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If this is wrong, it's definitely close to being on the right track so please double check the math and let me know if anyone gets something different. | If this is wrong, it's definitely close to being on the right track so please double check the math and let me know if anyone gets something different. | ||
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| + | I think this shows that we do not need to really worry about one showing up before the other. <math> B-H </math> is only valid for Obama showing up first, but taking <math> |(B-H)|</math> is saying we do not care who shows up first. | ||
| + | --[[User:Ahartnet|Ahartnet]] 16:36, 21 October 2008 (UTC) | ||
Revision as of 11:36, 21 October 2008
- If Z = B - H then the time T is $ T = |Z| = |(B-H)| $ because T is always positive.
- Therefore $ f_T(t) = f_z(t) + f_z(-t) $
- $ f_z(z) = f_B(b)\ast f_\tilde{H} (\tilde{h}) $ where $ f_\tilde{H} (\tilde{h}) = f_H(-h) $
- Then compute the convolution through the equation: $ f_z(z)= \int \limits_{-\infty}^{\infty} f_B(\tau)f_\tilde{H} (z-\tau) d\tau $
- There are two cases: 1) z < 0 2)z>0. The 1st case will have the limits of 0 to infinity and case 2 will have the limits of z to infinity
- i.e. the first integral will look something like this: $ f_z(z)= \int \limits_{0}^{\infty} \lambda e^{-\lambda \tau} \cdot \lambda e^{\lambda (z-\tau)} d\tau $
- Note, the second exponential DOES NOT have a negative sign in front of it because it's the negative of h for Hillary's time.
- I end up getting $ f_T(t)= \frac{\lambda e^{\lambda z}}{2} for z<0 $ and $ f_T(t)= \frac{\lambda e^{-\lambda z}}{2} for z>0 $ and 0 else.
If this is wrong, it's definitely close to being on the right track so please double check the math and let me know if anyone gets something different.
I think this shows that we do not need to really worry about one showing up before the other. $ B-H $ is only valid for Obama showing up first, but taking $ |(B-H)| $ is saying we do not care who shows up first. --Ahartnet 16:36, 21 October 2008 (UTC)
