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Darren Tully Eng 323 Page 1 of 3 04/09/00 BHW #11 Problem Statement The random variable Y = ln(X) has a normal distribution with a mean of 5 and a standard deviation of 1. Determine the mean, variance, standard deviation, median and skewness coefficient of X. Graph the PDFs of X and Y on separate graphs. Solution This problem statement is very straightforward for analysis. Given Y = ln(X), then Y ~ Normal(µ y =5, σy =1) Or ln(X) ~ Normal(µ ln x =5, σln x =1) Next is to determine what is to be solved for: 1. Mean = E(X) = µ x = α 2. Variance = V(X) = σ 2 x = β 2 3. Standard Deviation = σ x = V (X ) ~ 4. Median = µ X 5. Skewness coefficient of X = β1 All of these equations are found in the handout from class, on Lognormal Processes from the book Environmental Statistics and Data Analysis by Wayne Ott. Now to solve with simple plug and chug; 1. µx = Exp[µy + (σ2 y /2)] = Exp[5 + (1/2)] = 244.69 units 2. V(X) = Exp[2µy + σ2 y ]*(Exp[σ2 y ] – 1) = Exp[2*5 + 1]*(Exp[1] – 1) = 102880.65 units2 3. σx = √102880.65 = 320.75 units 4. By the definition of the median: ~ ) = 0.5 P(X ≤ µ X so then; ~ ⇒ P(X ≤ µ~x ) = 0.5; ⇒ P(ln (X) ≤ ln ( µ ~ )) = 0.5; ⇒ µ X X ~ ) - µy )/σy ) = 0.5; ⇒ P(Z ≤ (ln( µ ~ ) - µy )/σy ) = 0.5 P((Y - µy )/σy ≤ (ln µ X X and because Standard Normal distribution is symmetric about 0, then; P(Z ≤ 0) = 0.5 So, ~ provides; solving for µ X ~ ) - µy )/σy = 0 (ln( µ X Darren Tully Eng 323 Page 2 of 3 04/09/00 BHW #11 ~ )-5)/1 =0 (ln( µ X ~ )=5 ln( µ X ~ µ X = Exp[5] ~ = 148.41 µ X 6. β1 = (Exp[σ2 ] + 2)*√(Exp[σ2 ] – 1) = (Exp[1] + 2)*√(Exp[1]-1) = 6.185 f(Y) In Figure 1, the PDF of Y is distributed normally. The graph indicates the mean value in the middle of the bell curve, showing a nice distribution. 0.45 0.4 0.35 0.3 0.25 0.2 0.15 0.1 0.05 0 Mean = 5 0.0 5.0 10.0 Y = ln(x) Figure 1: Normal Density Function, Mean = 5, StDev = 1 ~ , Noting that the coefficient of skewness is a positive 6.185 and that µx > µ X indicates a positive skew in the PDF. Figure 2 shows the PDF of Y distributed lognormally showing the skew of the graph. Darren Tully Eng 323 Page 3 of 3 0.005 04/09/00 BHW #11 Median = 148.41 units f(X) 0.004 0.003 Mean = 244.69 units 0.002 0.001 0 0.0 200.0 400.0 600.0 800.0 X = e^Y Figure 2: Lognormal Density function 1000.0