hw7_sol - ECE 302 Division 2. Homework 7 Solutions. Problem...

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

View Full Document Right Arrow Icon

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

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: ECE 302 Division 2. Homework 7 Solutions. Problem 1. Lognormal random variable, its PDF, mean, and median. Suppose Y is a Gaussian random variable with mean and standard deviation . Let X = e Y . X is called a lognormal random variable since its log is a normal (Gaussian) random variable. (a) Find the PDF and the expectation of X . (b) The median of a continuous random variable Z is defined as such number d that P ( Z d ) = P ( Z d ) = 1 / 2. Find the median of X . Solution. From the definition of X it follows that P ( X 0) = 0. For x > 0, we have: F X ( x ) = P ( X x ) = P ( e Y x ) = P ( Y log x ) = F Y (log x ) (1) Therefore, for x > 0, f X ( x ) = F X ( x ) = d dx F Y (log x ) = 1 x f Y (log x ) = 1 x 2 e (log x ) 2 2 2 The mean of X can be found as follows: E [ X ] = E bracketleftbig e Y bracketrightbig = integraldisplay 1 2 e y e ( y ) 2 2 2 dy = integraldisplay 1 2 exp parenleftbigg- y 2- 2 y + 2- 2 y 2 2 2 parenrightbigg dy = integraldisplay 1 2 exp parenleftbigg- y 2- 2 y ( + 2 ) + 2 + 2 2 + 4- 2 2- 4 2 2 parenrightbigg dy = integraldisplay 1 2 exp parenleftbigg- ( y- - 2 ) 2- 2 2- 4 2 2 parenrightbigg dy = exp parenleftbigg + 2 2 parenrightbiggintegraldisplay 1 2 exp parenleftbigg- ( y- - 2 ) 2 2 2 parenrightbigg dy = exp parenleftbigg + 2 2 parenrightbigg , where we used the fact that the last integral is the integral of a Gaussian PDF with mean + 2 and standard deviation , and hence is equal to one. The median d of X satisfies F X ( d ) = 1 / 2. But, on the other hand, it was shown in Eq. (1) that F X ( d ) = F Y (log d ). Since a Gaussian PDF is symmetric about its mean, we know that F Y ( ) = 1 / 2. Hence, log d = , and d = e 1 0 30,000 60,000 90,000 120,000 0.5 1 1.5 2 2.5 3 3.5 x 10-5 Figure 1: Lognormal distribution with parameters = log(20000) 9 . 9 and = 1. Note that the lognormal distribution is skewed to the right (see Fig. 1) and hence its mean is larger than the median. Fig. 1 is a plot of the lognormal PDF with = log(20000) 9 . 9 and = 1. The mean and the median for this PDF are 20000 e 32974 and 20000, respectively. Problem 2. On exam taking strategies. This problem illustrates some probabilistic tools which may be used in analyzing and comparing various decision strategies. Suppose you are taking a course, Decisions 101, taught by Prof. Kind. The course has three tests. Normally, your final score S is computed as S = 1 3 ( T 1 + T 2 + T 3 ) , where T n is your score on test n , with n = 1 , 2 , 3. The final grade in the course is then computed as follows: You get an A if S 90 , B if 80 S < 90 C if 70 S < 80 D if 60 S < 70 F if S < 60 Your intrinsic knowledge of the subject is x . I.e., if the tests were perfectly designed and your scores....
View Full Document

This note was uploaded on 11/01/2009 for the course PSY 120 taught by Professor Donnely during the Spring '08 term at Purdue.

Page1 / 10

hw7_sol - ECE 302 Division 2. Homework 7 Solutions. Problem...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online