Answered: Part b) Find the second moment of Y…

Part b) Find the second moment of Y using E[Y²] =E, y°p(y) (finding the first moment using any other method will result in no credit, so do not use a moment generating function).

Algebra & Trigonometry with Analytic Geometry

13th Edition

ISBN:9781133382119

Author:Swokowski

Publisher:Swokowski

Chapter11: Topics From Analytic Geometry

Section11.4: Plane Curves And Parametric Equations

Problem 44E

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Question

Part b) Find the second moment of Y using E[Y²] = E, y²p(y) (finding the first moment using
any other method will result in no credit, so do not use a moment generating function).

Expert Solution

Step 1

The second order moment of the poisson distribution is given by,

$\begin{array}{rcl} E (y) & = & \sum_{y = 0}^{\infty} y^{2} p (y) \\ = & \sum_{y = 0}^{\infty} y^{2} e^{- λ} \frac{λ^{y}}{y!} \\ = & e^{- λ} \sum_{y = 1}^{\infty} y \frac{λ^{y}}{(y - 1)!} \\ = & e^{- λ} \sum_{y = 1}^{\infty} (y - 1 + 1) \frac{λ^{y}}{(y - 1)!} \\ = & e^{- λ} \sum_{y = 1}^{\infty} (y - 1) \frac{λ^{y}}{(y - 1)!} + e^{- λ} \sum_{y = 1}^{\infty} \frac{λ^{y}}{(y - 1)!} \\ = & e^{- λ} \sum_{y = 2}^{\infty} \frac{λ^{y}}{(y - 2)!} + e^{- λ} \sum_{y = 1}^{\infty} \frac{λ^{y}}{(y - 1)!} \\ = & e^{- λ} λ^{2} \sum_{y = 2}^{\infty} \frac{λ^{y - 2}}{(y - 2)!} + λ e^{- λ} \sum_{y = 1}^{\infty} \frac{λ^{y - 1}}{(y - 1)!} \end{array}$

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