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I have a vector $x = (x_1, \dots, x_n) \in \mathbb{R}^n$ from which I am 'dropping' $n-k$ dimensions. To be precise: Let $$ D(k) = \{y = (y_1, \dots, y_n) \mid y_1, \dots, y_n \in \{0,1\} \text{ and } \sum_{i=1}^n y_i = k\}. $$ For $y \in D(k),$ let $\text{diag}(y)$ denote the diagonal matrix with $y$ on its diagonal. Then $x\, \text{diag}(y)$ is the vector $x$ where $n-k$ dimensions have been dropped in the sense that the corresponding values of $x$ have been set to zero. My question: For a given $k$, what is the average norm $$ f(k) = \frac1{|D(k)|}\sum_{y \in D(k)} \lVert x \,\text{diag}(y)\rVert = \frac1{\binom{n}{k}}\sum_{y \in D(k)} \sqrt{\sum_{i=1}^n x_i^2 y_i} $$ of the dropped vectors in terms of $\lVert x\rVert$? I am looking for a solution of the form $$f(k) = (\text{something nice with $n$ and $k$}) \times \lVert x\rVert.$$

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There is no such formula.

Consider $n = 2, k=1$, and the vector $x = (1, 1)$ with length $s = \sqrt{2}$. Then the two projections are $(1, 0)$ and $(0, 1)$, and their average length is $1$.

Now consider $x' = (\sqrt{2}, 0)$, whose length is also $s$. The projections are $x'$ itself and the zero vector. The average length of those is $s/2$, which is different from $1$.

Hence the number $f(k)$ you seek must really be $f(k, x)$, which is probably not what you wanted to hear, but it's true.

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  • $\begingroup$ Math has a habit of telling me what I don't want to hear! $\endgroup$ – Elias Strehle Mar 11 '18 at 15:13
  • $\begingroup$ In case you are interested: The problem came up in a discussion about neural networks. $\endgroup$ – Elias Strehle Mar 11 '18 at 15:20
  • $\begingroup$ If you ask the slightly more (mathematically) natural question "what's the average, over all $x$ of norm $1$, under the uniform distribution" then there's a perfectly good answer; working out the cases $k = 1, 2$, $n = 3$ should let you see a general pattern. $\endgroup$ – John Hughes Mar 11 '18 at 15:24

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