# Normal and exponential-normal (?) distribution

mahtabhossain
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?

Mentor
2021 Award
What should the difference or even ##\dfrac{X}{Y}## be? These isn't a random variable anymore.

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

Mentor
2021 Award
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.

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
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?

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

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.

Mentor
2021 Award
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##