Difference between revisions of "7.4 Steven Streeter ECE302Fall2008sanghavi" - Rhea

Revision as of 12:36, 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)