| Line 6: | Line 6: | ||
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}{ | + | <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}{5}</math> |
Part c) | Part c) | ||
<math>E[Y] = \int_{-1}^{0} v*(1+v) \, dv + \int_{0}^{1} v*(v) \, dv \, = \frac{1}{6} </math> | <math>E[Y] = \int_{-1}^{0} v*(1+v) \, dv + \int_{0}^{1} v*(v) \, dv \, = \frac{1}{6} </math> | ||
Latest revision as of 12:21, 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}{5} $
Part c)
$ E[Y] = \int_{-1}^{0} v*(1+v) \, dv + \int_{0}^{1} v*(v) \, dv \, = \frac{1}{6} $
