the PDF of the exponential of a Gaussian random variable

[samuelandjw]

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!

[Dickfore]

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!

[samuelandjw]

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.

[Dickfore]

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?

[samuelandjw]

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.