How can I make the height of summation and fraction the same?

I was wondering if I could get some help with this. I want to align the heights of fraction and the summation. The summation is positioned too low compared to the fraction. How can I acheive this?

\documentclass[12pt,a4paper]{article}
\begin{document}
$\Lambda_{entry} =\frac{\psi/(1-\delta)}{\psi/(1-\delta) + m_{I}} + \frac{m_{I}(1-\Phi)}{\psi/(1-\delta) + m_{I}} \sum^{}_{(\tilde{g},\tilde{b}) \in \{(g,b)\}}\hspace{0.1em}f_{\tilde{g},\tilde{b}-1}$
\end{document}


• If you'd try it, your code snippet doesn't compile. Please add the missing minimal parts. Thank you. Jul 10 at 11:14
• Thank you very much for pointing that out. I edited. Jul 10 at 11:20
• @Anonymouslylost Can you make a sketch which shows how you would like to change the alignment? The central axis seems to be perfectly aligned: i.stack.imgur.com/o3GHx.png Jul 10 at 11:22
• I hope something like the highest point of the summation is aligned with the highest point of the fraction. The current version looks like someone hammered the summation to below. Jul 10 at 11:26

I hope something like the highest point of the summation [symbol] is aligned with the highest point of the fraction.

Be careful what you ask for...

\documentclass{article}
\usepackage{mathtools} % for '\mathclap' directive
\usepackage{graphicx}  % for '\scalebox' directive
\newcommand{\bigsum}{%
\mathop{\vcenter{\hbox{\scalebox{1.5}{$\displaystyle\sum$}}}}}

\begin{document}

$\Lambda_{\mathrm{entry}} = \frac{\psi/(1-\delta)}{\psi/(1-\delta) + m_{I}} + \frac{m_{I}(1-\Phi )}{\psi/(1-\delta) + m_{I}} \times \bigsum_{\mathclap{(\tilde{g},\tilde{b})\in\{(g,b)\}}} f^{}_{\tilde{g},\tilde{b}-1}$

\end{document}

• Thank you very much. It looks a lot better. Maybe I'm too sensitive. It's kind of annoying that the position of tilds are different between g and b and that -1 is bigger than b as well. Would there be a nice package to control these aspects? Perhaps I should open a new post. Jul 10 at 12:51
• @Anonymouslylost - I disagree with your claim that "-1 is bigger than b". The objective "the position of tild[e]s are different between g and b" is another one of those "be careful what you wish for" items.
– Mico
Jul 10 at 13:05
• Could you elaborate a bit more about what you meant? Do you mean I should be cautious in changing basic configurations? Jul 10 at 13:47
• @Anonymouslylost - IMNSHO, there are excellent typographical and aesthetic reasons for not placing the tilde symbols at the same height for \tilde{g}, \tilde{t}, and \tilde{b}. E.g., if you define \newcommand\xtilde[1]{\tilde{#1\mathstrut}}, compare the appearances of $\tilde{g}\ \tilde{t}\ \tilde{b}$ and $\xtilde{g}\ \xtilde{t}\ \xtilde{b}$. Incidentally, are you familiar with what Ralph Waldo Emerson had to say about [foolish] consistency?
– Mico
Jul 10 at 13:58

\documentclass[12pt,a4paper]{article}
% https://tex.stackexchange.com/a/546690
\newcommand\vc[1]{\vcenter{\hbox{#1}}}
\begin{document}
$\Lambda_{entry} = \frac{\psi/(1-\delta)}{\psi/(1-\delta) + m_{I}} + \frac{m_{I}(1-\Phi)}{\psi/(1-\delta) + m_{I}} \hspace{0.3em} \vc{\displaystyle\sum_{(\tilde{g},\tilde{b}) \in \{(g,b)\}} \hspace{-1em} f_{\tilde{g},\tilde{b}-1}}$
\end{document}

• I think the argument of \vc should be just the (suitably scaled) summation symbol, and should not include the range of summation.
– Mico
Jul 10 at 13:12
• I think this is exactly what I wanted. Thank you very much ! Jul 10 at 13:46