hw3sol - UCSD ECE 153 Handout#14 Prof Young-Han Kim...

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Unformatted text preview: UCSD ECE 153 Handout #14 Prof. Young-Han Kim Thursday, April 21, 2011 Solutions to Homework Set #3 (Prepared by TA Yu Xiang) 1. Time until the n-th arrival. Let the random variable N ( t ) be the number of packets arriving during time (0 ,t ]. Suppose N ( t ) is Poisson with pmf p N ( n ) = ( t ) n n ! e- t for n = 0 , 1 , 2 ,.... Let the random variable Y be the time to get the n-th packet. Find the pdf of Y . Solution: To find the pdf f Y ( t ) of the random variable Y , note that the event { Y t } occurs iff the time of the n th packet is in [0 ,t ], that is, iff the number N ( t ) of packets arriving in [0 ,t ] is at least n . Alternatively, { Y > t } occurs iff { N ( t ) < n } . Hence, the cdf F Y ( t ) of Y is given by F Y ( t ) = P { Y t } = P { N ( t ) n } = summationdisplay k = n ( t ) k k ! e- t . Differentiating F Y ( t ) with respect to t , we get the pdf f Y ( t ) as f Y ( t ) = summationdisplay k = n bracketleftbigg e- t ( t ) k k ! + e- t ( t ) k- 1 ( k 1)! bracketrightbigg = e- t ( t ) n- 1 ( n 1)! summationdisplay k = n e- t ( t ) k k ! + summationdisplay k = n +1 e- t ( t ) k- 1 ( k 1)! = e- t ( t ) n- 1 ( n 1)! for t > 0. Or we can use another way. Since we know that the time interval T between packet arrivals is an exponential random variable with pdf f T ( t ) = braceleftbigg e- t , if t , , otherwise . Let T i denote the i.i.d. exponential interarrival times, then Y = T 1 + T 2 + + T n . By convolving f T ( t ) with itself n 1 times, which can be also computed by its Fourier transform (characteristic function), we can show that the pdf of Y is given by f Y ( t ) = braceleftBigg e- t ( t ) n- 1 ( n- 1)! , if t , , otherwise . 1 2. Diamond distribution. Consider the random variables X and Y with the joint pdf f X,Y ( x,y ) = braceleftbigg c if | x | + | y | 1 / 2 0 otherwise , where c is a constant. (a) Find c . (b) Find f X ( x ) and f X | Y ( x | y ). (c) Are X and Y independent random variables? Justify your answer. Solution: (a) The integral of the pdf f X,Y ( x,y ) over < x < , < y < is c , and therefore by the definition of joint density c = 1 . (b) The marginal pdf is obtained by integrating the joint pdf with respect to y . For 0 x 1 2 , f X ( x ) = integraldisplay 1 2- x- 1 2 + x cdy = 2 parenleftbigg 1 2 x parenrightbigg , and for 1 2 x 0, f X ( x ) = integraldisplay 1 2 + x- 1 2- x cdy = 2 parenleftbigg 1 2 + x parenrightbigg . So the marginal pdf may be written as f X ( x ) = braceleftBigg 2 2 | x | | x | 1 2 otherwise....
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This note was uploaded on 09/16/2011 for the course ECE ECE153 taught by Professor Younghankim during the Spring '11 term at UCSD.

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hw3sol - UCSD ECE 153 Handout#14 Prof Young-Han Kim...

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