Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Fix positive $\lambda$ so that $\frac{\lambda}{n} \in [0,1]$ for all $n \ge 1$. Next, let $P_n(k)$ denote the probability of recording exactly $k \le n$ heads in an $n$-toss experiment, where we assume that the probability of obtaining heads in any single toss is $\frac{\lambda}{n}$. An explicit formula for $P_k(n)$ is given by

$$P_n(k) = \displaystyle {{n} \choose {k}}\Big(\frac{\lambda}{n}\Big)^k\Big(1 - \frac{\lambda}{n}\Big)^{n-k}.$$

Show that $\displaystyle \lim_{n \to \infty} P_n(k)$ exists and determine its value.

I expect that $P_n(k) \to 0$, independent of what $k$ is. This is because, as the number of tosses in the experiment approaches infinity, the probability of obtaining any one particular outcome should approach zero. I have tried to adapt the solutions to the following problem: Determine $\displaystyle \lim_{n \to \infty}{{n} \choose {\frac{n}{2}}}\frac{1}{2^n}$, where each $n$ is even. Is it also possible to apply Stirling's Approximation to this limit?

Hints or solutions are greatly appreciated.

share|cite|improve this question
Did's answer here may help… – Byron Schmuland Jan 31 '13 at 0:04
It will be $e^{-\lambda}\frac{\lambda^k}{k!}$, not $0$. – André Nicolas Jan 31 '13 at 0:04
up vote 3 down vote accepted

Note that $$\left(1 - \dfrac{\lambda}n\right)^{n-k} \sim \dfrac{e^{-\lambda}}{\left(1 - \dfrac{\lambda}n \right)^k}$$ Hence, $$\dbinom{n}k \left(\dfrac{\lambda}n \right)^k \left(1 - \dfrac{\lambda}n\right)^{n-k} \sim \dbinom{n}k \left(\dfrac{\lambda}{n - \lambda} \right)^k e^{-\lambda} = \dfrac{\lambda^k}{k!} e^{-\lambda} \dfrac{n(n-1)(n-2) \cdots(n-k+1)}{(n- \lambda)^k}$$ Now note that $$\dfrac{n(n-1)(n-2) \cdots(n-k+1)}{(n- \lambda)^k}$$ is nothing but $$\dfrac{1 \times \left(1 - \dfrac1n \right) \times \left(1 - \dfrac2n \right) \times \left(1 - \dfrac3n \right) \times \cdots \times \left(1 - \dfrac{k-1}n \right)}{\left(1 - \dfrac{\lambda}n \right)^k} \sim 1$$ Hence, your expression tends to $\dfrac{\lambda^k}{k!} e^{-\lambda} $.

share|cite|improve this answer

Here another way to proof $\lim\limits_{n\to \infty}{{n} \choose {k}}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k}= e^{-\lambda}\dfrac{\lambda^k}{k!}$

Note that $$\frac{\lambda ^k}{k!}\left(1- \frac{\lambda}{n}\right)^{n-k} \geq {{n} \choose {k}}\left(\frac{\lambda}{n}\right)^k\left(1 - \frac{\lambda}{n}\right)^{n-k} \geq \frac{\lambda ^k}{k!}\left(1- \frac{k}{n}\right)^k\left(1- \frac{\lambda}{n}\right)^{n-k}$$ As $n \to \infty$ both R.H.S and L.H.S equal to $e^{-\lambda}\dfrac{\lambda^k}{k!}$.

share|cite|improve this answer

Your Answer


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.