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Let $x_i$ be a series of vectors of equal length, and let $\beta$ be a constant vector of equal length to $x_i$'s

I have the following sum

$$\sum_{i=1}^p (x_i^T \beta)^2 = \sum_{i=1}^p x_i^T \beta \beta^T x_i = x_1^T \beta \beta^T x_1 + x_2^T \beta \beta^T x_2 + \dots + x_p^T \beta \beta^T x_p$$

In order to apply a statistical theorem, I need to factorise this into a form

$$\sum_{i=1}^d T_i (x) g_i (\beta)$$

where each $T_i (x): \mathbb{R}^n \to \mathbb{R}$ out puts a single scalar $x$. and where the lower the $d$ the better. ie, I want to find the simplist factorisation of the above sum such that the $x$ terms and $\beta$ terms are separated.

My attempt:

I tried writing out the matrix multiplication $x_i^T \beta$ as the sum $\sum_j x_{ij}\beta_j$ but this didn't get me anywhere since it leads me to

$$\sum_j \sum_j \beta_j \beta_k \sum_i x_{ij}x_{ik}$$

which gives a total of $d=p^2$ summands... which is terrible considering $\beta$ is only of length $p$.

Any help here finding a simpler factorisation is much appreciated, thank you.

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  • $\begingroup$ You can put the $x_i$ in a long vector after each other and then $\beta$ in another matrix $I_p \otimes \beta$ where $\otimes$ is Kronecker product. I can try write answer later today if you want. $\endgroup$ – mathreadler Oct 15 '18 at 5:50
  • $\begingroup$ wouldn't $(x_{i}^{t}\beta)$ be the inner product and be a scalar? $\endgroup$ – Shogun Oct 15 '18 at 5:52
  • $\begingroup$ hey @mathreadler thanks for introducing me to a new operation! Unfortunately I'm not sure that would work. The $T_i (x):\mathbb{R}^p \to \mathbb{R}$ are statistics and so must output a scalar, while in your form (if I understand correctly), $T_1 (x)$ would be a single vector of each $x$ end to end. This problem is related to sufficiency and the neyman factorisation theorem. I've editedmy post to clarify this $\endgroup$ – Xiaomi Oct 15 '18 at 7:03
  • $\begingroup$ @Xiaomi yes it is only part of solving the problem. like a hint that maybe can help you finish. i dont have time to write it all now as I am on work. I can write more later. $\endgroup$ – mathreadler Oct 15 '18 at 7:06
  • $\begingroup$ Oh, I see what you mean now. Thanks, I'll give that a think! $\endgroup$ – Xiaomi Oct 15 '18 at 7:07
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Let us give a small example for $$I_p \otimes \beta$$ will be if p=3 and $\beta = [1,2,3]$:

$$\left[\begin{array}{ccccccccc}1&2&3&0&0&0&0&0&0\\0&0&0&1&2&3&0&0&0\\0&0&0&0&0&0&1&2&3\end{array}\right]$$

We see that if we stuff $[x_1,x_2,x_3]^T$ into column vector we can do

$$(I_p \otimes \beta)[x_1,x_2,x_3]^T$$ and then we will get the 3 scalar products you have sought in resulting product vector. The only thing that remains is to take the squared two-norm of this vector.

$$\|(I_p \otimes \beta)[x_1,x_2,x_3]^T\|_2^2$$

It is known that $\|a\|_2^2=a^Ta$ so we can calculate this with for example:

$$([x_1,x_2,x_3](I_p \otimes \beta)^T)((I_p \otimes \beta)[x_1,x_2,x_3]^T)$$

And we have $\mathcal{FINISHED}$ :)

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  • $\begingroup$ that's pretty cool $\endgroup$ – Shogun Oct 15 '18 at 22:22
  • $\begingroup$ Very interesting! Thank you! $\endgroup$ – Xiaomi Oct 16 '18 at 7:25
  • $\begingroup$ Pff you didn't really think I finished, did you? It would be 16 times overkill at this rate. $\endgroup$ – mathreadler Oct 16 '18 at 8:48

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