10 - BIVARIATE DISTRIBUTIONS After some discussion of the Normal distribution, consideration is given to handling two continuous random variables. The Normal Distribution The probability density function f (x) associated with the general Normal distribution is: f(x) = V2TO2 1 e 2g2 (2-1)2 (10.1) The range of the Normal distribution is -oo to +oo and it will be shown that the total area under the curve is 1. It will also be shown that u is the mean and that 02 is the variance. A graphical representation of the Normal distribution is: X f(x) 1 ? -O x -> µ It is immediately clear from (10.1) that f(x) is symmetrical about x = u. This value of x is marked. When u = 0 the curve is symmetrical about the vertical axis. The value of 2 governs the width of the hump and it turns out that the inflexion points (one each side of the peak) are at x = u to where o = Vo2 is the standard deviation. When integrating expressions which incorporate the probability density function, it is essential to know the following standard result (a derivation is presented on page 10.10): Z +00 e dx = VT (10.2) This result makes it easy to check that the total area under the curve is 1: Total Area = 1 Z +00 00 e 202 dx (x-u)2 Let t = 1202 V2702 so dx = v202 dt. Then: Z V202 dt = - 1 -00 +00 -00 e-to dt VIT 1 .VT = 1 Total Area : 1 et2 V2102 - 10.1 - VTT
From the symmetry of the probability density function (10.1) it is clear that the expectation E(X) = u but to give a further illustration of the use of the standard result (10.2) the expectation is here evaluated by integration: E(X) = 1 Z +00 ? -? Let t = 1202 so x = V202 t + p V2702 -00 x.e 202 (x-p)2 dx and dx = V202 dt. Then: 1 Z to0 00 (V202+ +u).e-t /202 dt E(X) = V2TTO2 ? +00 -12 dt r - Z +00 202 -00 t.e-t dt + =7 0 = ? ? -0 ? µ VTT 202 1 0-12 -00 + VTT r - = ? 2 +00 ? r 202 --- [0- 0] + p = = µ 0 The variance, for once, is conveniently evaluated from the formula„V(2X) = E (X - u)2 : Let t = x - M 1202 0 1 Z +00 (x - p)2.e" 202 dx V(X) = E (X - u)2 V2102 so x - p = 202+ and dx = 202 dt. Then: = .00 V(X) = 1 Z +00 -00 202,2.e-12 /20-2 dt V 2 ]DO-2 Z t d = ? -121 e dt The integrand has been put into a Form ready for integration by parts: 92 V(X) = 0 02 0+00 + - : Z +00 e t dt -00 ? t.e-t2 -00 ? VTI 2 ?