HW10_Sol

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

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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 This study resource was shared via CourseHero.com

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(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 This study resource was shared via CourseHero.com
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

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