Mathematics, 28.07.2019 21:30 manarsadi6
Let y have a poisson distribution with mean λ. find e [y (y − 1)] and then use this to show that v(y) = λ.
Answers: 1
follows a Poisson distribution with mean 
, we know the PMF of 
![\mathbb E[Y(Y-1)]=\mathbb E[Y^2-Y]=\mathbb E[Y^2]-\mathbb E[Y]](/tpl/images/0144/2713/3e15b.png)
![\implies\mathbb E[Y(Y-1)]=\mathbb E[Y^2]-\lambda](/tpl/images/0144/2713/1654b.png)
![(\mathbb E[Y^2])](/tpl/images/0144/2713/60209.png)
, we can use the definition of the expectation of a function of a random variable:![\mathbb E[Y^2]=\displaystyle\sum_yy^2\,f_Y(y)=\sum_{y=0}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}](/tpl/images/0144/2713/7facb.png)
![\mathbb E[Y^2]=\displaystyle\sum_{y=1}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}](/tpl/images/0144/2713/be5c2.png)
![\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=1}^\infty y\frac{e^{-\lambda}\lambda^{y-1}}{(y-1)!}](/tpl/images/0144/2713/02766.png)
![\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=0}^\infty(y+1)\frac{e^{-\lambda}\lambda^y}{y!}](/tpl/images/0144/2713/fef8b.png)
![\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)](/tpl/images/0144/2713/d01e5.png)
![\implies\mathbb E[Y^2]=\lambda(\mathbb E[Y]+1)=\lambda^2+\lambda](/tpl/images/0144/2713/3b024.png)
![\mathbb E[Y(Y-1)]=\lambda^2+\lambda-\lambda=\lambda^2](/tpl/images/0144/2713/85c24.png)
![\mathbb V[Y]=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2](/tpl/images/0144/2713/02344.png)
![\mathbb V[Y]=\underbrace{\mathbb E[Y^2]-\mathbb E[Y]}_{\mathbb E[Y(Y-1)]}+\mathbb E[Y]-\mathbb E[Y]^2](/tpl/images/0144/2713/4b40e.png)
![\implies\mathbb V[Y]=\lambda^2+\lambda-\lambda^2](/tpl/images/0144/2713/71af0.png)
![\implies\mathbb V[Y]=\lambda](/tpl/images/0144/2713/2ac7f.png)
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Let y have a poisson distribution with mean λ. find e [y (y − 1)] and then use this to show that v(y...
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