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For part (a) finding the cdf $F(w)$ of $W$,my way to do it: $F(v)=F(g(v))= \begin{cases} 1/(5 \sqrt{2}) \int_{-\infty}^{-10} e^{-(-10)^2/50}\, dx & \text{$v<-10$} \\ 1/(5 \sqrt{2}) \int_{-10}^{10} e^{-(v)^2/50}\,dx & \text{$-10≤v≤10$}\\ 1/(5 \sqrt{2}) \int_{10}^{\infty} e^{-(10)^2/50}\, dx & \text{$v≥10$} \end{cases}$

I am not sure about my way to do it is correct. What do you think?

Michael
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1 Answers1

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Notice that for $-10\leq t < 10$, $w \leq t$ whenever $v \leq t$. In addition, $w$ is always less than or equal to $10$ and never less than $-10$. So $$P(w \leq t) = 0,\qquad t < -10$$ $$P(w \leq t) = P(v \leq t) = \frac{1}{5\sqrt{2\pi}}\int_{-\infty}^te^{-x^2/50}\;dx,\qquad t< 10$$ $$P(w \leq 10) = 1$$ We may write this in one equation using the step function $u$: \begin{align} F_W(w)&=u(w+10)\left(\frac{1-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w-10)\right) \\ &=\frac{u(w+10)-u(w+10)u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w+10)u(w-10) \\ &= \frac{u(w+10)-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50}\;dx + u(w-10) \end{align} We may take the derivative of $F_W$, remembering that $\frac{d}{dx}u(x) = \delta(x)$, where $\delta(x)$ is the Dirac delta: $$f_w(w)=\frac{\delta(w+10) - \delta(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^we^{-x^2/50}\;dx + \frac{u(w+10)-u(w-10)}{5\sqrt{2\pi}}e^{-w^2/50} + \delta(w-10)$$

bames
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  • cdf F(w) should be , 0 for w<-10 , the function in the question for -10<=w<=10 , 1 for w>10? @bames – Michael Feb 21 '18 at 03:09
  • @Michael The function in my answer works for all values of $w$, since it is zero for $w<-10$ and $1$ for $w\geq 10$. As defined, $w$ is never less than $-10$ and always $\leq 10$, which is why $F_W$ is the way it is. – bames Feb 21 '18 at 03:16
  • i only learn the way like that(https://i.imgur.com/GcKEWMS.png) to write the unit step function, i don't understand how your step function to be written. @bames – Michael Feb 21 '18 at 04:01
  • @Michael It is the same step function, I just used a different symbol. I changed it to $u$'s for you, so you can read it more easily. Let me know if there are any other questions you have! – bames Feb 21 '18 at 04:17
  • I can't get your function https://i.imgur.com/k9Vx6UA.jpg – Michael Feb 21 '18 at 04:43
  • @Michael That is the same as my function. Just note that $u(w+10)u(w-10) = u(w-10)$. So we get: \begin{align}u(w+10)\left(\frac{1-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50};dx + u(w-10)\right) &= \frac{u(w+10)-u(w+10)u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50};dx + u(w+10)u(w-10)\ &= \frac{u(w+10)-u(w-10)}{5\sqrt{2\pi}}\int_{-\infty}^w e^{-x^2/50};dx + u(w-10)\end{align} I'll add the simplification to my answer to make things more clear. – bames Feb 21 '18 at 04:48