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

I am trying to define an indexing $n$/$m$ and I am sending $m$ to infinity, but I get zero...not some relevant distribution. What is the technique or approach one must use here?!

share|cite|improve this question

Recall pmf of the geometric distribution: $\mathbb{P}(X = k) = p (1-p)^k$ for $k \geq 0$.

The geometric distribution has the interpretation of the number of failures in a sequence of Bernoulli trials until the first success.

Consider a regime when the probability of success is very small, such that $n p = \lambda$, and consider $x = \frac{k}{n}$. Then, in the large $n$ limit: $$ 1 = \sum_{k=0}^\infty \mathbb{P}(X = k) = \sum_{k=0}^\infty \lambda \left(\left(1 - \frac{\lambda}{n} \right)^{n \cdot k/n} \frac{1}{n} \right) \stackrel{n \to \infty}\rightarrow \int_0^\infty \lambda \mathrm{e}^{-\lambda x} \mathrm{d} x $$

Alternatively, you could look at the moment generating function for the geometric distribution: $$ \mathcal{M}(p, t) = \frac{p}{1-\mathrm{e}^t (1-p)} $$ To recover the mgf of the exponential distribution consider the limit: $$ \lim_{n \to \infty} \mathcal{M}\left( \frac{\lambda}{n}, \frac{t}{n} \right) = \lim_{n \to \infty} \frac{\lambda}{n - \mathrm{e}^{t/n}\left(n - \lambda\right)} = \lim_{n \to \infty} \frac{\lambda}{n \left(1 - \mathrm{e}^{t/n}\right) + \lambda \mathrm{e}^{t/n} } = \frac{\lambda}{\lambda-t} $$ which is the mgf of the exponential distribution with rate $\lambda$.

share|cite|improve this answer
there needs to be some change of variables in the sum and we're set. I guess the technique in finding continuous analogues is to sum over all k's instead of just trying to do something with one k. – Ramamurthy Shankar Dec 21 '11 at 4:52

It may be useful to consider things the other way round, that is, to start from a single exponential random variable $X_\lambda$ with density $\lambda\mathrm e^{-\lambda x}$ on $x\geqslant0$, for a given positive $\lambda$, and, for every positive $a$, to consider the integer part $Y_a$ of $X_\lambda/a$.

Then $[Y_a=n]=[na\leqslant X_\lambda\lt (n+1)a]$ hence $\mathrm P(Y_a=n)=\mathrm e^{-\lambda na}-\mathrm e^{-\lambda (n+1)a}$. One sees that $\mathrm P(Y_a=n)=(p_a)^n(1-p_a)$ with $$ p_a=\mathrm e^{-\lambda a}. $$ This proves that each $Y_a$ is geometric with parameter $p_a$. Note that, for a given $\lambda$, every parameter $p$ in $(0,1)$ is realized as $p=p_a$ for some $a$, hence this construction yields a whole family of geometric random variables $(Z_p)_{p\in(0,1)}$ such that each $Z_p$ is geometric with parameter $p$ and $Z_p\leqslant Z_q$ with full probability, for every $p\geqslant q$ in $(0,1)$: simply define each $Z_p$ by $Z_p=Y_{\alpha(p)}$, with $$ \alpha(p)=-\log(p)/\lambda. $$ Coming back to your question, $X_\lambda\leqslant aY_a\lt X_\lambda+a$ with full probability, hence $aY_a\to X_\lambda$ almost surely when $a\to0$, in particular $\alpha(p)Z_p\to X_\lambda$ when $p\to1$. Thus, $-\log(p)Z_p$ converges almost surely (hence in distribution) to the standard exponential random variable $X_1=\lambda X_\lambda$.

Since $-\log(p)\sim1-p$ when $p\to1$, this shows that $(1-p)Z_p$ converges almost surely (hence also in distribution) to $X_1$. Now, the convergence in distribution of some random variables does not depend on their realization hence, by the remarks above, for every family $(T_p)_{p\in(0,1)}$ of random variables such that each $T_p$ is geometrically distributed with parameter $p$, the random variables $(1-p)T_p$ converge in distribution to the standard exponential distribution when $p\to1$.

Note Such a simultaneous and almost sure construction of some geometric random variable $Z_p$ for every $p$ in $(0,1)$ from a single (exponentially distributed) random variable $X_\lambda$ is called a coupling. For more about this powerful idea, see here.

share|cite|improve this answer

The waiting time $T$ until the first success in a sequence of independent Bernoulli trials with probability $p$ of success in each one has a geometric distribution with parameter $p$: its probability mass function is $P(x) = p (1-p)^{x-1}$ and cumulative distribution function $F(x) = 1 - (1-p)^x$ for positive integers $x$ (note that some authors use a different convention where the random variable is $T-1$ rather than $T$, but that won't make a difference in the limit). The scaled version $p T$ converges in distribution as $p \to 0+$ to an exponential random variable with rate $1$, as for $x \ge 0$ $$ P(p T \le x) = F(x/p) = 1 - (1-p)^{\lfloor x/p\rfloor} \to 1 - e^{-x} $$

share|cite|improve this answer
Can you show more details about how $$(1-p)^{\lfloor x/p\rfloor} \to e^{-x}$$. – 81235 Jan 19 '15 at 15:08
Since $0 \le x/p - \lfloor x/p \rfloor \le 1$, $(1-p)^{\lfloor x/p \rfloor - x/p} \to 1$, while $\ln \left((1-p)^{x/p}\right) = x \dfrac{\ln(1-p)}{p} \to -x$ by definition of derivative of $\ln(x)$ at $x=1$. – Robert Israel Jan 19 '15 at 18:04

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.