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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?

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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 $q_{50}$.

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

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  • $\begingroup$ 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

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