Use the Central Limit Theorem to proof that...

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FRMST

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Hi guys, I've been trying to do this excercise, but I can't.

Use the Central Limit Theorem to proof that:

\(\displaystyle \[\mathop {\lim }\limits_{n \to \infty } \sum\limits_{k = 0}^n {\frac{{{e^{ - n}}{n^k}}}{{k!}} = 1/2} \]


\)
 
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HINT: Did you notice the similarity to the Poisson random variable?.

The sum looks like the Poisson distribution \(\displaystyle \frac{{\lambda}^{k}}{e^{\lambda}k!}\)

In your sum, n is the mean. \(\displaystyle \displaystyle \sum_{k=0}^{n}\frac{n^{k}}{e^{n}k!}\)


By the CLT, \(\displaystyle Z=\frac{x-n}{\sqrt{n}}\)

\(\displaystyle P(X\leq n)=P(Z\leq 0)\)

By the CLT, Z converges to the standard normal.

This means that \(\displaystyle P(Z\leq 0)\to \frac{1}{2}\)
 
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galactus said:
HINT: Did you notice the similarity to the Poisson random variable?.

The sum looks like the Poisson distribution \(\displaystyle \frac{{\lambda}^{k}}{e^{\lambda}k!}\)

In your sum, n is the mean. \(\displaystyle \displaystyle \sum_{k=0}^{n}\frac{n^{k}}{e^{n}k!}\)


By the CLT, \(\displaystyle Z=\frac{x-n}{\sqrt{n}}\)

\(\displaystyle P(X\leq n)=P(Z\leq 0)\)

By the CLT, Z converges to the standard normal.

This means that \(\displaystyle P(Z\leq 0)\to \frac{1}{2}\)
Click to expand...

thank you
 
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