(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. | ||
Revision as of 10:58, 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.
