(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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| − | <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>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> |
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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> | ||
Latest revision as of 11:21, 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} $
