# ass1sol - STAT455/855 Fall 2009 Applied Stochastic...

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STAT455/855 Fall 2009 Applied Stochastic Processes Assignment #1, Solutions Total Marks: 35 for 455 and 40 for 855. 1. Ross, Chapter 3 #28. (12 marks) Let us deﬁne Y i = ± 1 if the i th draw is red 0 if the i th draw is blue . Then X k = Y 1 + ... + Y k . (a) (1 mark) The expected value of X 1 is the expected value of Y 1 , which is just the probability that the ﬁrst draw is a red ball, that is, E [ X 1 ] = E [ Y 1 ] = r/ ( r + b ). (b) (2 marks) To get E [ X 2 ] we ﬁrst compute E [ Y 2 ]. Conditioning on X 1 , we have E [ Y 2 ] = E [ Y 2 ² ² X 1 = 0] P ( X 1 = 0) + E [ Y 2 ² ² X 1 = 1] P ( X 1 = 1) = P (2nd draw is red ² ² 1st draw is blue) b r + b + P (2nd draw is red ² ² 1st draw is red) r r + b = ³ r r + b + m ´³ b r + b ´ + ³ r + m r + b + m ´³ r r + b ´ = r ( r + b + m ) ( r + b )( r + b + m ) = r r + b . Therefore, E [ X 2 ] = E [ Y 1 ] + E [ Y 2 ] = r r + b + r r + b = 2 r r + b . (c) (3 marks) Similarly, to get E [ X 3 ] we ﬁrst compute E [ Y 3 ] and condition on X 2 to get E [ Y 3 ]. The conditional expectation of Y 3 given that X 2 = j , for j = 0 , 1 , 2, is E [ Y 3 ² ² X 2 = j ] = P (3rd draw is red ² ² j red in ﬁrst 2 draws) = r + jm r + b + 2 m .

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STAT 455/855 -- Assignment 1 Solutions p. 2 Therefore, conditioning on X 2 we can compute E [ Y 3 ] as E [ Y 3 ] = 2 X j =0 E [ Y 3 ± ± X 2 = j ] P ( X 2 = j ) = 2 X j =0 ² r + jm r + b + 2 m ³ P ( X 2 = j ) = r + mE [ X 2 ] r + b + 2 m = r + m 2 r/ ( r + b ) r + b + 2 m = r ( r + b + 2 m ) ( r + b )( r + b + 2 m ) = r r + b . Then we have E [ X 3 ] = E [ Y 1 ] + E [ Y 2 ] + E [ Y 3 ] = r r + b + r r + b + r r + b = 3 r r + b . (d) (3 marks) From parts (a), (b), and (c), clearly the pattern seems to be E [ X k ] = kr/ ( r + b ). We can verify this conjecture by showing that E [ Y i ] = r/ ( r + b ) for all i 1. We have already shown that it is true for i = 1 , 2 , 3. We will use an induction argument, so assume the pattern holds for i = 1 ,...,k - 1. After k - 1 draws, given X k - 1 = j the urn contains r + mj red balls and a total of r + b + ( k - 1) m balls. Therefore, E [ Y k ± ± X k - 1 = j ] = r + mj r + b +( k - 1) m , so conditioning on X k - 1 we have E [ Y k ] = k - 1 X j =0 E [ Y k ± ± X k - 1 = j ] P ( X k - 1 = j ) = k - 1 X j =0 r + mj r + b + ( k - 1) m P ( X k - 1 = j ) = r + mE [ X k - 1 ] r + b + ( k - 1) m = r + m ( E [ Y 1 ] + ... + E [ Y k - 1 ]) r + b + ( k - 1) m = r + m ( k - 1) r/ ( r + b ) r + b + ( k - 1) m = r r + b , where the ﬁfth equality follows from the induction hypothesis. (e) (3 marks) We deﬁne the types as suggested in the hint. The symmetry argument is that each type should evolve statistically identically so that the expected number of type i balls in the urn after k draws is the same as the expected number of
STAT 455/855 -- Assignment 1 Solutions p. 3 type j balls in the urn after k draws, for any i 6 = j and for all k 1. Since after k draws there are exactly r + b + km balls in the urn and there are r + b types the expected number of type i balls should be r + b + km r + b = 1 + km r + b . Since

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ass1sol - STAT455/855 Fall 2009 Applied Stochastic...

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