For x 1 f x x 1 since the dart always falls within

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For x > 1 , F X ( x ) = 1 since the dart always falls within the unit circle, and for x < 0 , F X ( x ) = 0 because X is non-negative (a distance is always non-negative). 3. (b) The random variable X is continuous because its CDF has no jumps. The PDF of X can be found by differentiating the CDF. For x [0 , 1] , we have f X ( x ) = dF X ( x ) dx = 2 x. For x < 0 and x > 1 , f X ( x ) = 0 . 3. (c) Note first that when y < 0 , the CDF of F Y ( y ) = 0 since the Y is non-negative. For y 1 , F Y ( y ) = 1 since the maximum value Y can take is 1. Notice also that Y = 0 if X 1 2 and that 1 4 Y 1 when X (1 / 2 , 1] . Consequently, for y [0 , 1 / 4] , we have F Y ( y ) = Pr( Y y ) = Pr( Y = 0) = Pr X 1 2 = F X 1 2 = 1 4 . For y [1 / 4 , 1] , we have, from Total Probability, F Y ( y ) = Pr( Y y ) = Pr Y y, X 1 2 + Pr Y y, X > 1 2 = Pr 0 y, X 1 2 + Pr X 2 y, X > 1 2 = Pr X 1 2 + Pr 1 2 < X p ( y ) = F X 1 2 + F X p ( y ) - F X 1 2 = y. 2

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3. (d) The random variable Y is neither continuous nor discrete. Y takes a continuum of values over the interval [1 / 4 , 1] and is therefore not discrete. Moreover, it takes the value 0 with positive probability so it is not continuous either. 4. (a) The function F is a valid CDF because it satisfies all the properties of a CDF (refer to Figure 4). Indeed, lim x →∞ F ( x ) = 1 and lim x →-∞ F ( x ) = 0 . Moreover we also have 0 F ( x ) 1 . The function F is also nondecreasing and right- continuous. This is true because F is a linear combination with non-negative coefficients of u ( x ) and u ( x - 2) , which are both non-decreasing and right-continuous. 4. (b) The function F is a CDF of a discrete random variable (we will call it X ) because it increases only by jumps and is constant between any two jumps. The CDF of X has jumps at x = 0 and x = 2 and is constant otherwise. This means that X can take two values 0 and 2 and P ( X = 2) = P ( X 2) - P ( X < 2) = F (2) - lim x 2 - F ( x ) = 1 - 1 3 = 2 3 . Similarly, P ( X = 0) = F (0) - lim x 0 - F ( x ) = 1 3 - 0 = 1 3 . Figure 5 depicts this PMF. 4. (c) Pr( X [0 , 2)) = Pr( X = 0) = 1 3 . Another way to establish this is the following: P ( X [0 , 2)) = P ( X < 2) - P ( X < 0) = P ( X < 2) = lim x 2 - F ( x ) = 1 3 .
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