I can't find the name of this symbol: enter image description here

It doesn't seem to be a Greek letter, and I have also tried using the \mathcal command. Does anyone know it?


The symbol used in a context: enter image description here

  • 4
    It seems like \Upsilon to me... (Greek capital letter)... – koleygr Oct 28 '17 at 15:46
  • @koleygr \Upsilon looks different.. – lala_12 Oct 28 '17 at 15:48
  • I think it's a \Upsilon in some non-starndard font. – Rmano Oct 28 '17 at 15:50
  • You can search here: tug.org/pracjourn/2006-1/hartke/hartke.pdf ; it's very similar to a slanted \Upsilon form anttor font. – Rmano Oct 28 '17 at 16:01
  • 1
    Maybe \otherUpsilon with \usepackage{fourier} – Troy Oct 28 '17 at 16:38

I think this is just the uppercase Y in mathpazo:

Let $i=1,\ldots,N$ denote the $N$ experimental units
and $Y_i$ the $i$'th response variable.
Now we also supose that we have a covariate $x_i$ for
each experimental unit.
The experiment can be with one or more factors, with or
without blocks, and almost any experimental design.
However, in order to illustrate the use of the covariate
we consider a single \linebreak
factor and a one-way analysis of a
variance model in the situation where the covariate was
not used.  Let $\mathtt{TREAT}$ be the factor in the
experiment with the $k$ levels $\mathtt{treat}_1$, $\ldots$,
$\mathtt{treat}_k$.  If $\mathtt{TREAT}_i$ denotes the
treatment of the $i$'th experimental unit, so that
$\mathtt{TREAT}_i$ is identical to one of the treatments
$\mathtt{treat}_1$, $\ldots$, $\mathtt{treat}_k$, we can write
the one-way analysis of variance model as
  Y_i = \alpha(\mathtt{TREAT}_i) + \epsilon_i,
supplemented with the usual assumptions that $\epsilon_1,
\ldots,\epsilon_N$ are independent and normally
distributed with the same variance $\sigma^2$.


Please, forgive my typo(s)... :-)

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.