Evluate the limit:
$$\lim_{n \to +\infty} \left[ {e}^{-n} \sum_{r=0}^{n} \frac{ n^r }{ r! } \right] = \frac{1}{2}$$
Solution
We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_n-n)/\sqrt{n}\) converges in distribution to the standard normal distribution. In particular, if \(N\) follows a standard normal distribution then \[ \lim_{n \to \infty}\Pr(Y_n \leqslant n) = \lim_{n \to \infty}\Pr\left( \frac{Y_n - n}{\sqrt{n}} \leqslant 0\right) = P(N \leqslant 0) = \frac{1}{2},\] from which we can now read the result.
$$\lim_{n \to +\infty} \left[ {e}^{-n} \sum_{r=0}^{n} \frac{ n^r }{ r! } \right] = \frac{1}{2}$$
Solution
We consider independent Poisson distributions \(X_1,X_2,\ldots\) with parameter 1. We know that \(Y_n = X_1 + \cdots + X_n\) is Poisson with parameter \(n\). From the central limit theorem, \((Y_n-n)/\sqrt{n}\) converges in distribution to the standard normal distribution. In particular, if \(N\) follows a standard normal distribution then \[ \lim_{n \to \infty}\Pr(Y_n \leqslant n) = \lim_{n \to \infty}\Pr\left( \frac{Y_n - n}{\sqrt{n}} \leqslant 0\right) = P(N \leqslant 0) = \frac{1}{2},\] from which we can now read the result.
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