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Fundamentals of Probability and Statistics for Engineers

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Expectations <strong>and</strong> Moments 89P ro<strong>of</strong> <strong>of</strong> P ropert y 4. to 2: show Property 4.2, let t <strong>and</strong> u be any real quantities<strong>and</strong> <strong>for</strong>m…t; u† ˆEf‰t…X m x †‡u…Y m Y †Š 2 gˆ 20 t 2 ‡ 2 11 tu ‡ 02 u 2 :Since the expectation <strong>of</strong> a nonnegative function <strong>of</strong> X <strong>and</strong> Y must be nonnegative,(t,u) is a nonnegative quadratic <strong>for</strong>m in t <strong>and</strong> u, <strong>and</strong> we musthave 20 02 2 11 0; …4:25†which gives the desired result.The normalization <strong>of</strong> the covariance through Equation (4.24) renders auseful substitute <strong>for</strong> 11. Furthermore, the correlation coefficient is dimensionless<strong>and</strong> independent <strong>of</strong> the origin, that is, <strong>for</strong> any constants a 1 ,a 2 ,b 1 ,<strong>and</strong> b 2with a 1 >0<strong>and</strong> a 2 > 0, we can easily verify that…a 1 X ‡ b 1 ; a 2 Y ‡ b 2 †ˆ…X; Y†:…4:26†P ropert y 4. 3. If X <strong>and</strong> Y are independent, then 11 ˆ 0 <strong>and</strong> ˆ 0: …4:27†P ro<strong>of</strong> <strong>of</strong> P ropert y 4. let 3: X <strong>and</strong> Y be continuous; their joint momentfound from11 is 11 ˆ EfXYg ˆZ 11Z 1xyf XY …x; y†dxdy:1If X <strong>and</strong> Y are independent, we see from Equation (3.45) that<strong>and</strong>f XY …x; y† ˆf X …x†f Y …y†; 11 ˆZ 11ˆ m X m Y :Z 1 Z 1 Z 1xyf X …x†f Y …y†dxdy ˆ xf X …x†dx yf Y …y†dy111Equations (4.23) <strong>and</strong> (4.24) then show that 11 ˆ 0 <strong>and</strong> ˆ 0. A similar resultcan be obtained <strong>for</strong> two independent discrete r<strong>and</strong>om variables.TLFeBOOK

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