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Solutions to Problems from Chapter 4 of Ross Exercise 1 (Ross 4.17). Suppose that the distribution function of X is given by 0 b<0 b 0≤b<1 4 1 b−1 F (b) = 2 + 4 1≤b<2 11 2≤b<3 2 1 3 < b. (a) Find P (X = i) for i = 1, 2, 3. (b) Find P (1/2 < X < 3/2). Solution. Recall that P (X < t) = lims↑t F (s). We denote this limit-from-the-left by F (t−). We have P (X = i) = P (X ≤ i) − P (X < i) = F (i) − f (i−) . So 1 1 1 − = , 2 4 4 11 3 1 P (X = 2) = F (2) − F (2−) = − = , 12 4 6 1 11 = P (X = 3) = F (3) − F (3−) = 1 − 12 12 P (X = 1) = F (1) − F (1−) = For the second part, P (1/2 < X < 3/2) = P (X < 3/2) − P (X ≤ 1/2) = F (3/2−) − F (1/2) = Exercise 2 (Ross 4.19). If the distribution 0 1 2 3 F (b) = 45 5 9 10 1 calculate the pmf of X. 1 function of X is given by b<0 0≤b<1 1≤b<2 2≤b<3 3 ≤ b < 3.5 b ≥ 3.5 5 1 1 − = . 8 8 2 Solution. The pmf satisfies p(a) = P (X = a) = F (a) − F (a−) . Thus p(a) > 0 only at the values t where F has a jump. So 1 2 1 3 1 P (X = 1) = F (1) − F (1−) = − = 5 2 10 4 3 1 P (X = 2) = F (2) − F (2−) = − = 5 5 5 4 1 9 − = P (X = 3) = F (3) − F (3−) = 10 5 10 9 1 P (X = 3.5) = F (3.5) − F (3.5−) = 1 − = . 10 10 P (X = 0) = F (0) − F (0−) = At all other values a, we have p(a) = 0. Exercise 3 (Ross 4.T3). Express P (X ≥ a) in terms of the d.f. of X. Solution. P (X ≥ a) = 1 − P (X < a) = 1 − F (a−) , where again F (a−) = lims↑a F (s). Exercise 4 (Ross 4.T4). If X has d.f. X, what is the d.f. of eX . Solution. Let Y = eX . The question asks for the d.f. of Y . We will find this in terms of the d.f. FX of X. FY (t) = P (Y ≤ t) = P (eX ≤ t) = P (X ≤ log(t)) = FX (log(t)) . Exercise 5 (Ross 4.T5). Find the d.f. for αX + β, where α 6= 0. 2 Solution. FαX+β (t) = P (αX + β ≤ t) ( P (X ≤ t−β ) if α > 0 α = t−β P (X ≥ α ) if α < 0 ( ) if α > 0 FX ( t−β α = t−β 1 − FX ( α −) if α < 0 3