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I want to prove why the mean and variance of a $\operatorname{Poisson}(\lambda)$, is different when the time index approaches infinite (it's approximated by the mean and variance of a Normal).

For example: $$ N_k = N_1 + (N_2 - N_1) + (N_3 - N_2) + ... + (N_k - N_{k-1}) $$

Using CLT: $\frac{N_k - k\lambda}{\sqrt{k\lambda}}$ is normally distributed (in the limit).

I want to answer why is that a Poisson R.V. characterized by $\mathbb{E}[X] = \lambda$ and $\operatorname{Var}[X] = \lambda$. Because when it approaches a normal distribution, $\mathbb{E}[Z] = \mu$ and $\operatorname{Var}[Z] = \sigma^2$.

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up vote 4 down vote accepted

The Poisson distribution does not approach the normal distribution, the centered Poisson distribution does. More precisely, if $X_\lambda$ is Poisson with parameter $\lambda$, then $Y_\lambda$ converges in distribution to a standard normal random variable $Z$, where $Y_\lambda=(X_\lambda-\lambda)/\sqrt{\lambda}$. In particular, for every $\lambda$, $E[Y_\lambda]=E[Z]=0$ and $\mathrm{var}(Y_\lambda)=\mathrm{var}(Z)=1$ (in your language, $\mu=0$ and $\sigma^2=1$).

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Note that the title of the question does not match its content. –  Did May 16 '13 at 15:56
    
But, isn't that case with any random variable? The CLT says (X - mean)/standard_dev is normally distributed? Am I confused with my concetps? –  Josh May 16 '13 at 16:15
1  
The CLT certainly does not say that Y = (X - mean)/standard_dev is normally distributed. Note that Y is normally distributed only when X itself is normally distributed. –  Did May 16 '13 at 16:17
    
So the difference between a centered Poisson and CLT is that CLT says that a sample mean is normally distributed over infinite trials and centered Poisson is over once instance. –  Josh May 16 '13 at 16:26
    
Ok, so I just worked through an example with moment generating function and was able to prove mathematically why a centered poisson is normally distributed, but how can I explain this in plain English? –  Josh May 16 '13 at 16:45

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