Revision as of 09:58, 21 October 2008 by Sstreete (Talk)

  • 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.

Alumni Liaison

Ph.D. on Applied Mathematics in Aug 2007. Involved on applications of image super-resolution to electron microscopy

Francisco Blanco-Silva