Can anyone help me with it: Using the central limit theorem for suitable Poisson random variables, prove that $$ \lim_{n\to\infty} e^{-n} \sum_{k=0}^{n} \frac{n^k}{k!}=1/2$$ Thanks!
Using Central Limit Theorem
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stochastic-processes
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0This question has been asked and answered [countless](http://math.stackexchange.com/questions/1297553) times. – 2015-12-07
1 Answers
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Hint: A Poisson$(n)$ random variable can be represented as the sum of $n$ i.i.d. Poisson$(1)$ rv's.
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0I'll give further hints if you wish. – 2011-02-13
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0@Shai Covo: Can you show me the steps? Thanks! – 2011-02-13
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1@kira: Hint 2: Move the exponent into the sum. – 2011-02-13
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1@kira: Find a random variable $X$ for which it holds ${\rm P}(X=n) = \sum\nolimits_{k = 0}^n {\frac{{e^{ - n} n^k }}{{k!}}} $. – 2011-02-14