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math-340:m340-f15-hw:hw-19 [2015/11/03 18:21]
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math-340:m340-f15-hw:hw-19 [2015/11/16 06:06] (current)
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   - $P(0 < X < 10) = 0.4774$, $P(X \ge 9) = 0.5793$, and $P(8 \le X < 14) = 0.4435$   - $P(0 < X < 10) = 0.4774$, $P(X \ge 9) = 0.5793$, and $P(8 \le X < 14) = 0.4435$
   - Let $G_W$ be the cumulative distribution function of $W$. Then, for $w > 0$, $G_W(w) = P(Z^2 \le w) = P(-\sqrt{w} \le Z \le \sqrt{w}) = \Phi(\sqrt{w}) - \Phi(-\sqrt{w})$. Hence $g_W(w) = \phi(\sqrt{w}) \cdot \frac{1}{2\sqrt{2}} + \phi(-\sqrt{w}) \cdot \frac{1}{2\sqrt{w}} = \frac{1}{\sqrt{w}}\phi(\sqrt{w})$,​ from which the result follows.   - Let $G_W$ be the cumulative distribution function of $W$. Then, for $w > 0$, $G_W(w) = P(Z^2 \le w) = P(-\sqrt{w} \le Z \le \sqrt{w}) = \Phi(\sqrt{w}) - \Phi(-\sqrt{w})$. Hence $g_W(w) = \phi(\sqrt{w}) \cdot \frac{1}{2\sqrt{2}} + \phi(-\sqrt{w}) \cdot \frac{1}{2\sqrt{w}} = \frac{1}{\sqrt{w}}\phi(\sqrt{w})$,​ from which the result follows.
-  - $E(Z^2) = E(W) = \int_0^\infty w \cdot \frac{1}{\sqrt{2\pi w}}e^{-\frac{w}{2}}dw = \frac{1}{\sqrt{2\pi}}\int_0^\infty \sqrt{w}e^{-\frac{w}{2}}dw$. Letting $u = \sqrt{w}$, or, equivalently,​ $w = u^2$, this becomes $E(W) = \sqrt{\frac{2}{\pi}}\int_0^\infty u^2e^{-\frac{u^2}{2}}du = \sqrt{\frac{2}{\pi}} \cdot \sqrt{\frac{\pi}{2}} = 1$, where the last integral was evaluated in class. Similarly, $E(Z^4) = E(W^2) = \int_0^\infty w^2 \cdot \frac{1}{\sqrt{2\pi w}}e^{-\frac{w}{2}}dw = \frac{1}{\sqrt{2\pi}}\int_0^\infty w^{\frac{3}{2}}e^{-\frac{w}{2}}dw$. Letting $u = \sqrt{w}$, or, equivalently,​ $w = u^2$, this becomes $E(W^2) = \sqrt{\frac{2}{\pi}}\int_0^\infty u^4e^{-\frac{u^2}{2}}du$. Using integration by parts (with $y = u^3$ and $dv = ue^{-\frac{u^2}{2}}du$,​ this becomes $E(W^2) = 3\sqrt{\frac{2}{\pi}}\int_0^\infty u^2e^{-\frac{u^2}{2}}du = 3$.+  - $E(Z^2) = E(W) = \int_0^\infty w \cdot \frac{1}{\sqrt{2\pi w}}e^{-\frac{w}{2}}dw = \frac{1}{\sqrt{2\pi}}\int_0^\infty \sqrt{w}e^{-\frac{w}{2}}dw$. Letting $u = \sqrt{w}$, or, equivalently,​ $w = u^2$, this becomes $E(W) = \sqrt{\frac{2}{\pi}}\int_0^\infty u^2e^{-\frac{u^2}{2}}du = \sqrt{\frac{2}{\pi}} \cdot \sqrt{\frac{\pi}{2}} = 1$, where the last integral was evaluated in class. Similarly, $E(Z^4) = E(W^2) = \int_0^\infty w^2 \cdot \frac{1}{\sqrt{2\pi w}}e^{-\frac{w}{2}}dw = \frac{1}{\sqrt{2\pi}}\int_0^\infty w^{\frac{3}{2}}e^{-\frac{w}{2}}dw$. Letting $u = \sqrt{w}$, or, equivalently,​ $w = u^2$, this becomes $E(W^2) = \sqrt{\frac{2}{\pi}}\int_0^\infty u^4e^{-\frac{u^2}{2}}du$. Using integration by parts (with $y = u^3$ and $dv = ue^{-\frac{u^2}{2}}du$), this becomes $E(W^2) = 3\sqrt{\frac{2}{\pi}}\int_0^\infty u^2e^{-\frac{u^2}{2}}du = 3$.