If the distribution of X is N(?, ?2), then We then say

QUESTION:

If the distribution of \(X\) is \(N\left(\mu, \sigma^{2}\right)\), then \(M(t)=E\left(e^{t X}\right)=\exp \left(\mu t+\sigma^{2} t^{2} / 2\right)\). We then say that \(Y=e^{X}\) has a lognormal distribution because \(X=\ln Y\).

(a) Show that the pdf of \(Y\) is

\(g(y)=\frac{1}{y \sqrt{2 \pi \sigma^{2}}} \exp \left[-(\ln y-\mu)^{2} / 2 \sigma^{2}\right], \quad 0<y<\infty\) .

(b) Using \(M(t)\), find (i) \(E(Y)=E\left(e^{X}\right)=M(1)\), (ii) \(E\left(Y^{2}\right)=E\left(e^{2 X}\right)=M(2)\), and (iii) \(\operatorname{Var}(Y)\).

Equation Transcription:

Text Transcription:

X  

N( mu,2)  

M(t)=E(e^tX)=exp (mu t+sigma^2 t^2/2)  

Y=e^X  

X=ln⁡Y  

Y  

g(y)=1/y sqrt 2 pi sigma^2 exp[⁡-(ln⁡y-mu)^2/2 sigma^2], 0<y<  infinity

M(t)  

E(Y)=E(e^X)=M(1)

E(Y^2)=E(e^2X)=M(2)  

Var⁡(Y)