The $ell_2$ norm of two stochastic vectors












0












$begingroup$


I would like to know why this is true:




Given an $n$-dimensional stochastic vector $v$ and another $n$-dimensional stochastic vector $u$ which distributes uniformly over ${1,dots,n}$ (meaning if the length of the vector $u$ is $n=3$ then $u = (1/3,1/3,1/3)$) then $lVert u-vrVert_2 leq 1$.




For example, let $v = (0.8,0.2)$ and $v = (0.5,0.5)$: $lVert u-vrVert_2 = lVert(0.3,-0.3)rVert_2=sqrt{2cdot0.3^2}$
which is less than 1.



I tried many different inputs and it works for all of them but I can’t figure out why?






inequality probability-distributions vectors stochastic-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:07










  • $begingroup$
    @herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:09






  • 1




    $begingroup$
    There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:17










  • $begingroup$
    @herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:20










  • $begingroup$
    If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
    $endgroup$
    – Clement C.
    Dec 17 '18 at 21:32


















0












$begingroup$


I would like to know why this is true:




Given an $n$-dimensional stochastic vector $v$ and another $n$-dimensional stochastic vector $u$ which distributes uniformly over ${1,dots,n}$ (meaning if the length of the vector $u$ is $n=3$ then $u = (1/3,1/3,1/3)$) then $lVert u-vrVert_2 leq 1$.




For example, let $v = (0.8,0.2)$ and $v = (0.5,0.5)$: $lVert u-vrVert_2 = lVert(0.3,-0.3)rVert_2=sqrt{2cdot0.3^2}$
which is less than 1.



I tried many different inputs and it works for all of them but I can’t figure out why?






inequality probability-distributions vectors stochastic-calculus







share|cite|improve this question











$endgroup$












  • $begingroup$
    Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:07










  • $begingroup$
    @herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:09






  • 1




    $begingroup$
    There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:17










  • $begingroup$
    @herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:20










  • $begingroup$
    If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
    $endgroup$
    – Clement C.
    Dec 17 '18 at 21:32
















0












0








0





$begingroup$


I would like to know why this is true:




Given an $n$-dimensional stochastic vector $v$ and another $n$-dimensional stochastic vector $u$ which distributes uniformly over ${1,dots,n}$ (meaning if the length of the vector $u$ is $n=3$ then $u = (1/3,1/3,1/3)$) then $lVert u-vrVert_2 leq 1$.




For example, let $v = (0.8,0.2)$ and $v = (0.5,0.5)$: $lVert u-vrVert_2 = lVert(0.3,-0.3)rVert_2=sqrt{2cdot0.3^2}$
which is less than 1.



I tried many different inputs and it works for all of them but I can’t figure out why?






inequality probability-distributions vectors stochastic-calculus







share|cite|improve this question











$endgroup$




I would like to know why this is true:




Given an $n$-dimensional stochastic vector $v$ and another $n$-dimensional stochastic vector $u$ which distributes uniformly over ${1,dots,n}$ (meaning if the length of the vector $u$ is $n=3$ then $u = (1/3,1/3,1/3)$) then $lVert u-vrVert_2 leq 1$.




For example, let $v = (0.8,0.2)$ and $v = (0.5,0.5)$: $lVert u-vrVert_2 = lVert(0.3,-0.3)rVert_2=sqrt{2cdot0.3^2}$
which is less than 1.



I tried many different inputs and it works for all of them but I can’t figure out why?






inequality probability-distributions vectors stochastic-calculus




inequality probability-distributions vectors stochastic-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 16 '18 at 18:52









Clement C.

50.9k33992




50.9k33992










asked Dec 16 '18 at 17:49









נירייב שמואלנירייב שמואל

195




195












  • $begingroup$
    Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:07










  • $begingroup$
    @herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:09






  • 1




    $begingroup$
    There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:17










  • $begingroup$
    @herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:20










  • $begingroup$
    If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
    $endgroup$
    – Clement C.
    Dec 17 '18 at 21:32




















  • $begingroup$
    Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:07










  • $begingroup$
    @herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:09






  • 1




    $begingroup$
    There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
    $endgroup$
    – herb steinberg
    Dec 16 '18 at 18:17










  • $begingroup$
    @herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
    $endgroup$
    – נירייב שמואל
    Dec 16 '18 at 18:20










  • $begingroup$
    If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
    $endgroup$
    – Clement C.
    Dec 17 '18 at 21:32


















$begingroup$
Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
$endgroup$
– herb steinberg
Dec 16 '18 at 18:07




$begingroup$
Your statement is too confusing. Are you asking about 2-d or 3-d vectors?
$endgroup$
– herb steinberg
Dec 16 '18 at 18:07












$begingroup$
@herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
$endgroup$
– נירייב שמואל
Dec 16 '18 at 18:09




$begingroup$
@herbsteinberg I am asking for n-d vectors. the 2-d and 3-d vectors where just an example in order to simplify the problem.
$endgroup$
– נירייב שמואל
Dec 16 '18 at 18:09




1




1




$begingroup$
There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
$endgroup$
– herb steinberg
Dec 16 '18 at 18:17




$begingroup$
There is other confusing stuff. For example U length 3 and U=(1/3,1/3,1/3) is wrong. In general $|v-u|le 1$ is not true without more descriptions for $u$ and $v$.
$endgroup$
– herb steinberg
Dec 16 '18 at 18:17












$begingroup$
@herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
$endgroup$
– נירייב שמואל
Dec 16 '18 at 18:20




$begingroup$
@herbsteinberg that's why I indicated that U distribute uniformly between 1 untill n.
$endgroup$
– נירייב שמואל
Dec 16 '18 at 18:20












$begingroup$
If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
$endgroup$
– Clement C.
Dec 17 '18 at 21:32






$begingroup$
If I may, why are there votes to close with no comment on it? Sure, the OP hasnt included their attempts (if any), but then the minimum here (for a new contributor) would be to point it out instead of just voting for closure... "נירייב שמואל is a new contributor to this site. Take care in asking for clarification, commenting, and answering."
$endgroup$
– Clement C.
Dec 17 '18 at 21:32












1 Answer
1






active

oldest

votes


















1












$begingroup$

First, you can show it will be at most 2 by the triangle inequality, and even a bit better: for $u,v$ two $n$-dimensional vectors in the probability simplex with $u$ being uniform,
$$
lVert u-vrVert_2 leq lVert urVert_2+ lVert vrVert_2 = frac{1}{sqrt{n}} + 1
$$



Now, for the stronger claim. Observe that
$$begin{align}
lVert u-vrVert_2^2 &= sum_{k=1}^n (u_k-v_k)^2 =
sum_{k=1}^n u_k^2 + sum_{k=1}^n v_k^2 - 2sum_{k=1}^n u_k v_k
= ncdot frac{1}{n^2} + lVert vrVert_2^2 - frac{2}{n}sum_{k=1}^n v_k\
&= frac{1}{n} + lVert vrVert_2^2 - frac{2}{n}
= lVert vrVert_2^2 - frac{1}{n} \
&leq 1 - frac{1}{n}
end{align}$$
where the last inequality comes from the fact that $$lVert vrVert_2^2 = sum_{k=1}^n v_k^2 leq sum_{k=1}^n v_k = 1$$
(more generally, $lVert vrVert_p leq lVert vrVert_q$ if $pgeq qgeq 1$).




Upshot: the squared $ell_2$ distance of a probability vector to uniformity is equal, up to an additive factor $1/n$, to its squared $ell_2$ norm:
$$
lVert u-vrVert_2^2 = lVert vrVert_2^2 - frac{1}{n}
$$
which can itself be interpreted as the collision probability
$$
lVert vrVert_2^2 = mathbb{P}_{v}{ X=Y}
$$
where $X,Y$ are independent random variables with pmf $v$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
    $endgroup$
    – Clement C.
    Dec 16 '18 at 18:49













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1 Answer
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1 Answer
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1












$begingroup$

First, you can show it will be at most 2 by the triangle inequality, and even a bit better: for $u,v$ two $n$-dimensional vectors in the probability simplex with $u$ being uniform,
$$
lVert u-vrVert_2 leq lVert urVert_2+ lVert vrVert_2 = frac{1}{sqrt{n}} + 1
$$



Now, for the stronger claim. Observe that
$$begin{align}
lVert u-vrVert_2^2 &= sum_{k=1}^n (u_k-v_k)^2 =
sum_{k=1}^n u_k^2 + sum_{k=1}^n v_k^2 - 2sum_{k=1}^n u_k v_k
= ncdot frac{1}{n^2} + lVert vrVert_2^2 - frac{2}{n}sum_{k=1}^n v_k\
&= frac{1}{n} + lVert vrVert_2^2 - frac{2}{n}
= lVert vrVert_2^2 - frac{1}{n} \
&leq 1 - frac{1}{n}
end{align}$$
where the last inequality comes from the fact that $$lVert vrVert_2^2 = sum_{k=1}^n v_k^2 leq sum_{k=1}^n v_k = 1$$
(more generally, $lVert vrVert_p leq lVert vrVert_q$ if $pgeq qgeq 1$).




Upshot: the squared $ell_2$ distance of a probability vector to uniformity is equal, up to an additive factor $1/n$, to its squared $ell_2$ norm:
$$
lVert u-vrVert_2^2 = lVert vrVert_2^2 - frac{1}{n}
$$
which can itself be interpreted as the collision probability
$$
lVert vrVert_2^2 = mathbb{P}_{v}{ X=Y}
$$
where $X,Y$ are independent random variables with pmf $v$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
    $endgroup$
    – Clement C.
    Dec 16 '18 at 18:49


















1












$begingroup$

First, you can show it will be at most 2 by the triangle inequality, and even a bit better: for $u,v$ two $n$-dimensional vectors in the probability simplex with $u$ being uniform,
$$
lVert u-vrVert_2 leq lVert urVert_2+ lVert vrVert_2 = frac{1}{sqrt{n}} + 1
$$



Now, for the stronger claim. Observe that
$$begin{align}
lVert u-vrVert_2^2 &= sum_{k=1}^n (u_k-v_k)^2 =
sum_{k=1}^n u_k^2 + sum_{k=1}^n v_k^2 - 2sum_{k=1}^n u_k v_k
= ncdot frac{1}{n^2} + lVert vrVert_2^2 - frac{2}{n}sum_{k=1}^n v_k\
&= frac{1}{n} + lVert vrVert_2^2 - frac{2}{n}
= lVert vrVert_2^2 - frac{1}{n} \
&leq 1 - frac{1}{n}
end{align}$$
where the last inequality comes from the fact that $$lVert vrVert_2^2 = sum_{k=1}^n v_k^2 leq sum_{k=1}^n v_k = 1$$
(more generally, $lVert vrVert_p leq lVert vrVert_q$ if $pgeq qgeq 1$).




Upshot: the squared $ell_2$ distance of a probability vector to uniformity is equal, up to an additive factor $1/n$, to its squared $ell_2$ norm:
$$
lVert u-vrVert_2^2 = lVert vrVert_2^2 - frac{1}{n}
$$
which can itself be interpreted as the collision probability
$$
lVert vrVert_2^2 = mathbb{P}_{v}{ X=Y}
$$
where $X,Y$ are independent random variables with pmf $v$.







share|cite|improve this answer











$endgroup$













  • $begingroup$
    Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
    $endgroup$
    – Clement C.
    Dec 16 '18 at 18:49
















1












1








1





$begingroup$

First, you can show it will be at most 2 by the triangle inequality, and even a bit better: for $u,v$ two $n$-dimensional vectors in the probability simplex with $u$ being uniform,
$$
lVert u-vrVert_2 leq lVert urVert_2+ lVert vrVert_2 = frac{1}{sqrt{n}} + 1
$$



Now, for the stronger claim. Observe that
$$begin{align}
lVert u-vrVert_2^2 &= sum_{k=1}^n (u_k-v_k)^2 =
sum_{k=1}^n u_k^2 + sum_{k=1}^n v_k^2 - 2sum_{k=1}^n u_k v_k
= ncdot frac{1}{n^2} + lVert vrVert_2^2 - frac{2}{n}sum_{k=1}^n v_k\
&= frac{1}{n} + lVert vrVert_2^2 - frac{2}{n}
= lVert vrVert_2^2 - frac{1}{n} \
&leq 1 - frac{1}{n}
end{align}$$
where the last inequality comes from the fact that $$lVert vrVert_2^2 = sum_{k=1}^n v_k^2 leq sum_{k=1}^n v_k = 1$$
(more generally, $lVert vrVert_p leq lVert vrVert_q$ if $pgeq qgeq 1$).




Upshot: the squared $ell_2$ distance of a probability vector to uniformity is equal, up to an additive factor $1/n$, to its squared $ell_2$ norm:
$$
lVert u-vrVert_2^2 = lVert vrVert_2^2 - frac{1}{n}
$$
which can itself be interpreted as the collision probability
$$
lVert vrVert_2^2 = mathbb{P}_{v}{ X=Y}
$$
where $X,Y$ are independent random variables with pmf $v$.







share|cite|improve this answer











$endgroup$



First, you can show it will be at most 2 by the triangle inequality, and even a bit better: for $u,v$ two $n$-dimensional vectors in the probability simplex with $u$ being uniform,
$$
lVert u-vrVert_2 leq lVert urVert_2+ lVert vrVert_2 = frac{1}{sqrt{n}} + 1
$$



Now, for the stronger claim. Observe that
$$begin{align}
lVert u-vrVert_2^2 &= sum_{k=1}^n (u_k-v_k)^2 =
sum_{k=1}^n u_k^2 + sum_{k=1}^n v_k^2 - 2sum_{k=1}^n u_k v_k
= ncdot frac{1}{n^2} + lVert vrVert_2^2 - frac{2}{n}sum_{k=1}^n v_k\
&= frac{1}{n} + lVert vrVert_2^2 - frac{2}{n}
= lVert vrVert_2^2 - frac{1}{n} \
&leq 1 - frac{1}{n}
end{align}$$
where the last inequality comes from the fact that $$lVert vrVert_2^2 = sum_{k=1}^n v_k^2 leq sum_{k=1}^n v_k = 1$$
(more generally, $lVert vrVert_p leq lVert vrVert_q$ if $pgeq qgeq 1$).




Upshot: the squared $ell_2$ distance of a probability vector to uniformity is equal, up to an additive factor $1/n$, to its squared $ell_2$ norm:
$$
lVert u-vrVert_2^2 = lVert vrVert_2^2 - frac{1}{n}
$$
which can itself be interpreted as the collision probability
$$
lVert vrVert_2^2 = mathbb{P}_{v}{ X=Y}
$$
where $X,Y$ are independent random variables with pmf $v$.








share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 17 '18 at 0:07

























answered Dec 16 '18 at 18:34









Clement C.Clement C.

50.9k33992




50.9k33992












  • $begingroup$
    Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
    $endgroup$
    – Clement C.
    Dec 16 '18 at 18:49




















  • $begingroup$
    Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
    $endgroup$
    – Clement C.
    Dec 16 '18 at 18:49


















$begingroup$
Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
$endgroup$
– Clement C.
Dec 16 '18 at 18:49






$begingroup$
Note: the reason for the last comment in the "upshot" is that this view of the $ell_2$ distance to uniform as directly related to the collision probability is the basis for most uniformity testing algorithms (given i.i.d. realizations from a discrete r.v., decide whether its distribution is uniform vs. far from uniform). I figured it was worth mentioning.
$endgroup$
– Clement C.
Dec 16 '18 at 18:49




















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Willebadessen

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Wappen Deutschlandkarte 51.633055555556 9.0330555555556 295 Koordinaten: 51° 38′  N , 9° 2′  O Basisdaten Bundesland: Nordrhein-Westfalen Regierungsbezirk: Detmold Kreis: Höxter Höhe: 295 m ü. NHN Fläche: 128,41 km 2 Einwohner: 8227 (31. Dez. 2017) [1] Bevölkerungsdichte: 64 Einwohner je km 2 Postleitzahlen: 34435, 34437, 34439 Vorwahlen: 05644, 05646, 05642 Kfz-Kennzeichen: HX Gemeindeschlüssel: 05 7 62 040 Stadtgliederung: 13 Stadtteile Adresse der Stadtverwaltung: Abdinghofweg 1 34439 Willebadessen Website: www.willebadessen.de Bürgermeister: Hans Hermann Bluhm (CDU) Lage der Stadt Willebadessen im Kreis Höxter Willebadessen ist eine Kleinstadt in Nordrhein-Westfalen und gehört zum Kreis Höxter im Regierungsbezirk Detmold. Die Stadt liegt im Osten des Regierungsbezirkes mitten im Naturpark Teutoburger Wald/Eggegebirge. Für den heu...
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Ida-Boy-Ed-Garten

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Die Lage des Ida-Boy-Ed-Gartens, rot markiert Der Ida-Boy-Ed-Garten Der Ida-Boy-Ed-Garten, Blick von der Wakenitzmauer durch das Tor in der Stadtmauer Der Ida-Boy-Ed-Garten ist eine Straße der Lübecker Altstadt. Inhaltsverzeichnis 1 Lage 2 Geschichte 3 Literatur 4 Weblinks Lage | Der etwa 90 Meter lange Ida-Boy-Ed-Garten befindet sich an der nördlichen Spitze der Altstadtinsel, vor dem Burgtor. Die in einem Bogen durch eine Grünanlage hangaufwärts führende Straße verbindet die Wakenitzmauer mit der Burgtorbrücke. Geschichte | Der heutige Ida-Boy-Ed-Garten wurde ursprünglich als Verlängerung der Wakenitzmauer angelegt. 1903 wurde in den erhaltenen Teil der mittelalterlichen Stadtmauer, an dem die Wakenitzmauer bei der Einmündung der Kaiserstraße begann, ein Tor mit einem großen Bogen für die Durchführung der Straße und zwei kleineren seitlichen Öffnungen für Fußgänger gebrochen, um für die neu errichteten Wohnhäuser in diesem Bereich ...
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Residenzschloss Arolsen

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Residenzschloss Arolsen Residenzschloss Arolsen Entstehungszeit: 1710–1728 Erhaltungszustand: Vollständig Ständische Stellung: Grafen Ort: Bad Arolsen Geographische Lage 51° 22′ 51″  N , 9° 1′ 19″  O 51.380833333333 9.0219444444444 293 Koordinaten: 51° 22′ 51″  N , 9° 1′ 19″  O Höhe: 293  m ü. NHN Das Residenzschloss Arolsen ist ein barockes Schloss in Bad Arolsen im Kreis Waldeck-Frankenberg in Nordhessen (Deutschland). Das Schloss wurde als dreiflügelige Anlage gebaut, an die sich ein englischer Garten anschließt. Zentrales landschaftsarchitektonisches Gestaltungselement ist ein ausgedehntes Rondell. Inhaltsverzeichnis 1 Geschichte 1.1 Vorgängerbau 1.2 Bau und Geschichte des neuen Schlosses 1.3 Bibliothek 1.4 Finanzielle und staatsrechtliche Konsequenzen 1.5 Heutige Nutzung 2 Innenausstattung 3 Umgebung 3.1 Park 3.2 Wirtschaftsgebäude 4 Literatur 5 Weblinks 6...
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