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Fundamentals of Probability and Statistics for Engineers

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136 <strong>Fundamentals</strong> <strong>of</strong> <strong>Probability</strong> <strong>and</strong> <strong>Statistics</strong> <strong>for</strong> <strong>Engineers</strong>Upon evaluating Y (t), the moments <strong>of</strong> Y are given by [Equation (4.52)]:EfY n gˆjn …n†Y …0†; nˆ1;2;...: …5:33†Ex ample 5. 9. Problem: a r<strong>and</strong>om variable X is discrete <strong>and</strong> its pmf is given inExample 5.1. Determine the mean <strong>and</strong> variance <strong>of</strong> Y where Y ˆ 2X ‡ 1.Answer: using the first <strong>of</strong> Equations (5.31), we obtainEfYg ˆEf2X ‡ 1g ˆX…2x i ‡ 1†p X …x i †ˆ…1† 1 2‡…1† 1 4‡…9† 1 8‡…25† 1 8ˆ 5;<strong>and</strong>iˆ… 1† 1 ‡…1† 1 ‡…3† 1 ‡…5† 1 …5:34†2 4 8 8ˆ 34 ;EfY 2 gˆEf…2X ‡ 1† gˆX2 …2x i ‡ 1† 2 p X …x i † i …5:35† 2 Y ˆ EfY 2 gE 2 fYg ˆ5 3 2ˆ 714 16 : …5:36†Following the second approach, let us use the method <strong>of</strong> characteristic functionsdescribed by Equations (5.32) <strong>and</strong> (5.33). The characteristic function <strong>of</strong> Y is Y …t† ˆXe jt…2xi‡1† p X …x i †i ˆ e jt 12 ‡ e jt 1‡ e 3jt 1‡ e 5jt 14 8 8ˆ 18 …4e jt ‡ 2e jt ‡ e 3jt ‡ e 5jt †;<strong>and</strong> we haveEfYg ˆjEfY 2 gˆ 1 …1† 1 jY …0† ˆj …84 ‡ 2 ‡ 3 ‡ 5† ˆ34 ; …2†Y…0† ˆ18 …4 ‡ 2 ‡ 9 ‡ 25† ˆ5: TLFeBOOK

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