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Let us say that $X$ is a normal random variable, given that its expected value is $0$ and variance is $1$ . How do we compute the density function of $Y= e^ X$?

What I think: Since expected value is $0$ and variance is $1$, $X$ is a standard normal variable, whose pdf is given as: $\frac{1}{\sqrt{2\pi}}e^{-x^2 / 2}$.

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    Could someone provide me a hint on how to proceed?2012-01-23
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    This is called a [log-normal distribution](http://en.wikipedia.org/wiki/Log-normal_distribution) (I'm not sure how to derive any formulas though).2012-01-23
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    Hint: What is the distribution function (cdf) of $X$? Can you use it to find the cdf of $Y$?2012-01-23
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    cdf of x would be : 1/2 * [1 + erf(x-mu/(sigma*sqrt(2))]2012-01-23
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    A method to answer this is explained in detail [here](http://math.stackexchange.com/a/43223/6179).2012-01-23
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    possible duplicate of [Distribution of Functions of Random Variables](http://math.stackexchange.com/questions/7605/distribution-of-functions-of-random-variables)2012-01-23
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    @Nate: As you know, this amounts to using a path that goes from the pdf of X to the cdf of X, then from the cdf of X to the cdf of Y, and finally from the cdf of Y to the pdf of Y, while there exists a way to go directly from the pdf of X to the pdf of Y.2012-01-23

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