Poisson theorem

Poisson 's theorem is a theorem in probability theory .

Wording

Let there be a series of Bernoulli test series. Let be $p_{n}$ ${\ displaystyle p_ {n}}$ - the probability of "success", $\mu _{n}$ ${\ displaystyle \ mu _ {n}}$ - the number of "successes".

Then if

$\lim _{n\to \infty }p_{n}=0;$ ${\ displaystyle \ lim _ {n \ to \ infty} p_ {n} = 0;}$
$\lim _{n\to \infty }np_{n}=\lambda ;$ ${\ displaystyle \ lim _ {n \ to \ infty} np_ {n} = \ lambda;}$
$\lambda >0;$ ${\ displaystyle \ lambda> 0;}$

then

\lim _{n\to \infty }P(\omega :\mu _{n}(\omega )=m)=e^{-\lambda }{\cfrac {\lambda ^{m}}{m!}}.

{\ displaystyle \ lim _ {n \ to \ infty} P (\ omega: \ mu _ {n} (\ omega) = m) = e ^ {- \ lambda} {\ cfrac {\ lambda ^ {m}} {m!}}.}

Proof

Using the Bernoulli formula, we obtain

\lim _{n\to \infty }P(\omega :\mu _{n}(\omega )=m)=C_{n}^{m}(p_{n})^{m}(1-p_{n})^{n-m}={\cfrac {n!}{m!(n-m)!}}{\bigg (}{\cfrac {\lambda }{n}}+o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}{\bigg )}^{m}{\bigg (}1-{\cfrac {\lambda }{n}}-o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}{\bigg )}^{n-m}=

{\ displaystyle \ lim _ {n \ to \ infty} P (\ omega: \ mu _ {n} (\ omega) = m) = C_ {n} ^ {m} (p_ {n}) ^ {m} (1-p_ {n}) ^ {nm} = {\ cfrac {n!} {M! (Nm)!}} {\ Bigg (} {\ cfrac {\ lambda} {n}} + o {\ bigg (} {\ cfrac {\ lambda} {n}} {\ bigg)} {\ bigg)} ^ {m} {\ bigg (} 1 - {\ cfrac {\ lambda} {n}} - o {\ bigg (} {\ cfrac {\ lambda} {n}} {\ bigg)} {\ bigg)} ^ {nm} =}

={\cfrac {1}{m!}}{\cfrac {(n-m+1)(n-m+2)\ldots n}{n^{m}}}{\bigg (}\lambda +o{\bigg (}\lambda {\bigg )}{\bigg )}^{m}{\bigg (}1-{\cfrac {\lambda }{n}}-o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}{\bigg )}^{n-m},

{\ displaystyle = {\ cfrac {1} {m!}} {\ cfrac {(n-m + 1) (n-m + 2) \ ldots n} {n ^ {m}}} {\ bigg (} \ lambda + o {\ bigg (} \ lambda {\ bigg)} {\ bigg)} ^ {m} {\ bigg (} 1 - {\ cfrac {\ lambda} {n}} - o {\ bigg (} {\ cfrac {\ lambda} {n}} {\ bigg)} {\ bigg)} ^ {nm},}

because

\lim _{n\to \infty }np_{n}=\lambda \;\Leftrightarrow \;p_{n}={\cfrac {\lambda }{n}}+o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}

{\ displaystyle \ lim _ {n \ to \ infty} np_ {n} = \ lambda \; \ Leftrightarrow \; p_ {n} = {\ cfrac {\ lambda} {n}} + o {\ bigg (} { \ cfrac {\ lambda} {n}} {\ bigg)}}

at

\lim _{n\to \infty }{\cfrac {o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}}{\cfrac {\lambda }{n}}}=0.

{\ displaystyle \ lim _ {n \ to \ infty} {\ cfrac {o {\ bigg (} {\ cfrac {\ lambda} {n}} {\ bigg)}} {\ cfrac {\ lambda} {n} }} = 0.}

But since

$\lim _{n\to \infty }{\cfrac {(n-m+1)(n-m+2)\ldots n}{n^{m}}}={\bigg (}\lim _{n\to \infty }{\cfrac {(n-m+1)}{n}}{\bigg )}\cdot {\bigg (}\lim _{n\to \infty }{\cfrac {(n-m+2)}{n}}{\bigg )}\cdot \ldots \cdot {\bigg (}\lim _{n\to \infty }{\cfrac {(n)}{n}}{\bigg )}=1;$ ${\ displaystyle \ lim _ {n \ to \ infty} {\ cfrac {(n-m + 1) (n-m + 2) \ ldots n} {n ^ {m}}} = {\ bigg (} \ lim _ {n \ to \ infty} {\ cfrac {(n-m + 1)} {n}} {\ bigg)} \ cdot {\ bigg (} \ lim _ {n \ to \ infty} {\ cfrac {(n-m + 2)} {n}} {\ bigg)} \ cdot \ ldots \ cdot {\ bigg (} \ lim _ {n \ to \ infty} {\ cfrac {(n)} {n} } {\ bigg)} = 1;}$
$\lim _{n\to \infty }(\lambda +o(\lambda ))^{m}=\lambda ^{m};$ ${\ displaystyle \ lim _ {n \ to \ infty} (\ lambda + o (\ lambda)) ^ {m} = \ lambda ^ {m};}$
$\lim _{n\to \infty }{\bigg (}1-{\cfrac {\lambda }{n}}-o{\bigg (}{\cfrac {\lambda }{n}}{\bigg )}{\bigg )}^{n-m}=e^{-\lambda },$ ${\ displaystyle \ lim _ {n \ to \ infty} {\ bigg (} 1 - {\ cfrac {\ lambda} {n}} - o {\ bigg (} {\ cfrac {\ lambda} {n}} { \ bigg)} {\ bigg)} ^ {nm} = e ^ {- \ lambda},}$

then the resulting equality turns into

\lim _{n\to \infty }P(\omega :\mu _{n}(\omega )=m)=e^{-\lambda }{\cfrac {\lambda ^{m}}{m!}}.

{\ displaystyle \ lim _ {n \ to \ infty} P (\ omega: \ mu _ {n} (\ omega) = m) = e ^ {- \ lambda} {\ cfrac {\ lambda ^ {m}} {m!}}.}

QED

Poisson theorem

Wording

Proof

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