I am not sure how to solve the following problem:

The probability density function of the Rayleigh distribution is,

$\ f(x;α) = \frac{x}{α^2} e^\frac{-x^2}{2α^2}, x ≥ 0, $

where α is the scale parameter of the distribution. Find the median of the Rayleigh distribution.

I need to derive the median of the distribution, but do not know how to do so. Thoughts?


The Rayleigh distribution has cumulative distribution function (CDF) $F_X(x) = 1-\text{e}^{\frac{-x^2}{2\alpha^2}}$.

Denote the median $q_{50}$.

Starting with the CDF...

$$\begin{align} 1-\text{e}^{\frac{-q_{50}^2}{2\alpha^2}} &= 0.5 \\ \text{e}^{\frac{-q_{50}^2}{2\alpha^2}} &= 0.5 \\ \frac{-q_{50}^2}{2\alpha^2} &= \text{ln}(0.5) \\ -q_{50}^2 &= 2\alpha^2 \text{ln}(0.5) \\ \\ q_{50} &=\alpha \sqrt{-2 \text{ln}(0.5)} \\ &= \alpha \sqrt{2\text{ln}(2)} \quad \quad \square \end{align}$$

See here or here for general quantiles.

Update: Based on comments amounting to "can I do this from the PDF," yes, it is possible but requires a little more effort (integration).

You can solve $\int_0^{q_{50}} f_X(x)dx = 0.5$ for $q_{50}$.

$$\begin{align}\int_0^{q_{50}} f_X(x)dx &= 0.5 \\ \int_0^{q_{50}} \frac{x}{\alpha^2}\text{e}^{-x^2/(2\alpha^2)} dx&= 0.5 \\ 1-\text{e}^{\frac{-q_{50}^2}{2\alpha^2}} &= 0.5 \\ &\text{Continue using CDF approach above} \end{align}$$

  • $\begingroup$ Right, because this is a CRV I should use the CDF. Thanks!! $\endgroup$ – Joe Ademo Oct 29 '18 at 20:31
  • $\begingroup$ Discrete random variables also have a CDF. I would rather you say since the median (50th-quantile) is an inverse of the CDF, it makes sense to start with the CDF and simply invert. That's how I think about this problem at least. $\endgroup$ – SecretAgentMan Oct 29 '18 at 20:34
  • $\begingroup$ My point was that since this is a CRV, it would be incorrect to use the PMF. However, I understand your point. Thanks for the help. $\endgroup$ – Joe Ademo Oct 29 '18 at 20:38
  • $\begingroup$ @JoeAdemo, I've added the approach starting from the PDF. $\endgroup$ – SecretAgentMan Nov 16 '18 at 15:32

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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