$d > \sigma$ ... (1)
$\exp^{-(\frac{d^2}{2\sigma^2})} < 10^{-0.5}$ ... (2)
Is (1) <==> (2) true ?
EDIT: > replaced by < in the 2nd expression
$d > \sigma$ ... (1)
$\exp^{-(\frac{d^2}{2\sigma^2})} < 10^{-0.5}$ ... (2)
Is (1) <==> (2) true ?
EDIT: > replaced by < in the 2nd expression
[EDITED replacing > with < in the 2nd equation]
This is the best I could come up with:
$(1) \iff (2)$ means that $(1) \implies (2) \wedge (1) \impliedby (2) \quad \forall d, \sigma $
This can be proven to be false since from $(2)$ you have that:
$log_e{\exp^{-{d^2\over{2\sigma^2}}}} \lt log_e{10^{-0,5}} \rightarrow -{d^2\over{2\sigma^2}} \lt log_e{10^{-0,5}} \rightarrow -{d^2\over{2\sigma^2}} \lt -0,5\cdot log_e{10}$
Assuming $\sigma \ne 0$, I can multiply both terms by $2\sigma^2$
$(2)$ $-d^2 \lt -0,5\cdot 2\sigma^2\cdot \log_e{10} \rightarrow -d^2 \lt -\sigma^2\cdot \log_e{10} \rightarrow d^2 \gt \sigma^2\cdot \log_e{10}$
By applying the square root operator, I get
$(2)$ $d \gt \sqrt{\log_e{10}}\cdot \sigma$
I assumed that $d$ and $\sigma$ are real numbers. To prove that $(1) \iff (2)$ does not hold you just need to find a single counterexample:
$\sigma = 1 \quad d = 1.1 \quad \rightarrow \sqrt{\log_e{10}}\cdot \sigma = 1.51 $
Does it makes sense to you? I tried. :)