Solution to homework for Section 5.3, page 366
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1. f_{R}(r) = r e^(-r^{2}/2);

   F_{R}(r) = 1 - e^{-r^{2}/2};

   E(R) = sqrt(pi/2)

   r = sqrt(2 ln 2);

   f(x,y) = (1/2pi) e^{-(x^2 + y^2)/2}


   a) F_{R}(1/2)

   b) (1/4) [F_{R}(2) - F_{R}(1)]

   c) E(|Y|) = (1/sqrt(2pi)) int_{-infty}^{infty} |y| e^(-y^2 / 2) dy
             
             = 2/sqrt(2pi) int_{0}^{infty} y e^(-y^2 / 2) dy

             = sqrt(2/pi) [-e^(-y^2 / 2]|_{0}^{infty}

             = sqrt(2/pi)

   d) P(-r < X < r) = 2 Phi(r) - 1 = 2(0.8810) - 1 = 0.762

   e) 1/2pi double int over square [e^{-(x^2 + y^2)}/2] dx dy

      = 4 [1/sqrt(2pi) int_{0}^{r} e^{-x^2 / 2} dx ]^2

      = 4 [P(0 < X < r)]^2

      = 4 [Phi(r) - Phi(0)]^2

      = 4(0.145161)

   f) Similar to part (e)

      = 4 P(0 < X < r/sqrt(2)) = 4 (0.08803)

   g) = (1/2) (probability found in part (e))
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