Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Question Is this supposed to mean, that I have to actually find Poisson distributed random variable that for a fixed binomial distributed random variable approximates it for the values for which their images coincide (note that a Poisson distributed random variable necesserily has as image $\mathbb{N}$ (with $0$) but a binomial distributed random variable has as image only $\left\{ 0,\ldots,n\right\} $) ?

How is this approximation to be understood (in the illustrative example I have formally written it out) on the level of random variables, since it seems to me, that this is just a "pointwise" (meaning the points in the image of the random variable) approximation...

Illustrative Example I have to calculate the using the binomial and Poisson distribution the probability of getting at most $4$ times the number $1$ in a series of $1000$ games played games, where in each game, randomly on number ist picked ot of the set $\left\{ 1,\ldots,50\right\} $.

I know, that if $X$ is the random variable that counts if how often a $1$ has come up, then the distribution of $X$ is the binomial distrubution, since each instance of a games is a Bernoulli trial (either a $1$ has come up - with probability $p=\frac{1}{50}$ - or it hasn't), so I only have to calculate $$ \sum_{k=0}^{4}P\left(X=k\right) $$

which is not a difficult task!

But for the Poisson distribution a problem arises: We proved in our course a theorem concerning the binomial and Poisson distribution, that says, that the latter approximates the former, if the number of Bernoulli trials is very big. Formal statement: If $p_{n}$ is a sequence in the interval $\left[0,1\right]$ and $np_{n}\rightarrow\lambda$, then $$ P\left(X=k\right)=\binom{n}{k}p_{n}^{k}(1-p_{n})^{n-k}\rightarrow e^{-\lambda}\frac{\lambda^{k}}{k!}\ \text{for}\ n\rightarrow\infty. $$

Now I could of course just say that $$ P\left(X=k\right)\approx e^{-\lambda}\frac{\lambda^{k}}{k!}\ \text{for}\ \lambda=np $$

and then just calculate $$ \sum_{k=0}^{4}e^{-np}\frac{(np)^{k}}{k!}. $$

But this just doesn't seems right, since there isn't any random variable involved, that actually is $"e^{-np}\frac{(np)^{k}}{k!}"$-distributed. And I thought I would have to exhibit such a random variable and then somehow use the above approximation, because their images have (as noted at the start) different cardinalities.

share|improve this question
    
This is "convergence in distribution". There are many notions of convergence in probability theory. –  Zhen Lin May 2 '12 at 9:45
    
There is a random variable which has Poisson distribution with parameter $np$. Just define $Y$ as having that distribution. Then, as was proved in your course, $P(X=k)\approx P(Y=k)$. The approximation is even good for $k>n$, since if $k>n$ then $P(X=k)=0$ while $P(Y=k)$ is close to $0$. As you know, for the Poisson approximation to be decent, we need $n$ "large" and $np$ "moderate." –  André Nicolas May 2 '12 at 10:15
    
@AndréNicolas I'm confused here, about what it means to define $Y$. For me, that means to describe $Y$ as a function $Y:\Omega \rightarrow \mathbb{R}$, where $\Omega$ is my probability model. But in the example above $\Omega$ is finite,so $Y(\Omega)$ has to be finite as well, so there is no way define $Y$ such that its probability mass function is Poisson distributed - since a Poisson distribution requires the image of the random variable that has this distribution to be countably infinite. From your comment, I assume, defining $Y$ means something different...But what ? –  user26698 May 2 '12 at 13:40
    
I see your point, you are being very careful, perhaps a little too much so. Let us change the $\Omega$ for your binomial by including the integers beyond $n$. Assign $0$ probability to these. Then we can compare directly, sample spaces are identical. –  André Nicolas May 2 '12 at 13:51
    
@AndréNicolas Ok! Thanks a lot, this finally cleared it for me! Is this the way (=enlarging the domains and assigning zero probability) it is generally done, when one distribution approximates the other, but the domains of the respective random variables are differen ? (Sadly, I can't accept your comment as an answer...) –  user26698 May 2 '12 at 13:57

1 Answer 1

The random variable is $X$. I understand their cardinalities are different but the core concept is that the probability distribution function of a binomial distribution $B(n,p)$ converges to that of a Poisson distribution $\pi(np)$ (as $n\rightarrow\infty,p\rightarrow0$). That being said, let $Y\sim\pi(np)$, the probability that $Y\ge n+1$ is negligible, such that ignoring them will not introduce any serious problem.

Please be advised that it is the probability distribution function of a binomial distribution that converges to the probability distribution function of a Poisson distribution (with conditions aforementioned, of course). This Poisson limit theorem doesn't say anything about "one distribution converges to another". See wiki: http://en.wikipedia.org/wiki/Poisson_limit_theorem.

share|improve this answer
    
How is my random variable $k$ ? Shouldn't it be $X$ ? Also, please see my two comment above for AndreNicolas, which further pinpoint my problem/question. –  user26698 May 2 '12 at 13:50
    
Sorry I meant $X$ (Since we are considering the probability that $X=k$... I admit I made a mistake.) Please have a look at my editted answer. –  wangdw May 2 '12 at 14:28

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.