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?

 

[fresh_42]

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

 

[WWGD]

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

 

[fresh_42]

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]

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]

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]

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.