This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: UCSD ECE 153 Handout #14 Prof. YoungHan Kim Thursday, April 21, 2011 Solutions to Homework Set #3 (Prepared by TA Yu Xiang) 1. Time until the nth 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 nth 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 &gt; t } occurs iff { N ( t ) &lt; 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 &gt; 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 &lt; x &lt; , &lt; y &lt; 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....
View
Full
Document
 Spring '11
 YoungHanKim

Click to edit the document details