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 $f_{\sigma}(x) = \dfrac{x}{\sigma^2} e^{-\dfrac{x^2}{2\sigma^2}}$, $x \ge 0$?

  • 2
    $\begingroup$ Hint: look at the $\chi_2^2$ distribution $\endgroup$
    – jcken
    Commented 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? $\endgroup$ Commented 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. $\endgroup$ Commented 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
    Commented 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
    Commented Dec 3, 2023 at 17:42

1 Answer 1


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

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.

  • $\begingroup$ Thanks for the answer. Can you please explain the reasoning for $\int_0^z$? My polar coordinates are a bit rusty. $\endgroup$ Commented 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. $\endgroup$ Commented Apr 14, 2021 at 11:57

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