HW10_Sol - ECEN 646 Homework 10 Solutions 1 All parts of...

Info icon This preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
ECEN 646 Homework 10 Solutions 1 All parts of this question can be solved by moving from cartesian co- ordinates to polar coordinates. Since X and Y are independent gaussian random variables, their joint density is given by the product of their indi- vidual densities. f X,Y ( x, y ) = 1 2 πσ 2 e x 2 + y 2 2 σ 2 (a) The probability that the dart falls within the circle can be found by integrating the above joint pdf over this circular region. This integral is most easily evaluated using polar coordinates. The dart falling within a circle of radius σ is the same as the event { x 2 + y 2 σ 2 } . Note that the Jacobian of this transformation is given by J ( x, y ) = radicalbig x 2 + y 2 . Therefore, we have f X,Y ( x, y ) dxdy = rf R, Θ ( r, θ ) drdθ = r 2 πσ 2 exp( - r 2 2 σ 2 ) Using this transformation, we have P ( x 2 + y 2 σ 2 ) = integraldisplay 2 π 0 integraldisplay σ 0 r 2 πσ 2 exp( - r 2 2 σ 2 ) dr = integraldisplay σ 0 r σ 2 exp( - r 2 2 σ 2 ) dr = 1 - e 1 / 2 (b) By symmetry the dart is equally likely to fall in any of the 4 quad- rants. Therefore, the probability that the dart falls in the first quadrant is 1 4 . The answer can also be found by evaluating the integral P (dart in 1 st quadrant) = integraldisplay π/ 2 0 integraldisplay 0 r 2 πσ 2 exp( - r 2 2 σ 2 ) dr = 1 4 1 https://www.coursehero.com/file/6709632/HW10-Sol/ This study resource was shared via CourseHero.com
Image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
(c) This probability can be evaluated using Bayes’ rule. Let A and B denote the events of the dart falling in the σ -radius circle and in the 1 st quadrant respectively. P ( A | B ) = P ( B | A ) P ( A ) P ( B ) P ( A ) and P ( B ) have been computed in the previous parts. It remains to find the probability that the dart falls in the 1 st quadrant given that it falls in the σ -radius circle. But by symmetry, this probability is just 1 4 . Substituting this in the above formula yields P ( A | B ) = 1 4 P ( A ) 1 / 4 = P ( A ) = 1 - e 1 / 2 The above derivation shows that the events A and B are independent. (d) P [0 r R, 0 θ Θ] = integraldisplay Θ 0 integraldisplay R 0 r 2 πσ 2 exp( - r 2 2 σ 2 ) dr = Θ 2 π (1 - exp( - R 2 2 σ 2 )) 0 Θ 2 π 0 R < This is the joint CDF of R and Θ. We can obtain the marginals by setting R to be or Θ to be 2 π . P [0 r R ] = Θ 2 π (1 - exp( - R 2 2 σ 2 )) | Θ=2 π = (1 - exp( - R 2 2 σ 2 )) f R ( r ) = r σ 2 exp( - r 2 2 σ 2 ) r 0 P [0 θ Θ] = Θ 2 π (1 - exp( - R 2 2 σ 2 )) | R = = Θ 2 π f θ (Θ) = 1 2 π 0 Θ 2 π 2 https://www.coursehero.com/file/6709632/HW10-Sol/ This study resource was shared via CourseHero.com
Image of page 2
The joint CDF is simply the product of the individual CDFs. This means that the random variables r and θ are independent. The distribution of
Image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern