handling whitespace in denominator of complex fraction

Question

I have a paper where I need to format a series of ugly 'fraction expressions' -- ugly because the term in the numerator is (often) a complex sum or series, and in order to get it to `fit', I usually have to break the expression in the numeratorover 2 or more lines. Typically, I accomplish this using \splitfrac. The denominator for these equations is often simple say, a single term, or single-line expression.

But, a challenge arises if I want (or, more often, need) to bracket the whole equation, using (say) \left( and \right), I end up with lots of extra whitespace around the denominator.

One approach tried was embedding everything in a pmatrix -- this gets rid of the whitespace in the denominator, but, I can't quite figure out how to centre a coefficient at the line separating the numerator and denominator (the vinculum, to be semantic about it ;-)

And that is what I'm hoping to be able to work out -- I have a number of equations that might require such 'centering'.

The following MWE shows both the 'ugly' and 'somewhat better' versions. Suggestions welcomed -- thanks in advance.

 \documentclass[11pt]{article}

 % set up basic package list
 \usepackage{amsmath,amstext,mathtools,amssymb}

 \begin{document}

 One form of the equation -- ugly white space in denominator

  \begin{equation*}
   D = -2 \left(\dfrac{
    \splitfrac{ \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \text{Pois}(y_i | \exp(\beta_0^{(t)}} 
 {+\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} x_{p,i})\bigr)}
     } {T}\right)
 \end{equation*}

 Better form (IMO) of the equation, using pmatrix `trick', but coefficient (-2) not centered 
 on horizontal line separating numerating and denominator:

 \begin{equation*}
 D=-2\begin{pmatrix} \dfrac{
    \splitfrac{ \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \text{Pois}(y_i | \exp(\beta_0^{(t)}} 
 {+\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} x_{p,i})\bigr)}
     } {T}
 \end{pmatrix}  
 \end{equation*}

 \end{document}

Answer 1

You could do the second (perhaps with a custom size bigger than Bigg) but I would do the third

\documentclass[11pt]{article}

 % set up basic package list
 \usepackage{mathtools,amssymb}

 \begin{document}

 One form of the equation -- ugly white space in denominator

  \begin{equation*}
   D = -2 \left(\dfrac{
    \splitfrac{ \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \mathrm{Pois}(y_i | \exp(\beta_0^{(t)}} 
 {+\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} x_{p,i})\bigr)}
     } {T}\right)
 \end{equation*}



  \begin{equation*}
   D = -2 \raisebox{10pt}{$\Biggl($}\dfrac{
    \splitfrac{ \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \mathrm{Pois}(y_i | \exp(\beta_0^{(t)}} 
 {+\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} x_{p,i})\bigr)}
     } {T}\raisebox{10pt}{$\Biggr)$}
 \end{equation*}




  \begin{equation*}
   D = -2
    \Bigl( \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \mathrm{Pois}(y_i | \exp(\beta_0^{(t)}
 +\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} x_{p,i})\bigr)\Bigr) /T 
  \end{equation*}



 \end{document}

Answer 2

I can't see a reason for not factoring the T term out of the big term. If you must split the big term across two rows (possibly because your document employs a two-column layout), I would still factor out the T term.

Do note the use of \bigm\vert instead of |. Not only is the vertical bar taller, it is also spaced better.

\documentclass[11pt]{article}
\usepackage{mathtools,amssymb}
\DeclareMathOperator{\Pois}{Pois}
\begin{document}

\begin{equation*}
D=-\frac{2}{T} \biggl(\,
   \sum_{t=1}^T \sum_{i=1}^n \log \bigl[ \Pois \bigl(\,y_i 
   \bigm\vert \exp(\beta_0^{(t)} +\beta_1^{(t)} x_{1,i} 
   + \dots + \beta_p^{(t)} x_{p,i}) \,\bigr) \bigr]
   \biggr)
\end{equation*}

\begin{equation*}
\begin{aligned}
D= -\frac{2}{T} \biggl(
   &\sum_{t=1}^T \sum_{i=1}^n \log \bigl[ \Pois \bigl(\,y_i 
   \bigm\vert \exp(\beta_0^{(t)} \\
   &+\beta_1^{(t)} x_{1,i} 
   + \dots + \beta_p^{(t)} x_{p,i}) \,\bigr) \bigr] \smash[t]{\biggr)}
   \end{aligned}
\end{equation*}
\end{document}

Answer 3

One possible answer/solution, which is a tweak of the 'raisebox' suggestion, might suffice...

   \begin{equation*}
    \raisebox{-12pt}{$D = -2$} \begin{pmatrix} \dfrac{
    \splitfrac{ \sum_{t=1}^T \sum_{i=1}^n \log \bigl( \text{Pois}(y_i | 
      \exp(\beta_0^{(t)}} {+\beta_1^{(t)} x_{1,i} + \ldots + \beta_p^{(t)} 
       x_{p,i})\bigr)}
        } {T}
  \end{pmatrix}
 \end{equation*}