(New page: *If Z = B - H then the time T is <math> T = |Z| = |(B-H)| </math> because T is always positive. *Therefore <math>f_T(t) = f_z(t) + f_z(-t)</math> *<math>f_z(z) = f_B(b)\ast f_\tilde{H} (\t...) |
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*Then compute the convolution through the equation: <math> f_z(z)= \int \limits_{-\infty}^{\infty} f_B(\tau)f_\tilde{H} (z-\tau) d\tau </math> | *Then compute the convolution through the equation: <math> f_z(z)= \int \limits_{-\infty}^{\infty} f_B(\tau)f_\tilde{H} (z-\tau) d\tau </math> | ||
*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 | *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: <math> f_z(z)= \int \limits_{0}^{\infty} \lambda e^{-\lambda | + | *i.e. the first integral will look something like this: <math> f_z(z)= \int \limits_{0}^{\infty} \lambda e^{-\lambda \tau} \cdot \lambda e^{\lambda (z-\tau)} d\tau </math> |
*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. | *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 <math> f_T(t)= \frac{\lambda e^{\lambda z}}{2} for z<0 </math> and <math> f_T(t)= \frac{\lambda e^{-\lambda z}}{2} for z>0 </math> and 0 else. | *I end up getting <math> f_T(t)= \frac{\lambda e^{\lambda z}}{2} for z<0 </math> and <math> f_T(t)= \frac{\lambda e^{-\lambda z}}{2} for z>0 </math> 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. | 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) | ||
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| + | Steven is correct. We have to take <math> |(B-H)|</math> into account because <math> B-H </math> is for the fact that only Obama showes up first. But we do need to take into account that Hillary may show up before him, and if we didn't take the abs. then we would be evaluating a negative number....which we totally cant do. | ||
| + | --[[User:Longja|Longja]] 21:00, 21 October 2008 (UTC) | ||
Latest revision as of 16:00, 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)
Steven is correct. We have to take $ |(B-H)| $ into account because $ B-H $ is for the fact that only Obama showes up first. But we do need to take into account that Hillary may show up before him, and if we didn't take the abs. then we would be evaluating a negative number....which we totally cant do. --Longja 21:00, 21 October 2008 (UTC)
