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Yes, I have to admit... I am having such a hard to write this formula in LaTeX, especially with \begin/\end{matrix}. This is what I have got so far. Any help will be greatly appreciated. Thanks.

F(\textbf{u};z)|Z(\textbf{u}^{{}'})\leq z^{{}'})=Prob\begin{Bmatrix}
Z(\textbf{u})\leq z,Z(\textbf{u}^{{}'})\leq z^{{}'}\
\end{Bmatrix}=\frac{ProbZ(\textbf{u})\leq z,Z(\textbf{u}^{{}'})\leq z^{{}'}}{ProbZ(\textbf{u}^{{}'})\leq z^{{}'}}=\frac{F(\textbf{u,u}{'};z,z{'})}{F(\textbf{u}^{{}'};z{'})}

enter image description here

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  • I've tried to format your code but it is not easy to parse it. There seems be be a backslash at the end of the second line. Is something missing? Also, this has nothing, as far as I can tell, to do with LaTeX 3. Could you tidy up your code and complete it so that we have a complete minimal example we can compile?
    – cfr
    Commented Dec 20, 2015 at 2:35

1 Answer 1

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There is no need for matrix constructions here:

enter image description here

\documentclass{article}

\usepackage{amsmath}
\DeclareMathOperator{\prob}{Prob}

\begin{document}

\begin{align*}
  F\bigl(\mathbf{u};z \mid Z(\mathbf{u}') \leq z'\bigr)
    &= \prob\bigl\{ Z(\mathbf{u}) \leq z \mid Z(\mathbf{u}') \leq z'\bigr\} \\
    &= \dfrac{\prob \bigl\{Z(\mathbf{u}) \leq z,Z(\mathbf{u}') \leq z'\bigr\}}
             {\prob \bigl\{Z(\mathbf{u}') \leq z'\bigr\}} \\
    &= \dfrac{F(\mathbf{u},\mathbf{u}';z,z')}
             {F(\mathbf{u}';z')}
\end{align*}

\end{document}

Use \{...\} for setting braces in math mode.

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  • I tend to assume I'm mistaken if it involves typesetting maths. Thanks.
    – cfr
    Commented Dec 20, 2015 at 2:48
  • @Mico: Much better!
    – Werner
    Commented Dec 20, 2015 at 7:10

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