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frequency function just given can be reexpressed as pXY (x, y) = pX |Y (x |y)pY (y) (the multiplication law of Chapter 1). This useful equation gives a relationship between the joint and conditional frequency functions. Summing both sides over all values of y, we have an extremely useful application of the law of total probability: pX (x) = ∑ y pX |Y (x |y)pY (y) 88 Chapter 3 Joint Distributions E X A M P L E B Suppose that a particle counter is imperfect and independently detects each incoming particle with probability p. If the distribution of the number of incoming particles in a unit of time is a Poisson distribution with parameter λ, what is the distribution of the number of counted particles? Let N denote the true number of particles and X the counted number. From the statement of the problem, the conditional distribution of X given N = n is binomial, with n trials and probability p of success. By the law of total probability, P(X = k) = ∞∑ n=0 P(N = n)P(X = k|N = n) = ∞∑ n=k λne−λ n! ( n k ) pk(1 − p)n−k = (λp) k k! e−λ ∞∑ n=k λn−k (1 − p)n−k (n − k)! = (λp) k k! e−λ ∞∑ j=0 λ j (1 − p) j j! = (λp) k k! e−λeλ(1−p) = (λp) k k! e−λp We see that the distribution of X is a Poisson distribution with parameter λp. This model arises in other applications as well. For example, N might denote the number of trafﬁc accidents in a given time period, with each accident being fatal or nonfatal; X would then be the number of fatal accidents. ■ 3.5.2 The Continuous Case In analogy with the deﬁnition in the preceding section, if X and Y are jointly contin- uous random variables, the conditional density of Y given X is deﬁned to be fY |X (y|x) = fXY (x, y)fX (x) if 0 < fX (x) < ∞, and 0 otherwise. This deﬁnition is in accord with the result to which a differential argument would lead. We would deﬁne fY |X (y|x) dy as P(y ≤ Y ≤ y + dy|x ≤ X ≤ x + dx) and calculate P(y ≤ Y ≤ y + dy|x ≤ X ≤ x + dx) = fXY (x, y) dx dyfX (x) dx = fXY (x, y) fX (x) dy Note that the rightmost expression is interpreted as a function of y, x being ﬁxed. The numerator is the joint density fXY (x, y), viewed as a function of y for ﬁxed x : you can visualize it as the curve formed by slicing through the joint density function 3.5 Conditional Distributions 89 perpendicular to the x axis. The denominator normalizes that curve to have unit area. The joint density can be expressed in terms of the marginal and conditional densities as follows: fXY (x, y) = fY |X (y|x) fX (x) Integrating both sides over x allows the marginal density of Y to be expressed as fY (y) = ∫ ∞ −∞ fY |X (y|x) fX (x) dx which is the law of total probability for the continuous case. E X A M P L E A In Example D in Section 3.3, we saw that fXY (x, y) = λ2e−λy, 0 ≤ x ≤ y fX (x) = λe−λx , x ≥ 0 fY (y) = λ2 ye−λy, y ≥ 0 Let us ﬁnd the conditional densities. Before doing the formal calculations, it is in- formative to examine the joint density for x and y, respectively, held constant. If x is constant, the joint density decays exponentially in y for y ≥ x ; if y is constant, the joint density is constant for 0 ≤ x ≤ y. (See Figure 3.7.) Now let us ﬁnd the conditional densities according to the preceding deﬁnition. First, fY |X (y|x) = λ 2e−λy λe−λx = λe−λ(y−x), y ≥ x The conditional density of Y given X = x is exponential on the interval [x,∞). Expressing the joint density as fXY (x, y) = fY |X (y|x) fX (x) we see that we could generate X and Y according to fXY in the following way: First, generate X as an exponential random variable ( fX ), and then generate Y as another exponential random variable ( fY |X ) on the interval [x,∞). From this representation, we see that Y may be interpreted as the sum of two independent exponential random variables and that the distribution of this sum is gamma, a fact that we will derive later by a different method. Now, fX |Y (x |y) = λ 2e−λy λ2 ye−λy = 1 y , 0 ≤ x ≤ y The conditional density of X given Y = y is uniform on the interval [0, y]. Finally, expressing the joint density as fXY (x, y) = fX |Y (x |y) fY (y) 90 Chapter 3 Joint Distributions we see that alternatively we could generate X and Y according to the density fXY by ﬁrst generating Y from a gamma density and then generating X uniformly on [0, y]. Another interpretation of this result is that, conditional on the sum of two independent exponential random variables, the ﬁrst is uniformly distributed. ■ E X A M P L E B Stereology In metallography and other applications of quantitative microscopy, aspects of a three- dimensional structure are deduced from studying two-dimensional cross sections. Concepts of probability and statistics play an important role (DeHoff and Rhines 1968). In particular, the following problem arises. Spherical particles are dispersed in a medium (grains in a metal, for example); the density function of the radii of the spheres can be denoted as fR(r). When the medium is sliced, two-dimensional, circular cross sections of the spheres are observed; let the density function of the radii of these circles be denoted by fX (x). How are these density functions related? x y H x r F I GUR E 3.13 A plane slices a sphere of radius r at a distance H from its center, producing a circle of radius x . To derive the relationship, we assume that the cross-sectioning plane is chosen at random, ﬁx R = r , and ﬁnd the conditional density fX |R(x |r). As shown in Figure 3.13, let H denote the distance from the center of the sphere to the planar cross section. By our assumption, H is uniformly distributed on [0, r ], and X = √r 2 − H 2. We can thus ﬁnd the conditional distribution of X given R = r : FX |R(x |r) = P(X ≤ x) = P( √ r 2 − H 2 ≤ x) = P(H ≥ √ r 2 − x2) = 1 − √ r 2 − x2 r , 0 ≤ x ≤ r 3.5 Conditional Distributions 91 Differentiating, we ﬁnd fX |R(x |r) = x r √ r 2 − x2 , 0 ≤ x ≤ r The marginal density of X is, from the law of total probability, fX (x) = ∫ ∞ −∞ fX |R(x |r) fR(r) dr = ∫ ∞ x x r √ r 2 − x2 fR(r) dr [The limits of integration are x and ∞ since for r ≤ x , fX |R(x |r) = 0.] This equation is called Abel’s equation. In practice, the marginal density fX can be approximated by making measurements of the radii of cross-sectional circles. Then the problem becomes that of trying to solve for an approximation to fR , since it is the distribution of spherical radii that is of real interest. ■ E X A M P L E C Bivariate Normal Density The conditional density of Y given X is the ratio of the bivariate normal density to a univariate normal density. After some messy algebra, this ratio simpliﬁes to fY |X (y|x) = 1 σY √ 2π(1 − ρ2) exp ⎛⎜⎜⎜⎝−12 [ y − μY − ρ σY σX (x − μX ) ]2 σ 2Y (1 − ρ2) ⎞⎟⎟⎟⎠ This is a normal density with mean μY + ρ(x −μX )σY /σX and variance σ 2Y (1 − ρ2). The conditional distribution of Y given X is a univariate normal distribution. In Example B in Section 2.2.3, the distribution of the velocity of a turbulent wind ﬂow was shown to be approximately normally distributed. Van Atta and Chen (1968) also measured the joint distribution of the velocity at a point at two different times, t and t + τ . Figure 3.14 shows the measured conditional density of the ve- locity, v2, at time t + τ , given various values of v1. There is a systematic departure from the normal distribution. Therefore, it appears that, even though the velocity is normally distributed, the joint distribution of v1 and v2 is not bivariate normal. This should not be totally unexpected, since the relation of v1 and v2 must con- form to equations of motion and continuity, which may not permit a joint normal distribution. ■ Example C illustrates that even when two random variables are marginally