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I am given a Rayleigh, distribution function:$$f(x)=\frac{1}{5}x\exp\left(\frac{-x^2}{10}\right)$$ with $x>0$ and asked to:

Use an appropriate random number generator algorithm to draw 500 samples from F(x).

What I thought on doing is using the Aceptance-rejection method :

  1. Generate a rv $Y$ distributed as $G$.

  2. Generate $U$ (independent from $Y$ ).

  3. If $U \leq \frac{f(Y)}{cg(Y)}$ , then set $X$ $=$ $Y$ (“accept”) ; otherwise go back to 1 (“reject”).

I thought that I will use the $\chi^2$ distribution as my $g(Y)$ with $k=1$ degrees of freedom. I made that decision on the fact that both functions have the same domain, namely $x\in(0,\infty)$ and their CDFs look "similar". Therefore : $$g(x)=\frac{x^{-\frac{1}{2}}\cdot\exp\left(-\frac{x}{2}\right)}{\sqrt{2}\Gamma\left(\frac{1}{2}\right)}$$ Then $$\frac{f(x)}{g(x)}=\frac{\sqrt{2}\Gamma\left(\frac{1}{2}\right)}{5}\cdot x ^{\frac{3}{2}}\cdot\exp\left(\frac{x}{2}-\frac{x^2}{10}\right)$$ And I found out that this function has a maximum at $$x=\frac{5+\sqrt{145}}{4}$$ which is around $4.26$. Hence $$\frac{f(x)}{g(x)}\leq c=4.26$$. But I also read that $c$ value has to be as "close" to 1 as possible and I think 4.26 is not "close" Is my calculation correct? Is it entirely wrong? Should I use different method for drawing that random samples? Thanks for any method

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    $\begingroup$ For your density function $f(x)$, the cumulative distribution function is $F(x)=1-e^{-\frac{\text{x}^2}{10}}$. Just generate a uniform random number $u$ on $[0,1]$ and solve for $x$ in $u=F(x)$. $\endgroup$ – JimB Jan 24 '18 at 18:31
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    $\begingroup$ With respect to your final comment / question: $c$ tells you how many uniform variates you'll have to generate on average in order to generate one "final" variate. It's not that a larger $c$ means your algorithm doesn't work, it just means that it will run more slowly, so you'd like $c$ smaller rather than larger. But for this distribution, @JimB 's suggestion is the way to go. $\endgroup$ – jbowman Jan 24 '18 at 18:52
  • $\begingroup$ See stats.stackexchange.com/q/184325/35989 $\endgroup$ – Tim Feb 3 '18 at 8:48
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The simplest way is to use the cumulative distribution function like in the title of your question. As pointed by Jim B., the CDF is:

$$F(x)=1-e^{-\frac{x^2}{10}}$$

The method is explained here: wikipedia or here: How does the inverse transform method work?

The Aceptance-rejection method is more complex, usually slower, and should not be the first choice. It's only useful when the CDF has no simple closed form.

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If you really want to do accept-reject I'd suggest that a $\chi^2_4$ density would be a considerably better choice as a proposal than a $\chi^2_1$. A $c$ of just under 1.5 will do (I'd just use 1.5, the difference is miniscule).

[With care, a well chosen gamma density would let you push it some way below 1.5. Probably not worth the effort though; you may be able to do considerably better with a mixture.]

However (as was suggested in comments), the inverse-cdf method is probably more convenient for the Rayleigh.

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