# Uniform Distributions Ratio [duplicate]

Let $$\xi$$ is $$U(0, 1)$$, $$\nu$$ is $$U(0, 1)$$. What type of distribution is $$g = \frac{\xi}{\nu}$$? I have build logarithm of $$g$$, here is the plot but it seems like it is not normal distribution: Distribution Hist

• Not normal.${}{}{}{}{}$
– mjw
Feb 24, 2021 at 15:21
• It is a ratio distribution : en.wikipedia.org/wiki/Ratio_distribution Feb 24, 2021 at 15:26
• @oliverjones thank you very much! Feb 24, 2021 at 15:28

Assming independence...

Consider the following system

$$\begin{cases} g=\frac{\xi}{\nu} \\ u=\xi \end{cases}$$

The Jacobian is $$|J|=\frac{\xi}{g^2}$$

$$f_G(g)=\frac{1}{2}\cdot\mathbb{1}_{[0;1)}(g)+\frac{1}{2g^2}\cdot\mathbb{1}_{[1;+\infty)}(g)$$

• How did you receive the last statement? Feb 24, 2021 at 15:46

Since $$\log 1/\nu$$ and $$\log 1/\xi$$ are exponentially distributed, your plot, $$\log g$$, follows the Difference Between Exponential Distributions.

How did you receive the last statement? –

after calculating the jacobian that is $$|J|=\frac{u}{g^2}$$ you conclude that the joint density is

$$f(u,g)=\frac{u}{g^2}$$

but in the system you see that

$$0<\nu=\frac{u}{g}<1$$

Now, when $$0 this means that $$0 thus in this interval to get $$f(g)$$ you have to integrate the joint density in

$$f(g)=\int_0^g \frac{u}{g^2}du=\frac{1}{2}$$

When $$g>1$$ there are no problem for $$\frac{u}{g}$$ to be less than one thus you integrate

$$f(g)=\int_0^1 \frac{u}{g^2}du=\frac{1}{2g^2}$$

... put all together and get the result

this is the drawing of your density

• Awesome!! Thank you very much Feb 25, 2021 at 9:04