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procedure of the preceding problem. 15.3. If we choose the quantities E and V as "independent" variables, then the probability distribution function (15.1.8) does not reduce to a form as simple as (15.1.13) or (15.1.15); it is marked instead by the presence, in the exponent, of a cross term proportional to the product AEAV. Consequently, the variables E and V are not statistically independent: (AEAV) 50. Show that av av (AEAV) = KTT +P at ӘР =kr{rG),+P), ,} verify as well expressions (15.1.14) and (15.1.18) for (AV)2 and (AE)2. [Note that in the case of a two-dimensional normal distribution, namely 1 xa (ax p(x,y) or exp{-}<<x* + 2xy + cy)} the quantities (x2).(xy), and (y2) can be obtained in a straightforward manner by carrying out a logarithmic differentiation of the formula i exp{-}\+2bey+c} dx dy 11 lar? 2л Viac – 62) -00-00 with respect to the parameters a, b, and c. This leads to the covariance matrix of the distribution, namely ((x2) (xy) 1 С (xx) (12) (ac- 62) b If b=0, then (72) = 1/a, (xy) = 0, (2) = 1/c.)22 a 15.4. A string of length l is stretched, under a constant tension F, between two fixed points A and B. Show that the mean square (fluctuational) displacement y(x) at point P, distant x from A, is given by KT (y(x)}2 = Fix(1 - x).