The PDF of the exponential of a Gaussian random variable

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

Use the properties of the Dirac delta function to evaluate this integral exactly!
 
Dickfore said:
The PDF for:
<br /> Y = \exp{\left(\alpha \, X\right)}<br />
where the PDF for X is:
<br /> \varphi_{X}(x) = \frac{1}{\sqrt{2 \, \pi} \, \sigma} \, e^{-\frac{(x - a)^{2}}{2 \, \sigma^{2}}}<br />
is given by:
<br /> \varphi_{Y}(y) = \int_{-\infty}^{\infty}{\delta(y - \exp(\alpha \, x)) \, \varphi_{X}(x) \, dx}<br />

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

Thank you, Dickfore.

I realize I can just use <br /> \varphi_{Y}(y)=\varphi_{X}(x) \frac{dx}{dy}<br />
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:
<br /> \left|\frac{d x}{d y}\right|<br />
expressed as a function of y. For what values of y do the above expressions make sense?
 
Dickfore said:
you had forgotten the Jacobian of the transofrmation:
<br /> \left|\frac{d x}{d y}\right|<br />
expressed as a function of y. For what values of y do the above expressions make sense?

Thx.
 
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