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

Using an appropriate probability distribution or otherwise show that

$$\lim_{n\to\infty} \int_0^n e^{-x}{x^{n-1}\over(n-1)!}dx =0.5$$

share|cite|improve this question
Hey if you are a new user please know that to get an answer you must first show/explain what you have tried. – Lyapunov Jul 16 '12 at 6:09
@POTUS Not really. But it is good etiquette to do so. – Pedro Tamaroff Jul 16 '12 at 6:11
Your question is related to this one:… – qoqosz Jul 16 '12 at 6:24
@POTUS Your assertion is empirically false. Unfortunately. – Did Jul 16 '12 at 6:51
Hint: work with a sample of independent exponential law, and use central limit theorem. – Davide Giraudo Jul 16 '12 at 8:53
up vote 2 down vote accepted

Let $\{X_n\}$ a sequence of independent identically distributed random variable of exponential law of mean $1$, that is, a density of $X_1$ is $$f(x):=e^{-x}\chi_{\{x\geq 0\}}.$$ We want to know a density $f_n$ of $S_n:=\sum_{k=1}^nX_k$. We can use induction: $f_n(x)=e^{-x}\frac{x^{n-1}}{(n-1)!}\chi_{\{x\geq 0\}}$. It's true for $n=1$ and if it's true for a $n$, we use convolution: \begin{align} f_{n+1}(x)&=\int_{\Bbb R}f_n(t)f_1(x-t)dt\\ &=\int_{\Bbb R}e^{-t}\frac{t^{n-1}}{(n-1)!}\color{green}{\chi_{(0,+\infty)}(t)}e^{-(x-t)}\color{red}{\chi_{(0,+\infty)}(x-t)}dt\\ &=e^{-x}\int_{\color{green}0}^{\color{red}x}\frac{t^{n-1}}{(n-1)!}dt\\ &=e^{—x}\frac{x^n}{n!}. \end{align} Denote $I_n:=\int_0^ne^{—x}\frac{x^{n-1}}{(n-1)!}dx$. We have, since $X_n\geq 0$ and the integrand is a density of $S_n$, that \begin{align} I_n&=P\left(\sum_{j=1}^nX_j\leq \color{red}n\right)\\ &=P\left(\sum_{j=1}^n(X_j\color{red}{-1})\leq 0\right)\\ &=P\left(\frac{\sum_{j=1}^nX_j\color{red}{-E[X_j]}}{\sqrt n}\leq 0\right), \end{align} the expectation of $X_1$ being $1$. Since the set $(\infty,0]$ has a boundary of measure $0$ and $\frac{\sum_{j=1}^nX_j-E[X_j]}{\sqrt n}$ converges in law to a normal law of mean $0$ and variance $1$, say $N$ (it's given by the central limit theorem), we have by portmanteau theorem, $$\lim_{n\to +\infty}I_n=P(N\leq 0)=1/2,$$ $N$ being symmetric.

share|cite|improve this answer
please explain your convolution.and how you get summation(1,n)X_j-E[X_j]. – Argha Jul 17 '12 at 10:29
@Ranabir I've added the details you wanted. – Davide Giraudo Jul 17 '12 at 10:45

How about:

$\mathcal{L}( \int_0^n e^{-x}{x^{n-1}\over(n-1)!}dx) = \frac{1}{s}\mathcal{L}(e^{-x}{x^{n-1}\over(n-1)!}) = \frac{1}{s(s + 1)^n}$

$\mathcal{L}^{-1}(\frac{1}{s(s + 1)^n}) = 1 - \frac{\Gamma(n,n)}{\Gamma(n)}$

$\lim_{n\to\infty} 1 - \frac{\Gamma(n,n)}{\Gamma(n)} = 1 - \frac{1}{2} = \frac{1}{2}$


The above hack could rightly be criticized as sort of "begging the question"; it's certainly not clear that evaluating the limit involving the upper incomplete gamma function is any easier than the original problem involving the lower! So, let's try this:


$P(n) = \frac{1}{\Gamma(n)}\int_0^n e^{-x}{x^{n-1}}dx$

$Q(n) = \frac{1}{\Gamma(n)}\int_n^\infty e^{-x}{x^{n-1}}dx$

So it should be fairly obvious that for all real n:

$P(n) + Q(n) = 1$


$\mathcal{L}(P(n) + Q(n)) = \mathcal{L}(1) \implies \frac{1}{s(s + 1)^n} + \mathcal{L}Q(n) = \frac{1}{s}$

$\implies \mathcal{L}Q(n) = \frac{1}{s} - \frac{1}{s(s + 1)^n}$

$\lim_{x\to\infty} P(n) = \lim_{s\to 0}s\mathcal{L}P(n) = 1$

$\lim_{x\to 0} Q(n) = \lim_{s\to\infty}s\mathcal{L}Q(n) = 1$


$\lim_{n\to\infty} P(n) + Q(n) = 1$

$\lim_{n\to\infty}P(n) - Q(n) = 0$,

from which the respective limits follow.

share|cite|improve this answer
How you get the result of first line. Please explain – Argha Jul 17 '12 at 10:30

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.