Mathematics: Let y have a poisson distribution with mean λ. find e [y (y − 1)] and then use this to show that v(y) = λ.

y + 7 = 1/2 ( x + 6 )Step-by-step explanation:y - y1 = m(x - x1)...

Let y have a poisson distribution with mean λ. find

Ответ:

madisonanders

15.03.2022

Mathematics

Since

follows a Poisson distribution with mean $\lambda$ , we know the PMF of

is

$f_Y(y)=\begin{cases}\dfrac{e^{-\lambda}\lambda^y}{y!}&\text{for }y\ge0\\\\0&\text{otherwise}\end{cases}$

Now,

$\mathbb E[Y(Y-1)]=\mathbb E[Y^2-Y]=\mathbb E[Y^2]-\mathbb E[Y]$
$\implies\mathbb E[Y(Y-1)]=\mathbb E[Y^2]-\lambda$

To compute the second moment $(\mathbb E[Y^2])$ , we can use the definition of the expectation of a function of a random variable:

$\mathbb E[Y^2]=\displaystyle\sum_yy^2\,f_Y(y)=\sum_{y=0}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}$
$\mathbb E[Y^2]=\displaystyle\sum_{y=1}^\infty y^2\frac{e^{-\lambda}\lambda^y}{y!}$
$\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=1}^\infty y\frac{e^{-\lambda}\lambda^{y-1}}{(y-1)!}$
$\mathbb E[Y^2]=\displaystyle\lambda\sum_{y=0}^\infty(y+1)\frac{e^{-\lambda}\lambda^y}{y!}$
$\mathbb E[Y^2]=\displaystyle\lambda\left(\sum_{y=0}^\infty y\frac{e^{-\lambda}\lambda^y}{y!}+\sum_{y=0}^\infty\frac{e^{-\lambda}\lambda^y}{y!}\right)$
$\implies\mathbb E[Y^2]=\lambda(\mathbb E[Y]+1)=\lambda^2+\lambda$

and so

$\mathbb E[Y(Y-1)]=\lambda^2+\lambda-\lambda=\lambda^2$

Recall that

$\mathbb V[Y]=\mathbb E[(Y-\mathbb E[Y])^2]=\mathbb E[Y^2]-\mathbb E[Y]^2$

so we have

$\mathbb V[Y]=\underbrace{\mathbb E[Y^2]-\mathbb E[Y]}_{\mathbb E[Y(Y-1)]}+\mathbb E[Y]-\mathbb E[Y]^2$
$\implies\mathbb V[Y]=\lambda^2+\lambda-\lambda^2$
$\implies\mathbb V[Y]=\lambda$

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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) = λ.

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