If m is the parameter of the Poisson distribution then the variance of the distribution is

Variance =${\displaystyle \sum _{r=0}^{\infty }{r}^{2}p\left(r\right)}-{\left\{{\displaystyle \sum _{r=0}^{\infty }rp\left(r\right)}\right\}}^2$

Now ${\displaystyle \sum _{r=0}^{\infty }{r}^{2}p\left(r\right)}=1^{2}p\left(1\right)+2^{2}p\left(2\right)+3^{2}p\left(3\right)+\mathrm{............}to\infty$

But $r^{2}p\left(r\right)=r^2.\frac{e^{-m}{m}^r}{\overline{)r}}=\frac{r}{\overline{)r-4}}{e}^{-m}.m^r=\frac{\left(r-1\right)+1}{\overline{)r-1}}{e}^{-m}.m^r$

                     $=\left(\frac{1}{\overline{)r-2}}+\frac{1}{\overline{)r-1}}\right)e^{-m}.m^r=e^{-m}{m}^2\frac{m^{r-2}}{\overline{)r-2}}+e^{-m}.m.\frac{m^{r-1}}{\overline{)r-1}}$

$\therefore {\displaystyle \sum r^{2}p\left(r\right)=e^{-m}}.n^2{\displaystyle \sum \frac{m^{r-2}}{\overline{)r-2}}}+e^{-m}m.{\displaystyle \sum \frac{m^{r-1}}{\overline{)r-1}}}$

                         $=e^{-m}.m^2{e}^m+e^{-m}.m.e^m=m^2+m$

Again, ${\displaystyle \sum rp\left(r\right)=1.p\left(1\right)}+2p\left(2\right)+3.p\left(3\right)+\mathrm{...........}to\infty$ but $rp\left(r\right)=r.\frac{e^{-m}.m^r}{\overline{)r}}=\frac{e^{-m}.m^r}{\overline{)r-1}}=e^{-m}.m\frac{m^{r-1}}{\overline{)r-1}}$

$\therefore {\displaystyle \sum rp\left(r\right)=e^{-m}m.{\displaystyle \sum \frac{m^{r-1}}{\overline{)r-1}}}}=e^{-m}.m.e^m=m$

$\therefore$ Variance =$\left(m^2+m\right)-{\left(m\right)}^2=m$ .