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

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

A First Course in Probability (10th Edition)

10th Edition

ISBN:9780134753119

Author:Sheldon Ross

Publisher:Sheldon Ross

Chapter1: Combinatorial Analysis

Section: Chapter Questions

Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...

See similar textbooks

Related questions

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,

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

Step by step

Solved in 2 steps

Knowledge Booster

$Statistical Power and Errors$

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, probability and related others by exploring similar questions and additional content below.

Similar questions