# “Convert” Rayleigh random variable into a Uniform random variable?

I have a nested question of sorts. My first question, is that I am wondering if it is possible to 'convert' a Rayleigh random variable into a uniform random variable, and how one may do this.

Strongly related and dependent on this question though, is: if such a conversion was possible, would 'corrupt data', being defined here as outliers in the original Rayleigh PDF, also remain 'corrupt' outliers in the uniform distribution?

• a) yes, just use the cdf of the chi2 with 2 degrees of freedom. b) No. They will no longer be outliers. – user603 Jul 1 '13 at 19:34
• @user603 Thank you. Is it a general principle, that outliers in one PDF, will never be outliers in the PDF to which they got converted to, or is this just something specific? – Creatron Jul 1 '13 at 19:36
• yes. An outlier has to be a able to drive an estimator to a specific (but arbitrary) value, regardless of the distribution of the majority of the data. A value drawn from a bounded distribution can't do that. – user603 Jul 1 '13 at 19:39
• No, that is not a general principle: @user603 is just assuming the outliers are not "extreme." They might or might not be outliers after the transformation. What can be said, though, is that if (say) $n$ values were drawn independently and randomly according to a $\chi^2$ distribution and an $n+1$st much larger value were thrown in, then--upon applying this probability transformation--as $n$ gets "large" it is highly likely (but not completely certain) that the $n+1$ resulting values would exhibit no apparent outliers. ("Large" means bigger than $2$ or so in this case. :-) – whuber Jul 1 '13 at 19:44
• With $N$ that large, for all practical purposes @user603 is absolutely right: you won't see any outliers among the transformed variables. (The chance of that happening is somewhere around $10^{-200}$ or so, depending on what outlier-detection algorithm you use.) – whuber Jul 1 '13 at 20:12

Your assertion that "a $\chi^{2}$ random variable (with 2 degrees of freedom; aka Rayleigh random variable)" is incorrect: a $\chi^{2}$ random variable with two degrees of freedom is an exponential random variable with mean $2$, and not a Rayleigh random variable.

• If $X$ is a $\chi^2$ random variable with two degrees of freedom, then its CDF is $F_X(x) = [1 - \exp(-x/2)]\mathbf 1_{x\in [0,\infty)}$ and so if $x_1, x_2, \ldots, x_n$ are samples of $X$, then $y_1, y_2, \ldots, y_n$ are samples of a $U[0,1)$ random variable where $y_i = 1-e^{-x_i/2}$, $1 \leq i \leq n$.
• If $Z$ is a Rayleigh random variable, its CDF is $F_Z(z) = [1 - \exp(-z^2/2)]\mathbf 1_{z\in [0,\infty)}$ and so if $z_1, z_2, \ldots, z_n$ are samples of $Z$, then $w_1, w_2, \ldots, w_n$ are samples of a $U[0,1)$ random variable where $w_i = 1-e^{-z_i^2/2}$, $1 \leq i \leq n$.
• $\sqrt{X}$ is a Rayleigh random variable, and so if we have samples $x_1, x_2, \ldots, x_n$ of $X$, then $\sqrt{x_1}, \sqrt{x_2}, \ldots, \sqrt{x_n}$ are samples of a Rayleigh random variable, and regardless of whether we choose to map $x_i \mapsto 1-e^{-x_i/2}$ or to map $\sqrt{x_i} \mapsto 1-e^{-\left(\sqrt{x_i}\right)^2/2}$, we get the same $y_1, y_2, \ldots, y_n$ as samples of a $U[0,1)$ random variable.
With regard to your question about outliers, note that the mean of $X$ is $2$ which maps to $1-e^{-1} \approx 0.632$ and so outliers, say any sample larger than $10$, are mapped into a narrow range $[0.99326\ldots, 1)$. Remember that sample values close to $0$ are not outliers, since the density of $X$ is a monotone decreasing function on $\mathbb R^+$ and so values close to $0$ are highly likely and not rare at all.
• Yes. For any continuous random variable $X$, $F(X)$ is uniformly distributed on the unit interval. – Dilip Sarwate Jul 2 '13 at 19:29