Let $X_1,X_2,\ldots,X_m$ be independent standard normal random variables. Then \\
\begin{align}
\textbf{E}(X_{1}^2+X_{2}^2+\ldots+X_{m}^2)^n
&= \sum_{substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\binom{n}{i_1,\ldots,i_m}\textbf{E}X_{1}^{2i_1}\ldots\textbf{E}X_{m}^{2i_m}\\
&= \sum_{substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\binom{n}{i_1,\ldots,i_m}(2i_{1}-1)!!\ldots(2i_{m}-1)!!\\
&= \sum_{substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\frac{n!\times2^{i_1}i_{1}!(2i_{1}-1)!!\ldots2^{i_m}i_{m}!(2i_{m}-1)!!}{2^{n}(i_{1}!)^2\ldots(i_{m}!)^2}\\
&= \sum_{substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\frac{n!}{2^n}\binom{2i_1}{i_1}\ldots\binom{2i_m}{i_m}.
\end{align}
答案1
\
您之前缺少一些内容\substack
,但请始终发布完整的文件:
\documentclass{article}
\usepackage{amsmath}
\begin{document}
Let $X_1,X_2,\ldots,X_m$ be independent standard normal random variables. Then \\
\begin{align}
\textbf{E}(X_{1}^2+X_{2}^2+\ldots+X_{m}^2)^n
&= \sum_{\substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\binom{n}{i_1,\ldots,i_m}\textbf{E}X_{1}^{2i_1}\ldots\textbf{E}X_{m}^{2i_m}\\
&= \sum_{\substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\binom{n}{i_1,\ldots,i_m}(2i_{1}-1)!!\ldots(2i_{m}-1)!!\\
&= \sum_{\substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\frac{n!\times2^{i_1}i_{1}!(2i_{1}-1)!!\ldots2^{i_m}i_{m}!(2i_{m}-1)!!}{2^{n}(i_{1}!)^2\ldots(i_{m}!)^2}\\
&= \sum_{\substack{i_1,i_2,\ldots,i_m\geq0\\i_1+\ldots+i_m=n}}\frac{n!}{2^n}\binom{2i_1}{i_1}\ldots\binom{2i_m}{i_m}.
\end{align}
\end{document}
以上没有错误,但是范围太广,如果使用
\textbf{E}(X_{1}^2+\ldots+X_{m}^2)^n
丢失X_2
术语会更合适。