- #1
mahtabhossain
- 2
- 0
Dear Users,
For normally distributed random variables x and y's p.d.f.:
[tex] \frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{- \frac{(x - \mu_x)^2}{2 \sigma_x^2}\right\} [/tex]
and
[tex] \frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{- \frac{(y - \mu_y)^2}{2 \sigma_y^2}\right\} [/tex]
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:
[tex] \frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{u - \frac{(e^u - \mu_x)^2}{2 \sigma_x^2}\right\} [/tex]
and
[tex] \frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{v - \frac{(e^v - \mu_y)^2}{2 \sigma_y^2}\right\} [/tex]
I am trying to find the p.d.f. of ln(x)-ln(y). Any suggestions?
For normally distributed random variables x and y's p.d.f.:
[tex] \frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{- \frac{(x - \mu_x)^2}{2 \sigma_x^2}\right\} [/tex]
and
[tex] \frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{- \frac{(y - \mu_y)^2}{2 \sigma_y^2}\right\} [/tex]
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:
[tex] \frac{1} {\sqrt{2\pi \sigma_x^2}}\exp\left\{u - \frac{(e^u - \mu_x)^2}{2 \sigma_x^2}\right\} [/tex]
and
[tex] \frac{1} {\sqrt{2\pi \sigma_y^2}}\exp\left\{v - \frac{(e^v - \mu_y)^2}{2 \sigma_y^2}\right\} [/tex]
I am trying to find the p.d.f. of ln(x)-ln(y). Any suggestions?