Download Problem Statement The random variable Y = ln(X) has a normal

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