# Normal and exponential-normal (?) distribution

Dear Users,

For normally distributed random variables x and y's p.d.f.:
$$\frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{- \frac{(x - \mu_x)^2}{2 \sigma_x^2}\right\}$$
and
$$\frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{- \frac{(y - \mu_y)^2}{2 \sigma_y^2}\right\}$$

What will be the p.d.f. of ln(x) - ln(y)? Is there any method to find it?

I think the followings are the p.d.f.s of u=ln(x) and v=ln(y) given x and y are normally distributed:
$$\frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{u - \frac{(e^u - \mu_x)^2}{2 \sigma_x^2}\right\}$$
and
$$\frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{v - \frac{(e^v - \mu_y)^2}{2 \sigma_y^2}\right\}$$

I am trying to find the p.d.f. of ln(x)-ln(y). Any suggestions?

## Answers and Replies

fresh_42
Mentor
What should the difference or even ##\dfrac{X}{Y}## be? These isn't a random variable anymore.

WWGD
Gold Member
There is a formula for the general distribution of a function of a random variable which is itself a random variable.

fresh_42
Mentor
But how we can achieve finite distribution integrals if one variable is in the denominator? And ##ln(X)-ln(Y)## can literally take any values.

Stephen Tashi
I think the followings are the p.d.f.s of u=ln(x) and v=ln(y) given x and y are normally distributed:

If x and y are normally distributed, they can take on negative values with non-zero probability. The expressions ln(x) and ln(y) are not defined for negative values.

WWGD
WWGD
Gold Member
Like Stephen said, ln(x) will not be defined at the left end. Maybe you're considering ln (|x|)-ln(|y|)? Or maybe your mean is large-enough that those values are far at the tail end to not worry about them?

member 587159
Or you can just add the assumption that ##X## and ##Y## are both positive.

member 587159
What should the difference or even ##\dfrac{X}{Y}## be? These isn't a random variable anymore.

Assuming ##Y\neq0## everywhere ##X/Y## is still a random variable.

fresh_42
Mentor
Assuming ##Y\neq0## everywhere ##X/Y## is still a random variable.
It is still the wrong direction and requires additional assumptions. Instead of asking what ##\log \dfrac{X}{Y}##means, one should start whether a random variable ##Z## with properties ##\ldots## can be described by a pdf ##\ldots##

WWGD