$$f(x)=\frac{e^x}{(1+e^x)^2}$$ $$g(x)=\frac{1}{2}e^{-\left | x \right |}$$ both for ($-\infty$$<x<$$\infty$).Why we can't use either f or g to sample from a Cauchy distribution. What's the reason for that ??And also whether either would provide a practically useful way of generating standard normal random variables??? Thanks in advance.
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There are constants $K_i$ such that $K_1f(x) \ge \varphi(x)$ and $K_2g(x) \ge \varphi(x),$ where $\varphi$ is the standard normal PDF. However, neither $f$ nor $g$ similarly 'majorizes' the Cauchy PDF because of the Cauchy's 'fat' tails.// If there is more to your question, please edit it to clarify. – BruceET Apr 23 '17 at 22:59
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For example with $g$ (Laplace), see this post under 'umbrella' heading. – BruceET Apr 23 '17 at 23:13