Since [tex]Y[/tex] follows a Poisson distribution with mean [tex]\lambda[/tex], we know the PMF of [tex]Y[/tex] is
[tex]f_Y(y)=\begin{cases}\dfrac{e^{-\lambda}\lambda^y}{y!}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}[/tex]
Now,
[tex]\mathbb E[Y(Y-1)]=\mathbb E[Y^2-Y]=\mathbb E[Y^2]-\mathbb E[Y][/tex]
[tex]\implies\mathbb E[Y(Y-1)]=\mathbb E[Y^2]-\lambda[/tex]
To compute the second moment [tex](\mathbb E[Y^2])[/tex], we can use the definition of the expectation of a function of a random variable:
[tex]\mathbb E[Y^2]=\displaystyle\sum_yy^2\,f_Y(y)=\sum_{y=0}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}[/tex]
[tex]\mathbb E[Y^2]=\displaystyle\sum_{y=1}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}[/tex]
[tex]\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=1}^\infty y\frac{e^{-\lambda}\lambda^{y-1}}{(y-1)!}[/tex]
[tex]\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=0}^\infty(y+1)\frac{e^{-\lambda}\lambda^y}{y!}[/tex]
[tex]\mathbb E[Y^2]=\displaystyle\lambda\left(\sum_{y=0}^\infty y\frac{e^{-\lambda}\lambda^y}{y!}+\sum_{y=0}^\infty\frac{e^{-\lambda}\lambda^y}{y!}\right)[/tex]
[tex]\implies\mathbb E[Y^2]=\lambda(\mathbb E[Y]+1)=\lambda^2+\lambda[/tex]
and so
[tex]\mathbb E[Y(Y-1)]=\lambda^2+\lambda-\lambda=\lambda^2[/tex]
Recall that
[tex]\mathbb V[Y]=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2[/tex]
so we have
[tex]\mathbb V[Y]=\underbrace{\mathbb E[Y^2]-\mathbb E[Y]}_{\mathbb E[Y(Y-1)]}+\mathbb E[Y]-\mathbb E[Y]^2[/tex]
[tex]\implies\mathbb V[Y]=\lambda^2+\lambda-\lambda^2[/tex]
[tex]\implies\mathbb V[Y]=\lambda[/tex]
