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I was told that the half-normal distribution, with density: $$ f(x)=\frac{1}{\sigma}\sqrt{\frac{2}{\pi}}\exp \left(-\frac{x^2}{2 \sigma^2} \right) \qquad \forall x \geq 0 $$ and the exponential distribution, with parameter $\lambda>0$ and density : $$ g(x) = \lambda \exp ( -\lambda x ) \qquad \forall x \geq 0 $$ cannot be parametrised (ie, for any $\lambda>0$, and $\sigma>0$) so that a constant $c > 0$ exists such that $g(x) \leq c f(x) \qquad \forall x \geq 0$

Is this true ? Is there a proof for this, especially one that I am likely to be able to understand (high school mathematics) ? If not, is there a counter-example ?

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up vote 1 down vote accepted

Since both $f(x)$ and $g(x)$ and non-negative for $x \geqslant 0$, the inequality $g(x) \leqslant c f(x)$ translates into $\log g(x) \leqslant \log(c) + \log f(x)$. That if such $c$ does exist, then there exists a $\kappa = \frac{c}{\lambda} \frac{1}{\sigma} \sqrt{\sqrt{2}{\pi}}$ such that $$ -\lambda x \leqslant -\frac{x^2}{2\sigma^2} + \log \kappa $$ That is $$ \frac{x^2}{2 \sigma^2 } - \lambda x = \frac{1}{2 \sigma^2} \left(x- \sigma^2 \lambda\right)^2 - \frac{\lambda^2 \sigma^2}{2} \leqslant \log \kappa $$ It is now clear such $\kappa$ can not exist, since $(x-\sigma^2 \lambda )^2$ is unbounded.

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