(New page: ==For part a== By definition... <math>P(|y| < \frac{1}{2}) \, = \int_{-\infty}^{\infty} f_Y (v) \, dv =\int_{-\frac{1}{2}}^{0} (1+v) \, dv + \int_{0}^{\frac{1}{2}} v \, dv = frac{1}{2}</...)
 
(For part a)
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By definition...
 
By definition...
  
<math>P(|y| < \frac{1}{2}) \, = \int_{-\infty}^{\infty} f_Y (v) \, dv =\int_{-\frac{1}{2}}^{0} (1+v) \, dv + \int_{0}^{\frac{1}{2}} v \, dv = frac{1}{2}</math>
+
<math>P(|y| < \frac{1}{2}) \, = \int_{-\infty}^{\infty} f_Y (v) \, dv =\int_{-\frac{1}{2}}^{0} (1+v) \, dv + \int_{0}^{\frac{1}{2}} v \, dv = \frac{1}{2}</math>

Revision as of 08:13, 16 October 2008

For part a

By definition...

$ P(|y| < \frac{1}{2}) \, = \int_{-\infty}^{\infty} f_Y (v) \, dv =\int_{-\frac{1}{2}}^{0} (1+v) \, dv + \int_{0}^{\frac{1}{2}} v \, dv = \frac{1}{2} $

Alumni Liaison

Sees the importance of signal filtering in medical imaging

Dhruv Lamba, BSEE2010