Mathematical Methods for Physics and Engineering - Matematica.NET

PROBABILITYIf X and Y are both discrete RVs thenp(z) = ∑ i,jf(x i ,y j ), (30.60)where the sum extends over all values of i and j for which Z(x i ,y j )=z. Similarly,if X and Y are both continuous RVs then p(z) is found by requiring that∫∫p(z) dz = f(x, y) dx dy, (30.61)dSwhere dS is the infinitesimal area in the xy-plane lying between the curvesZ(x, y) =z and Z(x, y) =z + dz.◮Suppose X and Y are independent continuous random variables in the range −∞ to ∞,with PDFs g(x) and h(y) respectively. Obtain expressions for the PDFs of Z = X + Y andW = XY .Since X and Y are independent RVs, their joint PDF is simply f(x, y) =g(x)h(y). Thus,from (30.61), the PDF of the sum Z = X + Y is given by∫ ∞∫ z+dz−xp(z) dz = dx g(x) dy h(y)−∞z−x(∫ ∞)= g(x)h(z − x) dx dz.−∞Thus p(z) istheconvolution of the PDFs of g and h (i.e. p = g ∗ h, see subsection 13.1.7).In a similar way, the PDF of the product W = XY is given by∫ ∞∫ (w+dw)/|x|q(w) dw = dx g(x) dy h(y)−∞w/|x|(∫ ∞= g(x)h(w/x) dx )dw ◭−∞|x|The prescription (30.61) is readily generalised to functions of n random variablesZ = Z(X 1 ,X 2 ,...,X n ), in which case the infinitesimal ‘volume’ element dS is theregion in x 1 x 2 ···x n -space between the (hyper)surfaces Z(x 1 ,x 2 ,...,x n )=z andZ(x 1 ,x 2 ,...,x n )=z + dz. In practice, however, the integral is difficult to evaluate,since one is faced with the complicated geometrical problem of determining thelimits of integration. Fortunately, an alternative (and powerful) technique existsfor evaluating integrals of this kind. One eliminates the geometrical problem byintegrating over all values of the variables x i without restriction, while shiftingthe constraint on the variables to the integrand. This is readily achieved bymultiplying the integrand by a function that equals unity in the infinitesimalregion dS and zero elsewhere. From the discussion of the Dirac delta function insubsection 13.1.3, we see that δ(Z(x 1 ,x 2 ,...,x n )−z) dz satisfies these requirements,and so in the most general case we have∫∫ ∫p(z) = ··· f(x 1 ,x 2 ,...,x n )δ(Z(x 1 ,x 2 ,...,x n ) − z) dx 1 dx 2 ...dx n ,(30.62)1154