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The text I'm reading doesn't offer an explanation beyond

"limit of binomial variance $np(1-p)$. So variance = $\lambda$"

I don't think what was written builds much understanding. The binomial distribution's variance is $np(1-p) = np - p^2$ which is $\lambda - p^2 \neq \lambda$.

I think what's supposed to be shown is that $\lim n\to \infty$ even with small $p$ produces $\lambda$ but $p^2$ is only epsilon greater than 0 so it disappears.

I've found proofs that use the variance shortcut formula as well as simplying stating it's a property of the distribution but I haven't found a proof that uses the idea the author of the book brought up. Is there one? Or is it not any more formal than what I wrote above?

financial_physician
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  • Check your formula $np(1-p) = np - p^2$ it should be $np(1-p) = np - n p^2$. That is $\lambda -\lambda p = \lambda (1-p)$ and in the limit that is $\lambda$. See also https://stats.stackexchange.com/questions/261119/intuitively-understand-why-the-poisson-distribution-is-the-limiting-case-of-the – kjetil b halvorsen Oct 06 '20 at 02:56
  • That’s a big algebra error on my part. Thank you. Why is $\lim n \to \infty \lambda(1-p) = \lambda$? Is it because p is close to 0? – financial_physician Oct 06 '20 at 03:11
  • Because the relevant limit is when $n \to\infty$ and $n p \to\lambda>0$, which implies $p\to 0$. – kjetil b halvorsen Oct 06 '20 at 03:14
  • So now you can fix that and self-answer? – kjetil b halvorsen Oct 06 '20 at 03:15
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    Respond to my own post? Yeah I can do that. I didn’t think I was supposed to haha – financial_physician Oct 06 '20 at 04:29
  • Somewhere you have that the mean is $\lambda=np$ or something similar, and this does not change as $n$ increases. This then gives $p=\frac{1}{n}\lambda$, so the variance is $np(1-p)=np-np^2=\lambda - \frac{1}{n}\lambda^2$ which tends $\to \lambda$ as $n$ increases – Henry Oct 06 '20 at 07:56

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thanks to kjetil, I now understand two things that were erroneous in my question.

  1. First off, $np(1-p) \neq np - p^2$, it's equal to $np - np^2$

  2. $np(1-p)=\lambda(1-p) = \lambda$. Why does $\lambda(1-p) = \lambda$? Because it's part of our assumptions that $p \to 0$ for the poisson distribution.

This provides some pretty great discussion on how to conceptualize the poisson.

The discussion in my own words.

The most enlightening to me was thinking of the equation $p = \lambda/n$. The probability $p$ of an event occurring is the number_of_occurrences_over_some_interval_or_space/ number_of_occurrences_for_smaller_partitions_of_that_interval_or_space.

Really what we're trying to ask is if our event in our probability space can show up anywhere on the space with the same probability. We call how much we expect to see it $\lambda$ and then we can scale that does to some arbitrarily small measure.

At this point, it doesn't make any sense to even think about $n$. It's just a variable we used to conceptualize what the poisson distribution really is. What's the probability that your "arbitrarily small sample" contains the event? 0. There's no way it could happen. How often do you expect it to happen if you were to scale up that epsilon large sample ($\epsilon=p=\lambda/n$ where $n\to\infty$) by n? $\lambda$ because that's what we defined it to be.

Now we have the poisson distribution.

You'll notice at this point, $n$ amd $p$ don't seem to have much to do with the poisson distribution. Well yeah... That's why they're not variables when we're describing it's density function.

financial_physician
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