0

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?

SecretAgentMan
  • 1,463
  • 10
  • 30
Mike S.
  • 1
  • 1
    Welcome to CV! Since you are new here, you may want to take a [tour](https://stats.stackexchange.com/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](https://stats.stackexchange.com/tags/self-study/info) if it is the case. – Márcio Augusto Diniz Oct 14 '17 at 21:11
  • The inverse *certainly* exists; not only does it *exist*, it's even possible to write in a simple closed form. – Glen_b Oct 15 '17 at 02:54
  • Could you help me in writing that then? I have the inverse, but it always ends in an imaginary number. – Mike S. Oct 15 '17 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](https://stats.stackexchange.com/help/on-topic) and in the [self-study tag-wiki](https://stats.stackexchange.com/tags/self-study/info) (noting in particular the encouragement to show what you tried) – Glen_b Oct 16 '17 at 23:32

2 Answers2

3

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

Márcio Augusto Diniz
  • 667
  • 1
  • 4
  • 11
  • 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](/review/low-quality-posts/163327) – kjetil b halvorsen Oct 14 '17 at 22:20
  • 2
    @Kjetil This answer conforms to our guidance concerning [tag:self-study] questions by providing a crucial idea without actually doing the work. – whuber Oct 14 '17 at 22:21
2

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.

SecretAgentMan
  • 1,463
  • 10
  • 30