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Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

EDIT:

Actually what's really of interest is the relation between $E\left(\min_{a\in C}\|Y-a\|\right)$ and $E\left(\min_{a\in C}\|X_0-a\|\right)$ where $C=Conv(X_1, \cdots, X_N)$ and $X_0$ is distributed as $X_i$. Ideally I would like to prove that the first is weakly bigger than the second (my intuition tells me it should be true, at least in most situations).

Right now I'm thinking along these lines. Let $\mu_X$ denote the mean of $X_i$, then let $S=\{x:\|x-\mu_X\|\leq r\}$ be a ball around $\mu_X$ with radius $r$ where we set $r=max_i\|X_i-\mu_X\|$. It then follows that for any point $x$ we have $x\in C\Rightarrow x\in S$, from which we have, for all $x$:

$$E\left(\min_{a\in C}\|x-a\|\right)\geq E\left(\min_{a\in S}\|x-a\|\right),$$

where $E\left(\min_{a\in S}\|x-a\|\right)=E(\max(\|x-\mu_X\|-r, 0))$.

Concerning $r$, if all elements in $X_i-\mu_X$ are independent and standard normal, $\|X_i-\mu_X\|$ will be chi distributed with $d$ degrees of freedom. But what about $max_i\|X_i-\mu_X\|$?

Even if not rigorous I'll would be happy just to compare the lower bounds of the expected distances of $Y$ and $X_0$. I guess a better way would be to construct a upper bound for $X_0$ (e.g. with a ball $S'$ s.t. $x\in S'\Rightarrow x\in C$) and then compare it with the lower bound for $Y$.

Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

EDIT:

Actually what's really of interest is the relation between $E\left(\min_{a\in C}\|Y-a\|\right)$ and $E\left(\min_{a\in C}\|X_0-a\|\right)$ where $C=Conv(X_1, \cdots, X_N)$ and $X_0$ is distributed as $X_i$. Ideally I would like to prove that the first is weakly bigger than the second (my intuition tells me it should be true when the central moments are the same).

Right now I'm thinking along these lines. Let $\mu_X$ denote the mean of $X_i$, then let $S=\{x:\|x-\mu_X\|\leq r\}$ be a ball around $\mu_X$ with radius $r$ where we set $r=max_i\|X_i-\mu_X\|$. It then follows that for any point $x$ we have $x\in C\Rightarrow x\in S$, from which we have, for all $x$:

$$E\left(\min_{a\in C}\|x-a\|\right)\geq E\left(\min_{a\in S}\|x-a\|\right),$$

where $E\left(\min_{a\in S}\|x-a\|\right)=E(\max(\|x-\mu_X\|-r, 0))$.

Concerning $r$, if all elements in $X_i-\mu_X$ are independent and standard normal, $\|X_i-\mu_X\|$ will be chi distributed with $d$ degrees of freedom. But what about $max_i\|X_i-\mu_X\|$?

Even if not rigorous I'll would be happy just to compare the lower bounds of the expected distances of $Y$ and $X_0$. I guess a better way would be to construct a upper bound for $X_0$ (e.g. with a ball $S'$ s.t. $x\in S'\Rightarrow x\in C$) and then compare it with the lower bound for $Y$.

Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

EDIT:

Actually what's really of interest is the relation between $E\left(\min_{a\in C}\|Y-a\|\right)$ and $E\left(\min_{a\in C}\|X_0-a\|\right)$ where $C=Conv(X_1, \cdots, X_N)$ and $X_0$ is distributed as $X_i$. Ideally I would like to prove that the first is weakly bigger than the second (my intuition tells me it should be true, at least in most situations).

Right now I'm thinking along these lines. Let $\mu_X$ denote the mean of $X_i$, then let $S=\{x:\|x-\mu_X\|\leq r\}$ be a ball around $\mu_X$ with radius $r$ where we set $r=max_i\|X_i-\mu_X\|$. It then follows that for any point $x$ we have $x\in C\Rightarrow x\in S$, from which we have, for all $x$:

$$E\left(\min_{a\in C}\|x-a\|\right)\geq E\left(\min_{a\in S}\|x-a\|\right).$$

Concerning $r$, if all elements in $X_i-\mu_X$ are independent and standard normal, $\|X_i-\mu_X\|$ will be chi-squared distributed with $d$ degrees of freedom. But what about $max_i\|X_i-\mu_X\|$?

In any case the idea is that $E\left(\min_{a\in S}\|x-a\|\right)$ is easier to derive than $E\left(\min_{a\in C}\|x-a\|\right)$. Even if not rigorous I'll would be happy just to compare the lower bounds of the expected distances of $Y$ and $X_0$. I guess a better way would be to construct a upper bound for $X_0$ (e.g. with a ball $S'$ s.t. $x\in S'\Rightarrow x\in C$) and then compare it with the lower bound for $Y$.

Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

EDIT:

Actually what's really of interest is the relation between $E\left(\min_{a\in C}\|Y-a\|\right)$ and $E\left(\min_{a\in C}\|X_0-a\|\right)$ where $C=Conv(X_1, \cdots, X_N)$ and $X_0$ is distributed as $X_i$. Ideally I would like to prove that the first is weakly bigger than the second (my intuition tells me it should be true, at least in most situations).

Right now I'm thinking along these lines. Let $\mu_X$ denote the mean of $X_i$, then let $S=\{x:\|x-\mu_X\|\leq r\}$ be a ball around $\mu_X$ with radius $r$ where we set $r=max_i\|X_i-\mu_X\|$. It then follows that for any point $x$ we have $x\in C\Rightarrow x\in S$, from which we have, for all $x$:

$$E\left(\min_{a\in C}\|x-a\|\right)\geq E\left(\min_{a\in S}\|x-a\|\right),$$

where $E\left(\min_{a\in S}\|x-a\|\right)=E(\max(\|x-\mu_X\|-r, 0))$.

Concerning $r$, if all elements in $X_i-\mu_X$ are independent and standard normal, $\|X_i-\mu_X\|$ will be chi distributed with $d$ degrees of freedom. But what about $max_i\|X_i-\mu_X\|$?

Even if not rigorous I'll would be happy just to compare the lower bounds of the expected distances of $Y$ and $X_0$. I guess a better way would be to construct a upper bound for $X_0$ (e.g. with a ball $S'$ s.t. $x\in S'\Rightarrow x\in C$) and then compare it with the lower bound for $Y$.

Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

Let $X_1, \cdots, X_N$ be i.i.d. $d$-dimensional random vectors, the exact distribution of $X$ is not very important in my application (as long as it continuous) so pick the one that works best, but ideally multivariate normal. Then let $Y$ be another $d$-dimensional random vector from a distribution that share the same central moments as the first distribution (i.e. the first distribution with shifted mean). Could one then find:

$$E\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|\right)\quad s.t. \sum_{i=1}^N \alpha_i = 1 \wedge \alpha_i\geq 0$$

or maybe bounds on it? Or maybe the probability $P\left(\min_{\alpha_1,\cdots,\alpha_N}\left\|Y-\sum_{i=1}^N\alpha_i X_i\right\|=0\right)$?

EDIT:

Actually what's really of interest is the relation between $E\left(\min_{a\in C}\|Y-a\|\right)$ and $E\left(\min_{a\in C}\|X_0-a\|\right)$ where $C=Conv(X_1, \cdots, X_N)$ and $X_0$ is distributed as $X_i$. Ideally I would like to prove that the first is weakly bigger than the second (my intuition tells me it should be true, at least in most situations).

Right now I'm thinking along these lines. Let $\mu_X$ denote the mean of $X_i$, then let $S=\{x:\|x-\mu_X\|\leq r\}$ be a ball around $\mu_X$ with radius $r$ where we set $r=max_i\|X_i-\mu_X\|$. It then follows that for any point $x$ we have $x\in C\Rightarrow x\in S$, from which we have, for all $x$:

$$E\left(\min_{a\in C}\|x-a\|\right)\geq E\left(\min_{a\in S}\|x-a\|\right).$$

Concerning $r$, if all elements in $X_i-\mu_X$ are independent and standard normal, $\|X_i-\mu_X\|$ will be chi-squared distributed with $d$ degrees of freedom. But what about $max_i\|X_i-\mu_X\|$?

In any case the idea is that $E\left(\min_{a\in S}\|x-a\|\right)$ is easier to derive than $E\left(\min_{a\in C}\|x-a\|\right)$. Even if not rigorous I'll would be happy just to compare the lower bounds of the expected distances of $Y$ and $X_0$. I guess a better way would be to construct a upper bound for $X_0$ (e.g. with a ball $S'$ s.t. $x\in S'\Rightarrow x\in C$) and then compare it with the lower bound for $Y$.