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.

  • $\begingroup$ This is "convergence in distribution". There are many notions of convergence in probability theory. $\endgroup$ – Zhen Lin May 2 '12 at 9:45
  • $\begingroup$ 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." $\endgroup$ – André Nicolas May 2 '12 at 10:15
  • $\begingroup$ @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 ? $\endgroup$ – MyCatsHat May 2 '12 at 13:40
  • $\begingroup$ 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. $\endgroup$ – André Nicolas May 2 '12 at 13:51
  • $\begingroup$ @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...) $\endgroup$ – MyCatsHat May 2 '12 at 13:57

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.

  • $\begingroup$ 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. $\endgroup$ – MyCatsHat May 2 '12 at 13:50
  • $\begingroup$ 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. $\endgroup$ – Dawei May 2 '12 at 14:28

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