# Rayleigh Distribution Quartiles

The Rayleigh distribution has PDF f(x) =xe−$\frac{x^2}{2}$, x >0. Let X have the Rayleigh distribution.

(a) Find P(1< X < 3).

(b) Find the first quartile, median, and third quartile of X.

Alright, so the first part is quite easy-- it's just the integral from 1 to 3 of f(x), but the second part is tricky. I know F(x) = 1 - e$^\frac{-x^2}{2}$, but the inverse is a function that doesn't exist. Any help?

• Welcome to CV! Since you are new here, you may want to take a tour, which has information for new users. It seems your question is from a textbook, therefore please add [self-study] tag and read its wiki if it is the case. Commented Oct 14, 2017 at 21:11
• The inverse certainly exists; not only does it exist, it's even possible to write in a simple closed form. Commented Oct 15, 2017 at 2:54
• Could you help me in writing that then? I have the inverse, but it always ends in an imaginary number. Commented Oct 15, 2017 at 18:32
• Show your work (i.e. how on earth do you get something imaginary?) -- it is likely we can help you identify your error. See our policy on homework style questions, e.g. as discussed in the help here and in the self-study tag-wiki (noting in particular the encouragement to show what you tried) Commented Oct 16, 2017 at 23:32

The inverse of $F$ exists: You have to use Naperian logarithm, i.e., $ln (e^a) = log_e (e^a) = a$.

• This does not provide an answer to the question. To critique or request clarification from an author, leave a comment below their post. - From Review Commented Oct 14, 2017 at 22:20
• @Kjetil This answer conforms to our guidance concerning self-study questions by providing a crucial idea without actually doing the work.
– whuber
Commented Oct 14, 2017 at 22:21

Since it has been a while, I think it is safe to put the full answer here.

The Rayleigh distribution has pdf $$f_X(x) = \frac{x}{\sigma^2}\text{e}^{-x^2/(2\sigma^2)}$$, for $$x\ge 0$$ with scale parameter $$\sigma > 0$$. The cdf is $$F_X(x) = 1 - \text{e}^{-x^2/(2\sigma^2)}$$.

(a) Find $$P(1< X <3)$$. There are several direct approaches.
$$P(1< X <3) = \int_1^3 f_X(x)dx = F_X(3) - F_X(1)$$

(b) The first quartile is $$q_{25} = F_X^{-1}(0.25)$$. Alternatively, consider the first quartile to be the solution to $$\int_0^{q_{25}} f_X(x)dx=0.25$$. Similarly, $$q_{50} = F_X^{-1}(0.5)$$ for the median (see here), and let $$q_{75}$$ be the third quartile.

Let $$q = q_p$$ be the quantile of interest such that $$q_p = F_X^{-1}(p)$$. Start with the CDF & invert (solve for $$q$$)...

\begin{align} 1-\text{e}^{\frac{-q^2}{2\alpha^2}} &= p \\ \text{e}^{\frac{-q^2}{2\alpha^2}} &= 1-p \\ \frac{-q^2}{2\alpha^2} &= \text{ln}(1-p) \\ -q^2 &= 2\alpha^2 \text{ln}(1-p) \\ \\ q &=\alpha \sqrt{-2 \text{ln}(1-p)} \quad \quad \square \end{align}

For the first quartile, set $$p=0.25$$, and so forth.