(New page: Part a) <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> Part b) <math>P...)
 
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Part b)
 
Part b)
  
<math>P(Y > 0|Y < \frac{1}{2}) \, = \frac{P(0 < Y < \frac {1}{2})}{P(Y < \frac{1}{2})} \,= \frac{\int_{0}^{\frac{1}{2}} v \, dv}{\int_{0}^{\frac{1}{2}} v \, dv + \int_{-1}^{0} (1+v) \, dv} \, = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{2}} \, = \frac{1}{6}
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<math>P(Y > 0|Y < \frac{1}{2}) \, = \frac{P(0 < Y < \frac {1}{2})}{P(Y < \frac{1}{2})} \,= \frac{\int_{0}^{\frac{1}{2}} v \, dv}{\int_{0}^{\frac{1}{2}} v \, dv + \int_{-1}^{0} (1+v) \, dv} \, = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{2}} \, = \frac{1}{6}</math>
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Part c)
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<math>E[Y] = \int_{-1}^{0} v*(1+v) \, dv + \int_{0}^{1} v*(v) \, dv \, = \frac{1}{6} </math>

Revision as of 12:20, 16 October 2008

Part a)

$ 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} $


Part b)

$ P(Y > 0|Y < \frac{1}{2}) \, = \frac{P(0 < Y < \frac {1}{2})}{P(Y < \frac{1}{2})} \,= \frac{\int_{0}^{\frac{1}{2}} v \, dv}{\int_{0}^{\frac{1}{2}} v \, dv + \int_{-1}^{0} (1+v) \, dv} \, = \frac{\frac{1}{8}}{\frac{1}{8} + \frac{1}{2}} \, = \frac{1}{6} $

Part c)

$ E[Y] = \int_{-1}^{0} v*(1+v) \, dv + \int_{0}^{1} v*(v) \, dv \, = \frac{1}{6} $

Alumni Liaison

Ph.D. 2007, working on developing cool imaging technologies for digital cameras, camera phones, and video surveillance cameras.

Buyue Zhang