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I have two sets of hypotheses in a multivariate setting that are said to be equivalent. I am wondering if someone can explain why they are as opposed to me just taking it to be true.

Version 1:

$H_0: \boldsymbol{\mu_1=...=\mu_K}$, where $\boldsymbol{\mu_i}$ are $p*1$ vectors.

Version 2:

$H_0: C\boldsymbol{\mu}=\boldsymbol{0}$, where $\boldsymbol{\mu}$ is a $pK*1$ vector created by stacking each of the $\boldsymbol{\mu_i}$, $\boldsymbol{0}$ is a $p(K-1)*1$ vector of $0$'s, and $C$ is a $p(K-1)*pK$ that meets the following criteria:

1) The rows of $C$ are linearly independent $(C$ has rank $p(K-1))$.

2) $C\boldsymbol{\nu=0}$, where $\boldsymbol{\nu}$ is a $pK*1$ vector of $1$'s, and $\boldsymbol{0}$ is a $p(K-1)*1$ vector of $0$'s.

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  • $\begingroup$ Version 1 of $H_0$ is k-1 independent equations of the p*1 vectors $\vec{\mu}_k$, and so is version 2. But maybe instead of writing $\vec{\mu}_1=\vec{\mu}_2$ and $\vec{\mu}_2=\vec{\mu}_3$, you can use an equivalent $-1\mu_{1,j} + 0.5\mu_{2,j}+0.5\mu_{3,j}=0$ and $\mu_{2,j}-1\mu_{3,j}=0$ (which is a transformation of the equations in version 1, but with the same meaning). Or you can use whatever other sum (which can differ for rows $j$) that is in line with $C\vec{\nu} = \vec{0}$ (the coefficients need to add to zero). $\endgroup$ – Sextus Empiricus Oct 25 '17 at 20:07
  • $\begingroup$ I have misread your question. I thought you had a $C\mu = 0$ with $\mu$ a p*K matrix from the vectors $\mu_k$. are you sure about this stacked pK*1 vector. I imagine myselve the matrix C below which I believe meets both criteria, but results in much different equations (namely $\mu_{1,k}=\mu_{1,k+1}$). $$C=\begin{bmatrix} 1 & -1 & 0 & \dots & 0 &0&0 \\ 0 & 1 & -1 & \dots & 0 &0&0\\ ...\\ 0 & 0 & 0 & \dots & 1 &-1&0\\ 0 & 0 & 0 & \dots & 0 &1&-1\\ \end{bmatrix}$$ $\endgroup$ – Sextus Empiricus Oct 25 '17 at 20:27
  • $\begingroup$ Hey Martijn, I believe the C you have proposed meets both criteria $\endgroup$ – user178237 Oct 25 '17 at 21:24
  • $\begingroup$ Although I believe it is true for any C that meets these criteria and I am hoping to show this. $\endgroup$ – user178237 Oct 25 '17 at 21:25
  • $\begingroup$ The C that I proposed is not the same as the hypothesis version 1 (it is stronger). So it is not true for any C. Therefore I wonder where you got this expression and whether you may have made copy error. $\endgroup$ – Sextus Empiricus Oct 25 '17 at 21:59

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