6
$\begingroup$

I have two i.i.d. $N(0, \sigma^2)$ random variables $X_1$ and $X_2$. Let $Z = \sqrt{X_1^2 + X_2^2}$. I am told that $R$ follows the Rayleigh distribution.

The Rayleigh distribution has PDF

$$f_{\sigma}(x) = \dfrac{x}{\sigma^2} e^{-\dfrac{x^2}{2\sigma^2}},$$

where $x \ge 0$.

My understanding is that, since $X_1$ and $X_2$ are independent $N(0, \sigma^2)$ random variables, we have that

$$Z = \sqrt{\left( \dfrac{X_1 - 0}{\sigma} \right)^2 + \left( \dfrac{X_2 - 0}{\sigma} \right)^2}$$

It is then said that this means that $Z$ has a Rayleigh distribution. But how does $Z = \sqrt{\left( \dfrac{X_1 - 0}{\sigma} \right)^2 + \left( \dfrac{X_2 - 0}{\sigma} \right)^2}$ imply that $Z$ has a Rayleigh distribution? (That is, what is the reasoning here?) And, in particular, how does this imply that $Z$ has density $f_{\sigma}(x) = \dfrac{x}{\sigma^2} e^{-\dfrac{x^2}{2\sigma^2}}$, $x \ge 0$?

$\endgroup$
5
  • 2
    $\begingroup$ Hint: look at the $\chi_2^2$ distribution $\endgroup$
    – jcken
    Apr 14, 2021 at 11:30
  • $\begingroup$ @jcken Thanks for that. Hmm, I see it here en.wikipedia.org/wiki/… . And we know that the sum of squares of independent standard normal random variables is chi-squared en.wikipedia.org/wiki/Chi-square_distribution#Definitions . But the problem here is that we don't have standard normal random variables $N(0, 1)$, but rather $N(0, \sigma^2)$. So what's going on here? $\endgroup$ Apr 14, 2021 at 11:35
  • $\begingroup$ @jcken I think Hunaphu pretty much answered my question, but, out of intellectual curiosity, can anyone clarify this point from jcken? I'm still not clear on this. $\endgroup$ Apr 14, 2021 at 12:24
  • 1
    $\begingroup$ Note that if $X \sim N(0, 1)$ then $\sigma X \sim N(0, \sigma^2)$. Equivalently, if $Y \sim N(0, \sigma^2)$ then $Y/\sigma \sim N(0,1)$. Essentially, we force the things in the squares to be $N(0,1)$ distributed by dividing by their standard deviation $\endgroup$
    – jcken
    Apr 14, 2021 at 14:02
  • $\begingroup$ Your understanding is not quite right: the correct formula has a factor of $\sigma$ missing from yours:$$Z =\sigma\, \sqrt{\left( \dfrac{X_1 - 0}{\sigma} \right)^2 + \left( \dfrac{X_2 - 0}{\sigma} \right)^2}.$$ $\endgroup$
    – whuber
    Dec 3, 2023 at 17:42

1 Answer 1

9
$\begingroup$

$$ P(Z \leq z) = P(\sqrt{X_1^2 + X_2^2} \leq z) = \int\int_A f_{X_1}f_{X_2}dx_1dx_2 $$

Where $A$ is the area where $Z$ is smaller than $z$ (but larger than zero since it is positive). Noting that $Z$ is a radius and switching to polar coordinates gives $X_1 = Z cos(\Theta)$ and $X_2 = Z \sin(\Theta)$ so $$ P(Z \leq z) = \frac{1}{2 \pi \sigma^2}\int_{0}^{2 \pi}\int_{0}^z\ r e^{-\tfrac{r^2}{2\sigma^2}}\,dr\,d\theta = (1 - e^{-\tfrac{z^2}{2\sigma^2}}). $$ And, $f_Z(z) = \tfrac{d}{dz}F_Z(z) = \frac{1}{\sigma^2}ze^{-\tfrac{z^2}{2\sigma^2}}$.

Were $2\pi\sigma^2 f_{X_1}(x_1)f_{X_2}(x_2) = e^{-\tfrac{x_1^2 + x_2^2}{2\sigma^2}}$ and $x_1^2 + x_2^2 = r^2(\cos^2 + \sin^2) = r^2$. Also, changing variables requires multiplication by the Jacobian giving the extra $\lvert r \rvert = r.$ The radius $r$ must be positive and since it is smaller than $z$ it is in the interval $0, z$. The angle $\theta$ is not restricted at all so it is allowed to be in $0, 2\pi$.

Even though the normal distribution arises naturally in many situations these formulas provide a very intuitive motivation with explanation for both the constant $\pi$ and the parameters for the normal distribution: It is the 1-dimensional version of something with exponential decay.

$\endgroup$
2
  • $\begingroup$ Thanks for the answer. Can you please explain the reasoning for $\int_0^z$? My polar coordinates are a bit rusty. $\endgroup$ Apr 14, 2021 at 11:51
  • $\begingroup$ Ahh, ok, that makes sense, since $\int \int_A \ dA = \int_{\theta = \alpha}^{\theta = \beta} \int_{r = g_1 (\theta)}^{r = g_2(\theta)} \ dr d\theta$ for polar coordinates. Thanks for the clarifying edit. $\endgroup$ Apr 14, 2021 at 11:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.