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Mathematical Methods for Physics and Engineering - Matematica.NET

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PROBABILITY30.5.3 Variance <strong>and</strong> st<strong>and</strong>ard deviationThe variance of a distribution, V [X], also written σ 2 , is defined byV [X] =E [ {∑(X − µ) 2] j=(x j − µ) 2 f(x j )∫(x − µ) 2 f(x) dx<strong>for</strong> a discrete distribution,<strong>for</strong> a continuous distribution.(30.48)Here µ has been written <strong>for</strong> the expectation value E[X] ofX. As in the case ofthe mean, unless the series <strong>and</strong> the integral in (30.48) converge the distributiondoes not have a variance. From the definition (30.48) we may easily derive thefollowing useful properties of V [X]. If a <strong>and</strong> b are constants then(i) V [a] =0,(ii) V [aX + b] =a 2 V [X].The variance of a distribution is always positive; its positive square root isknown as the st<strong>and</strong>ard deviation of the distribution <strong>and</strong> is often denoted by σ.Roughly speaking, σ measures the spread (about x = µ) of the values that X canassume.◮Find the st<strong>and</strong>ard deviation of the PDF <strong>for</strong> the distance from the origin of the electronwhose wavefunction was discussed in the previous two examples.Inserting the expression (30.47) <strong>for</strong> the PDF f(r) into (30.48), the variance of the r<strong>and</strong>omvariable R is given by∫ ∞V [R] = (r − µ) 2 4r2 e −2r/a 0dr = 4 (r 4 − 2r 3 µ + r 2 µ 2 )e −2r/a 0dr,0 a 3 0a 3 0 0where the mean µ = E[R] =3a 0 /2. Integrating each term in the integr<strong>and</strong> by parts weobtain∫ ∞V [R] =3a 2 0 − 3µa 0 + µ 2 = 3a2 04 .Thus the st<strong>and</strong>ard deviation of the distribution is σ = √ 3a 0 /2. ◭We may also use the definition (30.48) to derive the Bienaymé–Chebyshevinequality, which provides a useful upper limit on the probability that r<strong>and</strong>omvariable X takes values outside a given range centred on the mean. Let us considerthe case of a continuous r<strong>and</strong>om variable, <strong>for</strong> which∫Pr(|X − µ| ≥c) = f(x) dx,|x−µ|≥cwhere the integral on the RHS extends over all values of x satisfying the inequality1146

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