Question

If a variable takes values 0,1,2,3,........with frequencies proportional to$e^{-\lambda },e^{-\lambda }\lambda ,\frac{e^{-\lambda }{\lambda }^2}{2!},$ $\frac{e^{-\lambda }{\lambda }^3}{3!},$........then the mean of the distribution

Answer 1

Mean$=\frac{(0\times e^{-\lambda })+(1\times e^{-\lambda }\lambda )+\left(\frac{2\times e^{-\lambda }\times {\lambda }^2}{2!}\right)......}{e^{-\lambda }+e^{-\lambda }\lambda +\frac{e^{-\lambda }{\lambda }^2}{2!}+.......}$

Mean$=\frac{\lambda +\frac{{\lambda }^2}{1!}+\frac{{\lambda }^3}{2!}+\frac{{\lambda }^3}{3!}+.....}{e^{-\lambda }(1+\lambda +\frac{{\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}+.....)}$

Mean$=\frac{\lambda (1+\lambda +\frac{{\lambda }^2}{2!}+\frac{{\lambda }^3}{3!}+.......)}{e^{-\lambda }(1+\lambda +\frac{{\lambda }^2}{2!}+......)}=\frac{\lambda }{e^{-\lambda }}=e^{\lambda }\lambda$

Mean$=e^{\lambda }\lambda$