Timeline for Uniformly distributed random variable on unit circle.
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 29, 2020 at 17:55 | comment | added | Octavius | Understood. Thank you very much. | |
| Oct 29, 2020 at 17:55 | vote | accept | Octavius | ||
| Oct 29, 2020 at 16:23 | history | edited | tommik | CC BY-SA 4.0 | added 189 characters in body |
| Oct 29, 2020 at 16:15 | comment | added | tommik | @Octavius : ahah ... I edited the answer again with more and more details...your formula is correct but with non monotonic transformation it must be updated. in general you can pass via CDF...it is the general method. I wrote down the case of $|X|$. Reading the text you posted, they ask you the distribution function, thus my method is faster. | |
| Oct 29, 2020 at 16:10 | history | edited | tommik | CC BY-SA 4.0 | added 275 characters in body |
| Oct 29, 2020 at 16:05 | comment | added | Octavius | I see, so my CDF is $$F(x) = \frac{x}{2\sqrt{1-y_0^2}} + \frac{1}{2} \quad .$$ But how do I get the CDF for the random variables $x^4$, $|x|$, $-x$? The "transformation law" for the densities would be $$g(\tilde{x}) = \Big| \frac{\mathrm{d}h^{-1}}{\mathrm{d}\tilde{x}} \Big| \cdot f\big(h^{-1}(\tilde{x})\big)$$ where $f(x)$ is the density of the variable $X$, and $g(\tilde{x})$ is the density of the variable $\tilde{X} = h(X)$. With constant density as written above I would have density zero for the other variables. | |
| Oct 29, 2020 at 15:50 | comment | added | tommik | @Octavius : in the meantime you were writing the comment I edited my answer with more details. The fact that in your conditional density there is no $X$ is better for you....it's a Uniform density in the support that I showed. The X will appear as soon as you calculate the CDF | |
| Oct 29, 2020 at 15:48 | history | edited | tommik | CC BY-SA 4.0 | added 83 characters in body |
| Oct 29, 2020 at 15:42 | history | edited | tommik | CC BY-SA 4.0 | added 83 characters in body |
| Oct 29, 2020 at 15:42 | comment | added | Octavius | In this case I have to calculate: $$ f(x|y=y_0) = \frac{f(x,y)}{\tilde{f}(y)}$$, which is $$f(x|y=y_0) = \frac{1}{2\sqrt{1-y_0^2}}$$. But this term is $x$-independent. Wouldn't that mean, that the random variables $x^4$, $|x|$ and $-x$ have a density function of zero everywhere? | |
| Oct 29, 2020 at 15:18 | history | answered | tommik | CC BY-SA 4.0 |