Heads with fewer than n tosses it is clear that p t k

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heads with fewer than n tosses, it is clear that p T ( k ) = 0 for k < n . In addition, the probability of getting n heads in n tosses is q n so p T ( n ) = q n . Lastly, for k n + 1, we have T = k if there is no run of n heads in the first k - n - 1 tosses, followed by a tail, followed by a run of n heads, so p T ( k ) = P ( T > k - n - 1)(1 - q ) q n = summationdisplay i = k - n p T ( i ) (1 - q ) q n . (b) We use the PMF we obtained in the previous part to compute the moment generating function. Thus, M T ( s ) = E [ e sT ] = k = -∞ p T ( k ) e sk = q n e sn + (1 - q ) q n k = n +1 i = k - n p T ( i ) e sk . We observe that the set of pairs { ( i,k ) | k n +1 ,i k - n } is equal to the set of pairs { ( i,k ) | i 1 ,n + 1 k i + n } , so by reversing the order of the summations, we have M T ( s ) = q n e sn + (1 - q ) q n i =1 i + n k = n +1 p T ( i ) e sk = q n e sn parenleftBig 1 + (1 - q ) i =1 i k =1 p T ( i ) e sk parenrightBig = q n e sn parenleftBig 1 + (1 - q ) i =1 p T ( i ) e s - e s ( i +1) 1 - e s parenrightBig = q n e sn parenleftBig 1 + (1 - q ) e s 1 - e s i =1 p T ( i )(1 - e si ) parenrightBig . Now, since i =1 p T ( i ) = 1 and, by definition, i =1 p T ( i ) e si = M T ( s ), it follows that M T ( s ) = q n e sn parenleftbigg 1 + (1 - q ) e s 1 - e s (1 - M T ( s )) parenrightbigg . Rearrangement yields M T ( s ) = 1+ (1 - q ) e s 1 - e s 1 q n e sn + (1 - q ) e s 1 - e s = q n e sn ((1 - e s )+(1 - q ) e s ) 1 - e s +(1 - q ) q n e s ( n +1) = q n e sn (1 - qe s ) 1 - e s +(1 - q ) q n e s ( n +1) . (c) We have E [ T ] = d ds M T ( s ) vextendsingle vextendsingle vextendsingle s =0 = braceleftBig [1 - e s +(1 - q ) q n e s ( n +1) ][ nq n e sn (1 - qe s ) - qe s q n e sn ] (1 - e s +(1 - q ) q n e s ( n +1) ) 2 Page 2 of 7
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Massachusetts Institute of Technology Department of Electrical Engineering & Computer Science 6.041/6.431: Probabilistic Systems Analysis (Spring 2006) - q n e sn (1 - qe s )( - e s +( n +1)(1 - q ) q n e s ( n +1) ] (1 - e s +(1 - q ) q n e s ( n +1) ) 2 bracerightBigvextendsingle vextendsingle vextendsingle s =0 = (1 - q ) q n ( nq n (1 - q ) - q n +1 ) - q n (1 - q )( - 1+( n +1)(1 - q ) q n ) (1 - q ) 2 q 2 n = n (1 - q ) q n - q n +1 +1 - ( n +1)(1 - q ) q n (1 - q ) n q n = 1 - q n q n (1 - q ) . Note that for n = 1, this equation reduces to E [ T ] = 1 /q , which is the mean of a geometrically-distributed random variable, as expected. 5. We calculate f X | Y ( x | y ) using the definition of a conditional density. To find the density of Y , recall that Y is normal, so the mean and variance completely specify f Y ( y ). Y = X + N , so E [ Y ] = E [ X ] + E [ N ] = 0 + 0 = 0. Because X and N are independent, var( Y ) = var( X ) + var( N ) = σ 2 x + σ 2 n . So, f X | Y ( x | y ) = f X,Y ( x,y ) f Y ( y ) = f X ( x ) f N ( y - x ) f Y ( y ) = 1 2 πσ 2 x 1 2 πσ 2 n e - x 2 2 σ 2 x - ( y - x ) 2 2 σ 2 n 1 2 π ( σ 2 x + σ 2 n ) e - y 2 2( σ 2 x + σ 2 n ) = 1 radicalbigg 2 π σ 2 x σ 2 n σ
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  • Spring '06
  • Munther Dahleh
  • Computer Science, Electrical Engineering, Probability theory, Cumulative distribution function, Probabilistic Systems Analysis, Department of Electrical Engineering & Computer Science

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