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Let $$x = [1/n, 1/n,\dots, 1/n]$$ and $$y = [y_1, \dots, y_n]$$ be probability vectors on $n$ elements (i.e., $\sum x_i = \sum y_i = 1$ and $x_i,y_i\geq 0$).

Is it true that $x$ is always majorized by $y$?

My thought: yes. From what I can see on Wikipedia (the Geometry of majorization section), a vector $a$ is majorized by a vector $b$ iff $a$ is within the convex hull of vectors obtained from permuting all the elements of $b$. For probability vectors, the hull will be a subspace of the $n$ dimensional simplex. Intuitively, because $x$ is uniform it will always be at the `centre' of the simplex and therefore within the convex hull.

I've searched the internet for a while now and have not been able to obtain a reference to such a result. If one has a reference, I would be grateful too!

2 Answers 2

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Let $y_{(1)}, y_{(2)}, \cdots, y_{(n)}$ be a non-increasing order of $y_1, y_2, \cdots, y_n$. For $1 \leq k < n$, let $$ S_k = \sum_{i=1}^k y_{(i)} \quad \text{and} \quad T_k = \sum_{i=k+1}^n y_{(i)} $$ It is easy to see that $$ S_k + T_k = 1 \tag{$1$} $$ Moreover, we have $$ S_k \geq \sum_{i=1}^k y_{(k)} = ky_{(k)} \Rightarrow \frac{S_k}{k} \geq y_{(k)} \quad \text{and} \quad T_k \leq \sum_{i=k+1}^n y_{(k)} = (n - k)y_{(k)} \Rightarrow \frac{T_k}{n-k} \leq y_{(k)} $$ That is, $$ \frac{n-k}{k}S_k \geq T_k \tag{$2$} $$ Combining $(1)$ and $(2)$, we have $$ S_k + \frac{n-k}{k}S_k \geq S_k + T_k = 1 \Rightarrow S_k \geq \frac{k}{n} $$ Therefore, $x$ is majorized by $y$.

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Let $\{y_i^*\}_{i=1}^n$ be the decreasing (i.e., non-increasing) arrangement of the $y_i$'s. $x$ is majorized by $y$ if and only if $a_k=\frac{1}{k}\sum_{i=1}^ky_i^{*}\geq\frac{1}{n}$ for each $k=1,2,\dots n$. But $a_k$ is non-increasing, as can easily be checked, that is, $$a_1\geq a_2\cdots\geq a_n=\frac{1}{n},$$ so the assertion is proved.