91Ó°ÊÓ

A variable following Poisson distribution with mean λ.

We need to determine the mode or modes of the Poisson distribution with mean λ.

Let X be the variable following Poisson distribution with mean λ.

\(p\left( x \right) = \frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}\)where p(x) is the PMF of X

Now,

\(\frac{{p\left( x \right)}}{{p\left( {x - 1} \right)}} = \frac{{\frac{{{e^{ - \lambda }}{\lambda ^x}}}{{x!}}}}{{\frac{{{e^{ - \lambda }}{\lambda ^{x - 1}}}}{{\left( {x - 1} \right)!}}}} = \frac{\lambda }{x}\)

Case 1: When λ is not an integer

Let us suppose s is an integral part of λ, then λ=s+f,0<f<1

\(\begin{array}{l}\frac{{p\left( x \right)}}{{p\left( {x - 1} \right)}} = \frac{{s + f}}{x} > 1\,,\,x = 0,1,2,...,s\,\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\, < 1,x = s + 1,s + 2,...\\\frac{{p\left( 1 \right)}}{{p\left( 0 \right)}} > 1,\frac{{p\left( 2 \right)}}{{p\left( 1 \right)}} > 1,...,\frac{{p\left( s \right)}}{{p\left( {s - 1} \right)}} > 1,\frac{{p\left( {s + 1} \right)}}{{p\left( s \right)}} < 1,...\\{\rm{So,}}\,\,{\rm{p}}\left( 0 \right) < p\left( 1 \right) < ... < p\left( {s - 1} \right) < p\left( s \right) > p\left( {s + 1} \right) > ...\end{array}\)

Which shows p(s) is maximum. Hence, in this case, the distribution is unimodal, and the integral part of λ is the unique modal value.

Case 2: When λ is an integer

Let λ=k (an integer)

\(\begin{array}{l}\frac{{p\left( 1 \right)}}{{p\left( 0 \right)}} > 1,\frac{{p\left( 2 \right)}}{{p\left( 1 \right)}} > 1,...,\frac{{p\left( {k - 1} \right)}}{{p\left( {k - 2} \right)}} > 1,\frac{{p\left( k \right)}}{{p\left( {k - 1} \right)}} = 1,\frac{{p\left( {k + 1} \right)}}{{p\left( k \right)}} < 1,...\\{\rm{So}},\,\,p\left( 0 \right) < p\left( 1 \right) < ... < p\left( {k - 2} \right) < p\left( {k - 1} \right) = p\left( k \right) > p\left( {k + 1} \right) > p\left( {k + 2} \right) > ...\end{array}\)

In this case, we have two maximum values, p(k-1) and p(k), and thus, the distribution is bimodal, and the two modes are (k-1) and k or λ and (λ-1)