begin{align*}
tag{1}f(a)&=a^2+g(a)+h(a)\
tag{2}g'(a)&=-f(a)+frac2ag(a)\
tag{3}h'(a)&=f(a)-frac2{1-a}h(a)\
tag{4}f(1/2)&=1/2\
end{align*}
Because $0$ is a singular point, it doesn't make a good initial condition, but it is possible to solve this ODE with an arbitrary choice of $h(1/2)$, and determine the value of $h(1/2)$ such that $f(0)=0$. Doing this numerically, I find $h(1/2)=1/16$ (and hence $g(1/2)=3/16$) to high accuracy. Here is a picture of $f(a)$ (top), $g(a)$ (middle) and $h(a)$ (bottom) for $ain [0,1]$.
If you think it looks a lot like a cubic, you're right! It's indistinguishable from the unique cubic that passes horizontally through $(0,0)$ and $(1,1)$, namely $f(a)=3a^2-2a^3$. And lo and behold, if you plug this into the original integral equation, it is a solution. Go figure.
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
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$begingroup$
We can reparameterize the integral with $x=a-u$, $y=a+v$, and $t=frac u{u+v}$ to eliminate one of the integrals, and use $f(1-a)=1-f(a)$ to make the integral more symmetric:
begin{align*}
f(a)&=a^2+2int_0^aint_0^{1-a}fleft(frac{u}{u+v}right),dv,du\
&=a^2+2int_0^aint_{u/(1-a+u)}^1frac{u}{t^2}f(t),dt,du\
&=a^2+int_0^1frac{f(t)}{t^2}int_0^{min((1-a)t/(1-t),a)}2u,du,dt\
&=a^2+int_0^1frac{f(t)}{t^2}minleft(frac{(1-a)t}{1-t},aright)^2,dt\
&=a^2+int_a^1f(t)frac{a^2}{t^2},dt+int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt\
end{align*}
Let $g(a)=int_a^1f(t)frac{a^2}{t^2},dt$, $h(a)=int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt$, and $K=g(1)=int_0^1f(t)/t^2;dt$. Then we have the ODE:
begin{align*}
tag{1}f(a)&=a^2+g(a)+h(a)\
tag{2}g'(a)&=-f(a)+frac2ag(a)\
tag{3}h'(a)&=f(a)-frac2{1-a}h(a)\
tag{4}f(1/2)&=1/2\
end{align*}
Because $0$ is a singular point, it doesn't make a good initial condition, but it is possible to solve this ODE with an arbitrary choice of $h(1/2)$, and determine the value of $h(1/2)$ such that $f(0)=0$. Doing this numerically, I find $h(1/2)=1/16$ (and hence $g(1/2)=3/16$) to high accuracy. Here is a picture of $f(a)$ (top), $g(a)$ (middle) and $h(a)$ (bottom) for $ain [0,1]$.
If you think it looks a lot like a cubic, you're right! It's indistinguishable from the unique cubic that passes horizontally through $(0,0)$ and $(1,1)$, namely $f(a)=3a^2-2a^3$. And lo and behold, if you plug this into the original integral equation, it is a solution. Go figure.
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
We can reparameterize the integral with $x=a-u$, $y=a+v$, and $t=frac u{u+v}$ to eliminate one of the integrals, and use $f(1-a)=1-f(a)$ to make the integral more symmetric:
begin{align*}
f(a)&=a^2+2int_0^aint_0^{1-a}fleft(frac{u}{u+v}right),dv,du\
&=a^2+2int_0^aint_{u/(1-a+u)}^1frac{u}{t^2}f(t),dt,du\
&=a^2+int_0^1frac{f(t)}{t^2}int_0^{min((1-a)t/(1-t),a)}2u,du,dt\
&=a^2+int_0^1frac{f(t)}{t^2}minleft(frac{(1-a)t}{1-t},aright)^2,dt\
&=a^2+int_a^1f(t)frac{a^2}{t^2},dt+int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt\
end{align*}
Let $g(a)=int_a^1f(t)frac{a^2}{t^2},dt$, $h(a)=int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt$, and $K=g(1)=int_0^1f(t)/t^2;dt$. Then we have the ODE:
begin{align*}
tag{1}f(a)&=a^2+g(a)+h(a)\
tag{2}g'(a)&=-f(a)+frac2ag(a)\
tag{3}h'(a)&=f(a)-frac2{1-a}h(a)\
tag{4}f(1/2)&=1/2\
end{align*}
Because $0$ is a singular point, it doesn't make a good initial condition, but it is possible to solve this ODE with an arbitrary choice of $h(1/2)$, and determine the value of $h(1/2)$ such that $f(0)=0$. Doing this numerically, I find $h(1/2)=1/16$ (and hence $g(1/2)=3/16$) to high accuracy. Here is a picture of $f(a)$ (top), $g(a)$ (middle) and $h(a)$ (bottom) for $ain [0,1]$.
If you think it looks a lot like a cubic, you're right! It's indistinguishable from the unique cubic that passes horizontally through $(0,0)$ and $(1,1)$, namely $f(a)=3a^2-2a^3$. And lo and behold, if you plug this into the original integral equation, it is a solution. Go figure.
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
add a comment |
$begingroup$
We can reparameterize the integral with $x=a-u$, $y=a+v$, and $t=frac u{u+v}$ to eliminate one of the integrals, and use $f(1-a)=1-f(a)$ to make the integral more symmetric:
begin{align*}
f(a)&=a^2+2int_0^aint_0^{1-a}fleft(frac{u}{u+v}right),dv,du\
&=a^2+2int_0^aint_{u/(1-a+u)}^1frac{u}{t^2}f(t),dt,du\
&=a^2+int_0^1frac{f(t)}{t^2}int_0^{min((1-a)t/(1-t),a)}2u,du,dt\
&=a^2+int_0^1frac{f(t)}{t^2}minleft(frac{(1-a)t}{1-t},aright)^2,dt\
&=a^2+int_a^1f(t)frac{a^2}{t^2},dt+int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt\
end{align*}
Let $g(a)=int_a^1f(t)frac{a^2}{t^2},dt$, $h(a)=int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt$, and $K=g(1)=int_0^1f(t)/t^2;dt$. Then we have the ODE:
begin{align*}
tag{1}f(a)&=a^2+g(a)+h(a)\
tag{2}g'(a)&=-f(a)+frac2ag(a)\
tag{3}h'(a)&=f(a)-frac2{1-a}h(a)\
tag{4}f(1/2)&=1/2\
end{align*}
Because $0$ is a singular point, it doesn't make a good initial condition, but it is possible to solve this ODE with an arbitrary choice of $h(1/2)$, and determine the value of $h(1/2)$ such that $f(0)=0$. Doing this numerically, I find $h(1/2)=1/16$ (and hence $g(1/2)=3/16$) to high accuracy. Here is a picture of $f(a)$ (top), $g(a)$ (middle) and $h(a)$ (bottom) for $ain [0,1]$.
If you think it looks a lot like a cubic, you're right! It's indistinguishable from the unique cubic that passes horizontally through $(0,0)$ and $(1,1)$, namely $f(a)=3a^2-2a^3$. And lo and behold, if you plug this into the original integral equation, it is a solution. Go figure.
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
$endgroup$
We can reparameterize the integral with $x=a-u$, $y=a+v$, and $t=frac u{u+v}$ to eliminate one of the integrals, and use $f(1-a)=1-f(a)$ to make the integral more symmetric:
begin{align*}
f(a)&=a^2+2int_0^aint_0^{1-a}fleft(frac{u}{u+v}right),dv,du\
&=a^2+2int_0^aint_{u/(1-a+u)}^1frac{u}{t^2}f(t),dt,du\
&=a^2+int_0^1frac{f(t)}{t^2}int_0^{min((1-a)t/(1-t),a)}2u,du,dt\
&=a^2+int_0^1frac{f(t)}{t^2}minleft(frac{(1-a)t}{1-t},aright)^2,dt\
&=a^2+int_a^1f(t)frac{a^2}{t^2},dt+int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt\
end{align*}
Let $g(a)=int_a^1f(t)frac{a^2}{t^2},dt$, $h(a)=int_0^af(t)frac{(1-a)^2}{(1-t)^2},dt$, and $K=g(1)=int_0^1f(t)/t^2;dt$. Then we have the ODE:
begin{align*}
tag{1}f(a)&=a^2+g(a)+h(a)\
tag{2}g'(a)&=-f(a)+frac2ag(a)\
tag{3}h'(a)&=f(a)-frac2{1-a}h(a)\
tag{4}f(1/2)&=1/2\
end{align*}
Because $0$ is a singular point, it doesn't make a good initial condition, but it is possible to solve this ODE with an arbitrary choice of $h(1/2)$, and determine the value of $h(1/2)$ such that $f(0)=0$. Doing this numerically, I find $h(1/2)=1/16$ (and hence $g(1/2)=3/16$) to high accuracy. Here is a picture of $f(a)$ (top), $g(a)$ (middle) and $h(a)$ (bottom) for $ain [0,1]$.
If you think it looks a lot like a cubic, you're right! It's indistinguishable from the unique cubic that passes horizontally through $(0,0)$ and $(1,1)$, namely $f(a)=3a^2-2a^3$. And lo and behold, if you plug this into the original integral equation, it is a solution. Go figure.
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
answered Jan 29 at 11:33
Mario CarneiroMario Carneiro
18.6k34090
18.6k34090
add a comment |
add a comment |
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Required, but never shown
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I think it might be more useful to determine the distribution of $(x_1, y_1)$ in $[0,1]^2$, then use this information and the similarity you've noted between the distributions of $x^*$ on $[0,1]$ and it distribution on $[x_1, y_1]$ to deduce $f$.
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– Paul Sinclair
Jan 27 at 17:55
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@PaulSinclair I'm afraid I don't see how that is any different from what I did. The distribution of $(x_1,y_1)$ in $[0,1]^2$ after sorting is just uniform on the triangle $0le xle yle 1$, and the similarity is between the distribution of $x^*$ on $[0,1]$ and the marginal distribution of $x^*$ on $[x_1,y_1]$ once a choice of $x_1$ and $y_1$ is made. Unfortunately this doesn't directly translate to a self-similarity of the function so much as a similarity between $f$ and an integral over $f$, i.e. the exact integral equation I wrote down.
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– Mario Carneiro
Jan 29 at 7:01