Irwin-Hall 分布
对于 \(n\) 个均匀分布于 \([0,1]\) 的连续随机变量 \(X_1,X_2,\dots,X_n\)，其和的随机变量 \(X\) 满足：
\[P(X\le x)=\sum _{k=0}^{\lfloor x\rfloor}(-1)^k\binom nk\frac{(x-k)^n}{n!}
\]
证明：
设 \(f(x)\) 为 \(X_i\) 的概率密度函数。显然：
\[f(x)=\left\{\begin{matrix}
1 & (x\in [0,1])\\
0 & \text{otherwise}
\end{matrix}\right.
\]
那么
\[P(X\le x)=\int _{x_i\in [0,1],\sum x_i\le x}\prod f(x_i)\prod dx_i\\
=\int _{x_i\in [0,1],\sum x_i\le x}\prod dx_i\\
\]
采用归纳法证明。
\[\int _{x_i\in [0,1],\sum x_i\le x}\prod_{i=1}^{n+1} dx_i=\int_0^1\int _{x_i\in [0,1],\sum_{i=1}^n x_i\le x-x_{n+1}}\left(\prod_{i=1}^n dx_i\right) d{x_{n+1}}\\
=\int_0^1\sum_{k=0}^{\lfloor x-x_{n+1}\rfloor}(-1)^k\binom nk \frac{(x-k-x_{n+1})}{n!}dx_{n+1}\\
\]
记 \([x]=x-\lfloor x\rfloor\)，
\[\int_0^1\sum_{k=0}^{\lfloor x-x_{n+1}\rfloor}(-1)^k\binom nk \frac{(x-k-x_{n+1})}{n!}dx_{n+1}\\
=\int_0^{[x]}\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom nk \frac{(x-k-x_{n+1})}{n!}dx_{n+1}+\int_{[x]}^1\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^k\binom nk \frac{(x-k-x_{n+1})}{n!}dx_{n+1}\\
=\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom nk\frac{(x-k)^{n+1}-(x-k-[x])^{n+1}}{(n+1)!}+\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^k\binom nk\frac{(x-k-[x])^{n+1}-(x-k-1)^{n+1}}{(n+1)!}\\
=\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^k\binom nk\frac{(x-k)^{n+1}-(x-k-1)^{n+1}}{(n+1)!}+(-1)^{\lfloor x\rfloor}\binom n{\lfloor x\rfloor}\frac{[x]^{n+1}}{(n+1)!}\\
=\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^k\binom nk\frac{(x-k)^{n+1}}{(n+1)!}+\sum_{k=1}^{\lfloor x\rfloor}(-1)^k\binom n{k-1}\frac{(x-k)^{n+1}}{(n+1)!}+(-1)^{\lfloor x\rfloor}\binom n{\lfloor x\rfloor}\frac{[x]^{n+1}}{(n+1)!}\\
=\sum_{k=0}^{\lfloor x\rfloor-1}(-1)^k\binom {n+1}k\frac{(x-k)^{n+1}}{(n+1)!}+(-1)^{\lfloor x\rfloor}\binom n{\lfloor x\rfloor-1}\frac{[x]^{n+1}}{(n+1)!}+(-1)^{\lfloor x\rfloor}\binom n{\lfloor x\rfloor}\frac{[x]^{n+1}}{(n+1)!}\\
=\sum_{k=0}^{\lfloor x\rfloor}(-1)^k\binom {n+1}k\frac{(x-k)^{n+1}}{(n+1)!}
\]
不难发现 \(n=1\) 时命题成立，证毕。