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Feb 24, 2017 at 15:54 history edited mdewey
Replaced randomization with random-allocation
Sep 21, 2012 at 14:08 vote accept user697473
Sep 17, 2012 at 21:41 comment added user697473 @whuber -- thanks. That was a typo. I fixed it.
Sep 17, 2012 at 21:40 history edited user697473 CC BY-SA 3.0
Fixed typo (i.e., changed \mathbf{z} to \mathbf{Z}).
Sep 17, 2012 at 21:05 comment added whuber I believe @mpiktas was really wondering what $\mathbf{z}$ meant (as opposed to $\mathbf{Z}$): I suspect it's a typographical error.
Sep 17, 2012 at 19:17 answer added Charlie timeline score: 2
Sep 17, 2012 at 18:01 history tweeted twitter.com/#!/StackStats/status/247757074501623808
Sep 17, 2012 at 17:52 history edited user697473 CC BY-SA 3.0
Replaced article link with link to a better-looking PDF of the same article
Sep 17, 2012 at 17:31 comment added user697473 @mpiktas, $Z_i$ is the $i$th element of $\mathbf{Z}$. $\mathbf{Z}, \mathbf{c},$ and $\mathbf{c'}$ are vectors of zeroes and ones; all of these vectors have the same length. $\iota$ is a "counting vector" -- all ones -- so $\iota^T\mathbf{c}$ is just the number (i.e., the sum) of ones in $\mathbf{c}$. And a similar definition holds for $\iota^T\mathbf{c'}$. I hope that this helps.
Sep 17, 2012 at 17:28 comment added user697473 @StasK, thanks for the suggestion. I added a link to the original post.
Sep 17, 2012 at 17:27 history edited user697473 CC BY-SA 3.0
Added link to article mentioned in post
Sep 17, 2012 at 17:03 comment added StasK @user697473, please give a link to the full text of the paper you cite. Respect the time of the people answering this; or at least assume they will be too lazy even to Google it (I certainly was).
Sep 17, 2012 at 17:02 answer added StasK timeline score: 8
Sep 17, 2012 at 12:59 answer added Michael R. Chernick timeline score: 3
Sep 17, 2012 at 12:39 comment added mpiktas Please explain the definitions in the second paragraph more clearly, i.e. whats is $\mathbf{z}$, $\mathbf{c}$ and $\mathbf{c}'$?
Sep 17, 2012 at 12:38 comment added Peter Flom No. If you rule out some assignment vectors a priori you no longer have random assignment, strictly speaking.
Sep 17, 2012 at 12:37 history edited mpiktas CC BY-SA 3.0
edited title
Sep 17, 2012 at 12:18 history asked user697473 CC BY-SA 3.0