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math-340:m340-f15-hw:hw-25 [2015/11/10 08:49]
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math-340:m340-f15-hw:hw-25 [2015/11/12 12:16] (current)
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 Answers: Answers:
  
-  - Part of assignment due 11 November+  - (a) $\frac{8}{15}$ (b) $\frac{2}{3}$ (%%c%%) $\frac{1}{3}$
   - For any $w > 0$, $P(W \le w) = \int\int_B \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy$,​ where $B = \{(x, y) \ | \ x^2 + y^2 \le w^2\}$. Switching to polar coordinates,​ $P(W \le w) = \frac{1}{2\pi}\int_0^w \int_0^{2\pi} re^{-\frac{r^2}{2}}d\theta dr = 1 - e^{-\frac{w^2}{2}}$. Hence the probability density function of $W$ is $f_W(w) = \begin{cases}we^{-\frac{w^2}{2}},&​ \text{if } w > 0, \\ 0,& \text{otherwise}.\end{cases}$   - For any $w > 0$, $P(W \le w) = \int\int_B \frac{1}{2\pi}e^{-\frac{x^2+y^2}{2}}dxdy$,​ where $B = \{(x, y) \ | \ x^2 + y^2 \le w^2\}$. Switching to polar coordinates,​ $P(W \le w) = \frac{1}{2\pi}\int_0^w \int_0^{2\pi} re^{-\frac{r^2}{2}}d\theta dr = 1 - e^{-\frac{w^2}{2}}$. Hence the probability density function of $W$ is $f_W(w) = \begin{cases}we^{-\frac{w^2}{2}},&​ \text{if } w > 0, \\ 0,& \text{otherwise}.\end{cases}$
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     - $f_W(w) = \begin{cases}-\frac{1}{2}\log\left(\frac{w}{2}\right),&​ \text{if } 0 < w < 2, \\ 0,& \text{otherwise}.\end{cases}$     - $f_W(w) = \begin{cases}-\frac{1}{2}\log\left(\frac{w}{2}\right),&​ \text{if } 0 < w < 2, \\ 0,& \text{otherwise}.\end{cases}$
     - $f_W(w) = \begin{cases}1,&​ \text{if } 0 < w \le  \frac{1}{2},​ \\ \frac{1}{4w^2},&​ \text{if } w > \frac{1}{2},​ \\ 0,& \text{otherwise}.\end{cases}$     - $f_W(w) = \begin{cases}1,&​ \text{if } 0 < w \le  \frac{1}{2},​ \\ \frac{1}{4w^2},&​ \text{if } w > \frac{1}{2},​ \\ 0,& \text{otherwise}.\end{cases}$