{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

hmwk6sol2027

# hmwk6sol2027 - Solutions to Homework 6 ISyE 2027 Fall 2006...

This preview shows pages 1–3. Sign up to view the full content.

Solutions to Homework 6, ISyE 2027 Fall 2006 Problem 1 (a) geometric (b) Bernoulli (c) binomial (d) Poisson Problem 2 (a) We know that 1 = 4 k =1 P ( X = k ), so 1 = 4 k =1 P ( X = k ) = c 4 k =1 k = 10 c so c = 1 / 10. (b) The expected value (or mean) of X is just E [ X ] = 4 k =1 kP ( X = k ) = (1 / 10) 4 k =1 k 2 = 30 / 10 = 3 . (c) To compute the probability mass function of Y , we first need to determine the range of Y , or the set of possible values Y can take. It’s clear that the range of Y is { 1 , 4 , 9 , 16 } . Therefore, the probability mass function of Y is P ( Y = 1) = P ( X = 1) = 1 / 10 P ( Y = 4) = P ( X = 2) = 2 / 10 P ( Y = 9) = P ( X = 3) = 3 / 10 P ( Y = 16) = P ( X = 4) = 4 / 10 (d) The expected value of Y = X 2 is just E [ Y ] = (1) P ( Y = 1) + (4) P ( Y = 4) + (9) P ( Y = 9) + (16) P ( Y = 16) = 1 / 10 + 8 / 10 + 27 / 10 + 64 / 10 = 100 / 10 = 10 . (e) Using the “law of the unconscious statistician”, we see that E [ X 2 ] = 4 k =1 k 2 P ( X = k ) = 4 k =1 (1 / 10) k 3 = 10 . (f) Here E [ X ( X - 1)] = E [ X 2 - X ] = E [ X 2 ] - E [ X ] = 10 - 3 = 7 . (g) The variance of X is just V ar ( X ) = E [ X 2 ] - ( E [ X ]) 2 = 10 - 9 = 1 . (h) The standard deviation is just the square root of the variance, so in this case it’s 1. (i) If we condition on the event that { X 2 } , then X is either 1 or 2. 1

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
2 P ( X = 1 | X 2) = P ( X = 1 , X 2) P ( X 2) = P ( X = 1) P ( X 2) = (1 / 10) (3 / 10) = 1 / 3 . Similarly, P ( X = 2 | X 2) = P ( X = 2) P ( X 2) = 2 / 3 . (j) The conditional expectation of X , given { X 2 } is E [ X | X 2] = (1) P ( X = 1 | X 2) + (2) P ( X = 2 | X 2) = 1 / 3 + 4 / 3 = 5 / 3 .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}