# The PDF of the exponential of a Gaussian random variable

What is the PDF of the exponential of a Gaussian random variable?

i.e. suppose W is a random variable drawn from a Gaussian distribution, then what is the random distribution of exp(W)?

Thank you!

## Answers and Replies

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The PDF for:
$$Y = \exp{\left(\alpha \, X\right)}$$
where the PDF for $X$ is:
$$\varphi_{X}(x) = \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, e^{-\frac{(x - a)^{2}}{2 \, \sigma^{2}}}$$
is given by:
$$\varphi_{Y}(y) = \int_{-\infty}^{\infty}{\delta(y - \exp(\alpha \, x)) \, \varphi_{X}(x) \, dx}$$

Use the properties of the Dirac delta function to evaluate this integral exactly!

The PDF for:
$$Y = \exp{\left(\alpha \, X\right)}$$
where the PDF for $X$ is:
$$\varphi_{X}(x) = \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, e^{-\frac{(x - a)^{2}}{2 \, \sigma^{2}}}$$
is given by:
$$\varphi_{Y}(y) = \int_{-\infty}^{\infty}{\delta(y - \exp(\alpha \, x)) \, \varphi_{X}(x) \, dx}$$

Use the properties of the Dirac delta function to evaluate this integral exactly!
Thank you, Dickfore.

I realize I can just use $$\varphi_{Y}(y)=\varphi_{X}(x) \frac{dx}{dy}$$
and the result is almost log-normal distribution pdf, but your method looks quite interesting.

I do have a question, in your last integral, suppose $$y$$ is constant in the integral, then the result would be $$\varphi_{Y}(y)=\varphi_{X}(x=\frac{1}{\alpha}\ln{y})$$, which is not quite the same as the log-normal. I'm not sure if I have gotten it wrong.

you had forgotten the Jacobian of the transofrmation:
$$\left|\frac{d x}{d y}\right|$$
expressed as a function of y. For what values of y do the above expressions make sense?

you had forgotten the Jacobian of the transofrmation:
$$\left|\frac{d x}{d y}\right|$$
expressed as a function of y. For what values of y do the above expressions make sense?
Thx.