How do you put this long equation in latex?

Question

\documentclass[a4paper,12pt]{report}
\usepackage[latin1]{inputenc}
\usepackage[T1]{fontenc}
\usepackage[round]{natbib}
\usepackage{pgf}
\usepackage{url}
\usepackage[english]{babel}
\usepackage{multirow}
\usepackage{fancyhdr}
\usepackage{anysize}
\usepackage{amsmath, mathtools}

\begin{document}
\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:

\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[6]  P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) \\ P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)= \\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)= \\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]} \\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]} \\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}= \\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}
\end{document}

I write this in LaTeX and I have an error.

Gives me error because I have other split before?

Can I help me to solve equation?

Answer 1

As already pointed out by David Carlisle, the immediate source of the syntax error in the second equation environment is that a } was incorrectly written as ].

In addition, there seem to be a few more errors of mathematical content, such as incorrectly placed square brackets. I've tried my best to clean up the appearance of both equations. Hopefully, you can use this code as a half-way point towards cleaning up the remaining issues.

Observe that I replaced some of the \dots directives with \dotsb ("dots binary"), as they (i.e., the typographic ellipses) would appear to indicate multiplicative elision.

\documentclass[a4paper,12pt]{report} 
\usepackage[latin1]{inputenc} 
\usepackage[T1]{fontenc} 
%% Commented out unneeded \usepackage statements:
%\usepackage[round]{natbib} 
%\usepackage{pgf} 
%\usepackage{url} 
%\usepackage[english]{babel} 
%\usepackage{multirow} 
%\usepackage{fancyhdr} 
%\usepackage{anysize} 
\usepackage{mathtools} % 'mathtools' loads 'amsmath' automatically
\begin{document}
\setcounter{equation}{19} % just for this example

\begin{equation}\label{eq:20}
\begin{split}
\MoveEqLeft[3] 
\bigl\{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n \bigr\} \\
{}=\bigl\{&N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
  & N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
  & N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0 \bigr\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independent increments of the Poisson Process $N$, we obtain:
\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[3]  
P\bigl(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\bigr) \\ 
{}={}&P(N(0,x_{1}]=0)\,
      P(N(x_{1},x_{1}+h_{1}]=1)\,
      P(N(x_{1}+h_{1},x_{2}]=0) \\
     &P(N(x_{2},x_{2}+h_{2}]=1) \dotsb 
      P(N(x_{n-1}+h_{n-1},x_{n}]=0) \\
     &P(N(x_{n}, x_{n}+h_{n}]=1)\,
      P(N(x_{n}+h_{n},t]=0) \\
{}={}&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]\,
      e^{-\mu(x_{1},x_{1}+h_{1}]}]\,
      e^{-\mu(x_{1}+h_{1},x_{2}]}[\mu(x_{2},x_{2}+h_{2}] \\
     &e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dotsm 
      e^{-\mu(x_{n-1}+h_{n-1},x_{n}]}[\mu(x_{n}, x_{n}+h_{n}] \\
     &e^{-\mu(x_{n}, x_{n}+h_{n}}]\,
      e^{\mu(x_{n}+h_{n},t]} \\
{}={}&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dotsb \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}
\end{document}

Answer 2

The line &[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}= miss a \} (second to last line of the second paragraph).

Here is the working LaTeX, but beware, it looks horrible :

\begin{equation}\label{eq:20}
\begin{split} \
\MoveEqLeft[3] \{T_{1} \in (x_{1},x_{1}+h_{1}],\dots,T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n\} = \\
&\{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, \\
& N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, \\
& N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{split}
\end{equation}

Taking probabilities on both sides and using the property of independen increments of the Poisson Process N, we obtain:

\begin{equation} \label{eq:21}
\begin{split} 
\MoveEqLeft[6]  P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) 
\\ 
P(N(0,x_{1}]=0)P(N(x_{1},x_{1}+h_{1}]=1)P(N(x_{1}+h_{1},x_{2}]=0)= 
\\
& P(N(x_{2},x_{2}+h{2}]=1) \dots P(N(x_{n-1}+h_{n-1},x_{n}]=0) 
\\
&P(N(x_{n}, x_{n}+h_{n}]=1) P(N(x_{n}+h_{n},t]=0)= 
\\
&e^{-\mu(x_{1}}[\mu(x_{1},x_{1}+h_{1}]e^{-\mu(x_{1},x_{1}+h_{1}]}]e^{-\mu(x_{1}+h_{1},x_{2}]} 
\\
&[\mu(x_{2},x_{2}+h_{2}]e^{-\mu(x_{2}, x_{2}+h_{2}]}] \dots e^{-\mu(x_{n-1}+h_{n-1},x_{n}]} 
\\
&[\mu(x_{n}, x_{n}+h_{n}]e^{-\mu(x_{n}, x_{n}+h_{n}]]e^{\mu(x_{n}+h_{n},t]}}= 
\\
&e^{-\mu(t)} \mu(x_{1}, x_{1}+h_{1}] \dots \mu(x_{n}, x_{n}+h_{n}] 
\end{split}
\end{equation}

But I must recommend you to use \begin{align} or \begin{align*} instead of \begin{equation}\begin{split} as it is the norm now.

Here is an example :

\begin{align*}
    &P(T_{1} \in (x_{1},x_{1}+h_{1}], \dots, T_{n} \in (x_{n},x_{n}+h_{n}], N(t)=n) = 
    \\
    & \hspace{1cm} \{N(0,x_{1}]=0, N(x_{1},x_{1}+h_{1}] = 1, N(x_{1}+h_{1},x_{2}]=0, 
    \\  
    & N(x_{2}, x_{2}+h_{2}]=1,\dots, N(x_{n-1}+h_{n-1},x_{n}]=0, 
    \\
    & N(x_{n},x_{n}+h_{n}]=1, N(x_{n}+h_{n},t]=0\}
\end{align*}

Note \hspace to make a ligne start just a bit later in order to make it look like a line break instead of a new line.