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\begin{align*}
\text{Mean: }\\
E[Y] & = f(y) \cdot 0 + \sum_{y=1}^{\infty} f(y)\cdot y\\
& = [p_i + (1-p_i)e^{-\lambda_i}] \cdot 0 + \sum_{y=1}^{\infty} (1-p_i)\frac{e^{-\lambda_i}\lambda_i^y}{y!} \cdot y\\
& = 0 + \sum_{y=1}^{\infty} (1-p_i)\frac{e^{-\lambda_i}\lambda_i^y}{y!} \cdot y\\
& = (1-p_i) e^{-\lambda_i} [\lambda_i+\frac{2\lambda_i^2}{2!}+\frac{3\lambda_i^3}{3!}+....]\\
& = (1-p_i) e^{-\lambda_i} \lambda_i [1+\frac{\lambda_i}{1!}+\frac{\lambda_i^2}{2!}+....]\\
& = (1-p_i)\lambda_i e^{-\lambda_i}e^{\lambda}\\
& = (1-p_i)\lambda.\\
\text{Variance: }\\
E[Y^2] & = f(y) \cdot 0 + \sum_{y=1}^{\infty} f(y)\cdot y^2\\
& = [p_i + (1-p_i)e^{-\lambda_i}] \cdot 0 + \sum_{y=1}^{\infty} (1-p_i)\frac{e^{-\lambda_i}\lambda_i^y}{y!} \cdot y^2\\
& = 0 + \sum_{y=1}^{\infty} (1-p_i)\frac{e^{-\lambda_i}\lambda_i^y}{y!} \cdot y^2\\
& = (1-p_i) e^{-\lambda_i} [\lambda_i+\frac{4\lambda_i^2}{2!}+\frac{9\lambda_i^3}{3!}+\frac{16\lambda_i^4}{4!}+\frac{25\lambda_i^5}{5!}+....]\\
& = (1-p_i) e^{-\lambda_i}[\lambda_i+(\lambda_i^2+\lambda_i^2)+(\frac{\lambda_i^3}{2!}+\lambda_i^3)+(\frac{\lambda_i^4}{3!} \\
+\frac{\lambda_i^4}{2!}+)+(\frac{\lambda_i^5}{4!}+\frac{\lambda_i^5}{3!})+....]\\
& = (1-p_i)e^{-\lambda_i} [\lambda_i(1+\lambda_i+\frac{\lambda_i^2}{2!} + \frac{\lambda_i^3}{3!}+.....)+\lambda_i^2(1+\lambda_i+\frac{\lambda_i^2}{2!} + \frac{\lambda_i^3}{3!}+.....)]\\
& = (1-p_i) e^{-\lambda_i} [\lambda_ie^{\lambda_i}+\lambda_i^2 e^{\lambda}]\\
& = (1-p_i) e^{-\lambda_i} e^{\lambda_i} (\lambda_i+\lambda_i^2) \\
& = (1-p_i)(\lambda_i+\lambda_i^2)\\
V[Y]= E[y^2]-(E[y])^2 = (1-p_i)(\lambda_i+\lambda_i^2) - ((1-p_i)\lambda)^2.
\end{align*}
答案1
这是一个猜测(因为您还没有告诉我们问题是什么):
\documentclass{article}
\usepackage{mathtools}
\begin{document}
\noindent Mean:
\begin{align*}
E[Y]
& = f(y) \cdot 0 + \sum_{y=1}^\infty f(y) \cdot y\\
& = \left[p_i + (1 - p_i) e^{-\lambda_i}\right] \cdot 0 + \sum_{y=1}^{\infty}(1 - p_i) \frac{e^{-\lambda_i} \lambda_i^y}{y!} \cdot y\\
& = 0 + \sum_{y=1}^\infty (1 - p_i) \frac{e^{-\lambda_i} \lambda_i^y}{y!} \cdot y\\
& = (1 - p_i) e^{-\lambda_i} \left[\lambda_i + \frac{2\lambda_i^2}{2!} + \frac{3\lambda_i^3}{3!}+\dots\right]\\
& = (1 - p_i) e^{-\lambda_i} \lambda_i \left[1 + \frac{\lambda_i}{1!} + \frac{\lambda_i^2}{2!}+\dots\right]\\
& = (1 - p_i) \lambda_i e^{-\lambda_i} e^{\lambda}\\
& = (1 - p_i) \lambda.
\intertext{Variance:}
E\left[Y^2\right]
& = f(y) \cdot 0 + \sum_{y=1}^\infty f(y) \cdot y^2\\
& = \left[p_i + (1 - p_i) e^{-\lambda_i}\right] \cdot 0 + \sum_{y=1}^\infty (1 - p_i) \frac{e^{-\lambda_i} \lambda_i^y}{y!} \cdot y^2\\
& = 0 + \sum_{y=1}^\infty (1 - p_i) \frac{e^{-\lambda_i} \lambda_i^y}{y!} \cdot y^2\\
& = (1 - p_i) e^{-\lambda_i} \left[\lambda_i + \frac{4\lambda_i^2}{2!} + \frac{9\lambda_i^3}{3!} + \frac{16\lambda_i^4}{4!} + \frac{25\lambda_i^5}{5!} + \dots\right]\\
& = (1 - p_i) e^{-\lambda_i} \biggl[\lambda_i + \left(\lambda_i^2 + \lambda_i^2\right) + \left(\frac{\lambda_i^3}{2!} + \lambda_i^3\right)\\
&\hphantom{{}= (1 - p_i) e^{-\lambda_i} \biggl[} + \left(\frac{\lambda_i^4}{3!} + \frac{\lambda_i^4}{2!}\right) + \left(\frac{\lambda_i^5}{4!} + \frac{\lambda_i^5}{3!}\right) + \dots\biggr]\\
& = (1 - p_i) e^{-\lambda_i} \biggl[\lambda_i \left(1 + \lambda_i + \frac{\lambda_i^2}{2!} + \frac{\lambda_i^3}{3!} + \dots\right)\\
&\hphantom{{}= (1 - p_i) e^{-\lambda_i} \biggl[} + \lambda_i^2 \left(1 + \lambda_i + \frac{\lambda_i^2}{2!} + \frac{\lambda_i^3}{3!} + \dots\right)\biggr]\\
& = (1 - p_i) e^{-\lambda_i} \left[\lambda_i e^{\lambda_i} + \lambda_i^2 e^{\lambda}\right]\\
& = (1 - p_i) e^{-\lambda_i} e^{\lambda_i} \left(\lambda_i + \lambda_i^2\right)\\
& = (1 - p_i) \left(\lambda_i + \lambda_i^2\right).
\end{align*}
So we have
\begin{align*}
V[Y]
& = E\left[y^2\right] - (E[y])^2\\
& = (1 - p_i) \left(\lambda_i + \lambda_i^2\right) - ((1 - p_i)\lambda)^2.
\end{align*}
\end{document}
答案2
我认为主要有两个问题:
&
您在最后一行等式中缺少一个对齐符号 ( ),以 开头;请在第一个 之前V[Y]= E[y^2]-(E[y])^2
插入以修复它。&
=
- 在 之前有一个非常糟糕且可能不必要的换行符
+\frac{\lambda_i^4}{2!}
;我建议删除该换行符。
您可能应该将非常长的环境拆分\align*
为两个单独的较短环境,分别用于均值和方差表达式。在此过程中,请将标题行(Mean:
和Variance
)从align*
环境中完全移除。
另外,您可能希望用 替换所有那些.....
和....
实例\cdots
。此外,您可能应该在很多地方增加方括号的大小,例如,明智地使用\Bigl[
和\Bigr]
。我不建议使用\left[
和 ,\right]
因为这会导致“栅栏”符号(从印刷上讲)相对于其周围环境太大。