# 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
Science Advisor
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
Science Advisor
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
Science Advisor
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
Science Advisor
Gold Member
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##
Why? Use that ## P(Z:=X/Y < w )=P(X< yw) ; y \in Y## so that ##\int_{- \infty}^{\infty}ydy\int _{-\infty}^{yw} xdx ##defines a distribution for ##Z ## under some reasonable conditions like ## y \neq 0 ## and others. Instead of ##yw## as an integration limit, you can use any function of either, including log. Please double-check, I am on my phone here.

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