Question

11. The recurrence relation between P(x) and P(x+1) in a Poisson distribution is given by
a) P(x+1)-m P(x) = 0
b) m P(x+1) - P(x) = 0
c)(x+1) P(x+1)- m P(x) = 0
d) (x+1) P(x) - x P(x+1)=0​

Answer 1

$\underline{\textbf{Given:}}$

$\textsf{P(x) is the probability function of Poisson distribution}$

$\underline{\textbf{To find:}}$

$\textsf{The relation between P(x) and P(x+1)}$

$\underline{\textbf{Solution:}}$

$\underline{\textsf{Concept used:}}$

$\textsf{The probabillity function of Poisson distribution is}$

$\boxed{\mathsf{P(x)=\dfrac{e^{-m}m^x}{x!}}}$

$\textsf{We derive relation between P(x) and P(x+1)}$

$\mathsf{Consider,}$

$\mathsf{P(x+1)=\dfrac{e^{-m}m^{x+1}}{(x+1)!}}$

$\mathsf{Using,}\;\boxed{\mathsf{(n+1)!=(n+1}\,n!}}$

$\implies\mathsf{P(x+1)=\dfrac{e^{-m}m^x\,m^1}{(x+1)\,x!}}$

$\implies\mathsf{P(x+1)=\dfrac{e^{-m}m^x}{x!}{\times}\dfrac{m}{x+1}}$

$\textsf{This can be written as}$

$\mathsf{\dfrac{(x+1)}{m}P(x+1)=\dfrac{e^{-m}m^x}{x!}}$

$\mathsf{\dfrac{(x+1)}{m}P(x+1)=P(x)}$

$\mathsf{(x+1)\;P(x+1)=m\;P(x)}$

$\implies\boxed{\mathsf{(x+1)\;P(x+1)-m\;P(x)=0}}$

$\underline{\textbf{Answer:}}$

$\mathsf{Option\;(c)\;is\;correct}$