Putting formula on the left side

Question

Consider the following

\documentclass{article}

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}

\begin{document}

\begin{lem}
    \begin{align}%
        \begin{split}%
            &\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) =  \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\
            & = \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}- \sum\limits_{k=u+1}^{u+s-z}  \sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j}   p^{*(j-u)}_j p_{u+s+z-k}
        \end{split}\nonumber
    \end{align}
\end{lem}

\end{document}

Answer 1

This uses a further line, because the final formula is very wide.

If your text width allows it, you can move left the alignment point in the first line and perhaps avoid the need to split the last formula.

\documentclass{article}
%\usepackage{geometry}
\usepackage{amsthm,amsmath,amssymb}

\newtheorem{lem}{Lemma}

\begin{document}

\begin{lem}
\begin{equation}
\hspace{0pt}
\begin{aligned}[b]
\mathbb{P}_u(\tau^0=s&, -R_{\tau^0}=z)
=\sum_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k}
\\
&= \sum_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}
\\
&\qquad-\sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j}
    p^{*(j-u)}_j p_{u+s+z-k}
\end{aligned}
\hspace{10000pt minus 1fil}
\end{equation}
\end{lem}

\end{document}

I used a trick explained in exercise 19.8 of the text book. It's necessary to have a glob of zero glue at the beginning, in order that the alignment is flush left. The \hspace{100000pt minus 1fil} tricks TeX into thinking that the alignment is obnoxiously wide (but there is a lot of shrinkability to actually allow the typesetting).

See also https://tex.stackexchange.com/a/348745/4427

Answer 2

If you want to have an effect like \raggedright you have to put the alignment character & at the first place of every line inside the split environment.

\documentclass{article}

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}

\begin{document}

\begin{lem}
    \begin{align}%
        \begin{split}%
            &\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) =  \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\
            & = \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}- \sum\limits_{k=u+1}^{u+s-z}  \sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j}   p^{*(j-u)}_j p_{u+s+z-k}
        \end{split}\nonumber
    \end{align}
\end{lem}

\end{document}

Answer 3

In your case I would consider the multline* math environment (defined by amsmath):

\documentclass{article}

\usepackage{amsthm}
\usepackage{amsmath}
\usepackage{amssymb}
\newtheorem{lem}{Lemma}

\begin{document}
\begin{lem}
    \begin{multline*}
\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) =  \sum\limits_{k=0}^{u+s-z} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k}   \\
    = \sum\limits_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}-
      \sum\limits_{k=u+1}^{u+s-z} 
      \sum\limits_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j}   p^{*(j-u)}_j p_{u+s+z-k}
    \end{multline*}
\end{lem}
\end{document}

Answer 4

With the help of geometry and mathtools, I propose these two variants:

\documentclass{article}
\usepackage[showframe]{geometry}
\usepackage{amsthm, amssymb}
\newtheorem{lem}{Lemma}

\usepackage{mathtools}

\begin{document}

\begin{lem}
\begin{multline} \mathbb{P}_u(\tau^0=s , -R_{\tau^0}=z) = \smash{\sum_{k=0}^{u+s-z}} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k} \\ = \sum_{k=0}^{u+s-z} p^{*(s-1)}_k p_{u+s+z-k}-\!\! \sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
\end{multline}
\end{lem}
\vskip2em

\begin{lem}$\mathbb{P}_u(\tau^0=s, -R_{\tau^0}=z) = \displaystyle\smash[b]{\smashoperator{\sum_{k=0}^{u+s-z}}} \mathbb{P}( \tau^0>s-1, S_{s-1}=k)p_{u+s+z-k}$
  \begin{equation} =\smashoperator{\sum_{k=0}^{u+s-z}} p^{*(s-1)}_k p_{u+s+z-k} -\!\! \sum_{k=u+1}^{u+s-z} \sum_{j=u+1}^{k} = \frac{s-1+u-k}{s-1+u-j}p^{*(s-1+u-j)}_{k-j} p^{*(j-u)}_j p_{u+s+z-k}
 \end{equation}
\end{lem}

\end{document}