오늘의 계산 50 - Limit of a Probabilistic Sum

Enjoy this calculation and have a good day!

Problem. For $\alpha > 0$, examine the following limit \begin{equation}\label{eqn:wts} \lim_{n\to\infty} e^{-\alpha\sqrt{n}} \sum_{k=0}^{n-1} 2^{-n-k} \binom{n-1+k}{k} \sum_{m=0}^{n-1-k}\frac{(\alpha\sqrt{n})^m}{m!}. \end{equation}

Solution. Let $A_n$ denote the formula inside the limit \eqref{eqn:wts}. By noting that the double summation is taken for those non-negative integers $k, m$ with $l := k+m \leq n-1$, by changing the order of summation, \begin{align*} A_n &= e^{-\alpha\sqrt{n}} \sum_{m=0}^{n-1}\sum_{k=0}^{n-1-m} \binom{n-1+k}{k} 2^{-n-k}\frac{(\alpha \sqrt{n})^m}{m!} \\ &= \frac{e^{-\alpha\sqrt{n}}}{(n-1)!} \sum_{m=0}^{n-1} \left( \sum_{k=0}^{n-1-m} \frac{1}{k!} \frac{(n-1+k)!}{2^{n+k}} \right) \frac{(\alpha \sqrt{n})^m}{m!} \\ &= \frac{e^{-\alpha\sqrt{n}}}{(n-1)!} \sum_{m=0}^{n-1} \left( \sum_{k=0}^{n-1-m} \frac{1}{k!} \int_{0}^{\infty} x^{n+k-1}e^{-2x}\;dx \right) \frac{(\alpha \sqrt{n})^m}{m!} \\ &= \frac{e^{-\alpha\sqrt{n}}}{(n-1)!} \int_{0}^{\infty} \left( \sum_{m=0}^{n-1} \sum_{k=0}^{n-1-m} \frac{x^k}{k!} \frac{(\alpha \sqrt{n})^m}{m!} \right) x^{n-1} e^{-2x}\;dx \\ &= \frac{e^{-\alpha\sqrt{n}}}{(n-1)!} \int_{0}^{\infty} \left( \sum_{l=0}^{n-1} \sum_{k=0}^{l} \frac{x^k}{k!} \frac{(\alpha \sqrt{n})^{l-k}}{(l-k)!} \right) x^{n-1} e^{-2x}\;dx \\ &= \frac{e^{-\alpha\sqrt{n}}}{(n-1)!} \int_{0}^{\infty} \left( \sum_{l=0}^{n-1} \frac{(x+\alpha\sqrt{n})^l}{l!} \right) x^{n-1} e^{-2x}\;dxdx \\ &= \int_{0}^{\infty} \left( \sum_{l=0}^{n-1} \frac{(x+\alpha\sqrt{n})^l}{l!} e^{-(x+\alpha\sqrt{n})} \right) \frac{x^{n-1} e^{-x}}{(n-1)!}\;dx. \end{align*} Also, observe that \begin{align*} &\sum_{l=0}^{n-1} \frac{(x+\alpha\sqrt{n})^l}{l!} e^{-(x+\alpha\sqrt{n})} \\ &\hspace{5em}= \frac{1}{(n-1)!}\sum_{l=0}^{n-1} \binom{n-1}{l} (x+\alpha\sqrt{n})^l e^{-(x+\alpha\sqrt{n})} \int_{0}^{\infty} t^{n-1-l}e^{-t} \; dt \\ &\hspace{5em}= \frac{1}{(n-1)!} \int_{0}^{\infty} (t+x+\alpha\sqrt{n})^{n-1} e^{-(t+x+\alpha\sqrt{n})} \; dt \\ &\hspace{5em}= \frac{1}{(n-1)!} \int_{x+\alpha\sqrt{n}}^{\infty} t^{n-1}e^{-t} \; dt. \end{align*} Thus if we define $G_n(x)$ as \begin{equation*} G_n(x) = \int_{x}^{\infty} \frac{x^{n-1}}{(n-1)!}\,e^{-x} \; dx, \end{equation*} the above calculation shows that we can write \begin{equation}\label{eqn:a_n} A_n = - \int_{0}^{\infty} G_n(x+\alpha\sqrt{n})G_n'(x)\;dx. \end{equation}

Now let $\{ X_1, X_2, \cdots, Y_1, Y_2, \cdots \}$ be a family of independent random variables each having an exponential distribution of parameter $1$. Then for its partial sums $S_n = X_1 + \cdots + X_n$ and $T_n = Y_1 + \cdots + Y_n$, it is easy to see that \begin{equation*} F_{S_n}(x) = \Bbb{P}(S_n \leq x) = 1 - G_n (x), \end{equation*} and likewise for $F_{T_n}(x) = \Bbb{P}(T_n \leq x)$. Thus \eqref{eqn:a_n} reduces to \begin{align*} A_n &= \int_{0}^{\infty} \Bbb{P}\left(T_n > x+\alpha\sqrt{n}\right) \; dF_{S_n}(x) \\ & = \Bbb{P}\left(T_n > S_n +\alpha\sqrt{n}\right) = \Bbb{P}\left(\frac{T_n - n}{\sqrt{n}} > \frac{S_n - n}{\sqrt{n}} +\alpha\right). \end{align*} Since $\Bbb{E}S_n = \Bbb{E}S_n = n$ and $\Bbb{V}S_n = \Bbb{V}T_n = n$, Central Limit Theorem yields \begin{equation*} \lim_{n\to\infty} A_n = \Bbb{P}\left(Z_2 > Z_1 +\alpha\right), \end{equation*} where $Z_i \sim N(0, 1)$ are independent random variables each having a standard normal distribution. Therefore we have \begin{equation*} \lim_{n\to\infty} A_n = \frac{1}{2}\left[ 1 - \mathrm{erf}\left(\frac{\alpha}{2}\right)\right]. \end{equation*}

References

1. Help evaluating a limit in "Math StackExchange"