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$d > \sigma$ ... (1)

$\exp^{-(\frac{d^2}{2\sigma^2})} < 10^{-0.5}$ ... (2)

Is (1) <==> (2) true ?

EDIT: > replaced by < in the 2nd expression

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What? Could you please clearify your question... –  draks ... Jun 4 '12 at 12:50
    
@draks I've edited it –  shn Jun 4 '12 at 14:13
    
If $r>0$, this is true. What makes you worry? There is just one step missing... –  draks ... Jun 4 '12 at 17:18
    
@draks The program code that I made does not return the same result with the two equations. I've edited my post, can you check again ? –  shn Jun 5 '12 at 12:49
1  
If $d=100$ and $\sigma=1$ then (1) is certainly true and (2) is certainly false. –  Gerry Myerson Jun 5 '12 at 12:56

1 Answer 1

[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. :)

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