What is the distribution of $Y = e^X$ when $X$ is normally distributed?
Am I supposed to use characteristics function of normal random variable ?
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2You might find some information here: http://en.wikipedia.org/wiki/Log-normal_distribution – Julian Rosen Jul 21 '13 at 19:07
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2There are several different methods. What is stopping you here? – Did Jul 21 '13 at 19:11
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I don't know how to start! If it is using characteristic function its better, but I think the easiest one is using CDF of $X$ ? – Mahdi Khosravi Jul 21 '13 at 19:13
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1This is the easiest approach for your case: http://en.wikipedia.org/wiki/Random_variable#Functions_of_random_variables – Alex Jul 21 '13 at 19:20
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1Have you not covered how to determine the density of a function of a continuous random variable? – Stephen Herschkorn Jul 21 '13 at 19:41
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In case you still haven't sorted it out: apply the formula in the Wiki link I gave you to get for $X \sim N(0,1)$ $$ X'_{y}=\frac{1}{y}\\ f_{X}(\varphi^{-1}(y))=\frac{1}{\sqrt{2 \pi}}e^{-\frac{\log^2 y}{2}} $$ Hence the distribution of $Y$ is $$ f_Y(y)=\frac{1}{\sqrt{2 \pi}y}e^{-\frac{\log^2 y}{2}} $$ for $y>0$. Here I don't use the absolute value for $y$ because the pdf is defined for strictly positive values since $\log y$ doesn't exist for $y \leq 0$.
Alex
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