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
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$\begingroup$ Not normal.${}{}{}{}{}$ $\endgroup$– mjwFeb 24, 2021 at 15:21
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1$\begingroup$ It is a ratio distribution : en.wikipedia.org/wiki/Ratio_distribution $\endgroup$– oliverjonesFeb 24, 2021 at 15:26
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$\begingroup$ @oliverjones thank you very much! $\endgroup$– Кирилл ВолковFeb 24, 2021 at 15:28
3 Answers
Assming independence...
Consider the following system
$$\begin{cases} g=\frac{\xi}{\nu} \\ u=\xi \end{cases}$$
The Jacobian is $|J|=\frac{\xi}{g^2}$
thus your density is
$$f_G(g)=\frac{1}{2}\cdot\mathbb{1}_{[0;1)}(g)+\frac{1}{2g^2}\cdot\mathbb{1}_{[1;+\infty)}(g)$$
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$\begingroup$ How did you receive the last statement? $\endgroup$ 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 $0<g<1$ this means that $0<u<g$ 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