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'Bernoulli distribution'
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| Title of object: |
Bernoulli distribution |
| Canonical Name: |
BernoulliDistribution2 |
| Type: |
Definition |
| Created on: |
2002-09-13 22:46:55.44697-04 |
| Modified on: |
2002-09-13 23:34:42.530665-04 |
| Classification: |
msc:60E05 |
| Synonyms: |
Bernoulli distribution=binomial distribution |
Preamble:
\usepackage{graphicx}
%\usepackage{xypic}
\usepackage{bbm}
\newcommand{\Z}{\mathbbmss{Z}}
\newcommand{\C}{\mathbbmss{C}}
\newcommand{\R}{\mathbbmss{R}}
\newcommand{\Q}{\mathbbmss{Q}}
\newcommand{\mathbb}[1]{\mathbbmss{#1}}
\newcommand{\figura}[1]{\begin{center}\includegraphics{#1}\end{center}}
\newcommand{\figuraex}[2]{\begin{center}\includegraphics[#2]{#1}\end{center}} |
Content:
Consider an experiment with two possible outcomes (success and failure), which happen randomly. Let $p$ be the probability of success. If the experiment is repeated $n$ times, the probability of having exactly $x$ successes is
$$f(x)=\left({n\atop x}\right)p^x(1-p)^{(n-x)}.$$
The distribution function determined by the probability function $f(x)$ is called a \emph{Bernoulli distribution} or \emph{binomial distribution}.
Here are some plots for $f(x)$ with $n=20$ and $p=0.3$, $p=0.5$.
\figuraex{binom10p3}{scale=0.75}
\figuraex{binom10p5}{scale=0.75}
The corresponding distribution function is
$$F(n)=\sum_{k\leq x}\left({n\atop k}\right)p^k q^{n-k}$$
where $q=1-p$. Notice that if we calculate $F(n)$ we get the binomial expansion for $(p+q)^n$, and this is the reason for the distribution being called binomial.
We will use the moment generating function to calculate the mean and variance for the distribution. The mentioned function is
$$G(t)=\sum_{x=0}^n e^{tx}\left({n\atop x}\right)p^x q^{n-x}$$
wich simplifies to
$$G(t)=(pe^t+q)^n.$$
Differentiating gives us
$$G'(t)=n(pe+q)^{n-1}pe^t$$
and therefore the mean is
$$\mu = E[X]=G'(0)=np.$$
Now for the variance we need the second derivative
$$G''(t)=n(n-1)(pe^t+q)^{n-2} + n(pe^t+q)^{n-1}pe^t$$
so we get $$E[X^2]=G''(0)=n(n-1)p^2 + np$$
and finally the variance (recall $q=1-p$):
$$\sigma^2 = E[X^2] - E[X]^2 = npq.$$ |
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