Originally posted by BillHoyt
Re-read your source. With understanding, please. That is not a description of the conditions under which Poisson is valid. Neither is it a description of Poisson as "Binomial lite". If you have real trouble understanding this, I will help you. That is an honest offer, if you approach this honestly. Unfortunately, your approach is part of the problem for me.
Please help me. Nothing Lurker wrote leads me to believe that he misunderstood his
source. What do you think is his misunderstanding?
Now let me give you an exercise from Hogg & Craig's Introduction to Mathematical Statistics, third edition. It is from page 98. If you don't have a copy, just ask anyone who's been through Cherry's Stat 231 course here at _____'s exotic dance bar. She used to hold it late in the evenings, or early morning's, depending on your perspective.
"3.23 Let X have a Poisson distribution with mu=100. Use Chebyshev's inequality to determine a lower bound for Pr(75 < X < 125)."
Chebyshev's inequality places an upper bound on the probability that a random variable is far from its mean. Specifically, letting mu be the mean of
X,
V be the variance of
X, and
d be the distance that we're interested in, it says that<blockquote>Pr( |
X-mu| >=
d ) <=
V/d<sup>2</sup>.</blockquote>The probability of the complementary event, namely that
X is within the distance
d of its mean, is therefore greater than 1 -
V/d<sup>2</sup>.
In this exercise,
d is 25, the mean is 100, and, since the variance of a Poisson random variable equals its mean,
V is also 100. So we get<blockquote>Pr(75 <
X < 125) > 1 - 100/25<sup>2</sup> = 21/25.</blockquote>Chebyshev's inequality is nice because it applies to any random variable regardless of its distribution, provided we know its mean and variance. (Well, any random variable that has a mean and variance, anyway. But that's most of them.) However, if we know the exact distribution, as we do here, we can often do much better. Chebyshev's inequality here yields a lower bound of 21/25 = 0.84. The actual value of the probability is about 0.986.
Poisson with mu=100? What justifies that?
I don't understand. It's a textbook exercise. It says that
X has a Poisson distribution with mu = 100. So by definition
X does have that distribution. What kind of justification are you looking for?
I don't see the relevance of a textbook exercise about Chebyshev's inequality, even one which happens to involve a Poisson distribution, to the question of when in real life it is appropriate to use a Poisson distribution to model a situation we're faced with.