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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

  • 2
    What? Could you please clearify your question...2012-06-04
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    @draks I've edited it2012-06-04
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    If $r>0$, this is true. What makes you worry? There is just one step missing...2012-06-04
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    @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 ?2012-06-05
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    If $d=100$ and $\sigma=1$ then (1) is certainly true and (2) is certainly false.2012-06-05
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    @GerryMyerson but $d > \sigma$ <=> $d^2 > -2\sigma^2 log(10^{-0.5})$ <=> $\exp^{-(\frac{d^2}{2\sigma^2})} < 10^{-0.5}$, this is not true ?2012-06-05
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    well I mean < not > in the second expression2012-06-05

1 Answers 1

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. :)