This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: HW #7 Assignment Due Thurs Oct 27, 2011 Problem #1: Let X take the values 0, 1, 2, 3, 4 and 5 with probabilities p , p 1 , p 2 , p 3 , p 4 and p 5 . Find the probability mass function of ( ) Y X 2 = 2 Y takes on the values 4,1,0,1,4,9 corresponding to each X value. The probability of Y=0 is P2, Y=1 is P1 + P3, Y=4 is P0 + P4, and Y=9 is P5. Problem #2: If X is uniform on (0,1), find a) P ( x > 1 / 2 )= 1 / 2 = 1 / 2 1 1 dx b) P ( y = sin X 2 > 1 / 3 ) y = 1/3 when x = 2 arcsin ( 1 / 3 ) 0.216 and P ( x > .216 )= 1 .216 = 0.216 1 1 dx = 0.784 c) The density of Y X = = 1 (since x = x for x in (0,1) ) d) the density of X Y e = For X from 0 to 1 the values of Y are from 1 to e, and e^x is a monotonic increasing function so it has an inverse. Which is ln(x). Consider an infinitesimal slice of probability in y of thickness dy. It is transformed as follows: P ( Y [ y , y + dy ])= p y ( y ) dy = P ( X [ ln ( y ) , ln ( y )+ dx ])= p x ( ln ( y )) dx = 1 dx In the last step, since the density of X is uniform p x ( ln ( y ))= 1 . The absolute value is because even if the function increases or decreases we care only about the size of the interval in x. The only thing that remains is to divide by dy to solve for p y ( y )=( 1 dx )/ dy = 1 /( dy / dx ) ....
View Full
Document
 '11
 MTODOROVSKA

Click to edit the document details