# BIVARIATE CONTINUOUS RANDOM VARIABLES

1.3.2 Bivariate Density Function

We may loosely say that the bivariate continuous random variable is a variable that takes a continuum of values on the plane according to the rule determined by a joint density function defined over the plane. The rule is that the probability that a bivariate random variable falls into any region on the plane is equal to the volume of the space under the density function over that region. We shall give a more precise definition similar to Definition 3.3.1, which was given for a univariate continuous random variable.

DEFINITION 3.4.1 If there is a nonnegative function f(x, y) defined over the whole plane such that

(3.4.1)

for any xi, x-2, y, У2 satisfying x ^ xg, y ^ y^, then (X, Y) is a bivariate continuous random variable and f(x, y) is called the joint density function.

In order for a function /(x, y) to be a joint density function, it must be nonnegative and the total volume under it must be equal to 1 because of the probability axioms. The second condition may be mathematically expressed as

(3.4.2)

If f(x, y) is continuous except possibly over a finite number of smooth curves on the plane, in addition to satisfying the above two conditions, it will satisfy (3.4.1) and hence qualify as a joint density function. For such a function we may change the order of integration so that we have

(3.4.3)

We shall give a few examples concerning the joint density and the evalu­ation of the joint probability of the form given in (3.4.1).

EXAMPLE 3.4.1 Iff(x, y) = xye (3C+;> x > 0, у > 0, and 0 otherwise, what is P(X > 1, Y < 1)?

By (3.4.1) we have

(3.4.4) P(X > 1, T < 1) = J I xye ^x+y)dxdy

[ xe Xdx • [ ye ydy.

J і Jo

To evaluate each of the single integrals that appear in (3.4.4), we need the following formula for integration by parts:

tb dU [b dV

^ Vdx = U(b)V(b) – Uia)V(a) – U ^ dx,

a J a where U and V are functions of x. Putting U = — e x, V = x, a = 1, and b = °° in (3.4.5), we have

(3.4.6) J xe Xdx = [~xe *]“ + J e Xdx= e 1 + [— e *]“ = 2e 1. Putting U = — e y, V = y, a = 0, and b = 1 in (3.4.5) we have

Therefore from (3.4.4), (3.4.6), and (3.4.7) we obtain (3.4.8) P(X > 1, Y < 1) = 2e~l – 2Г1).

If a function f(x, y) is a joint density function of a bivariate random variable (X, Y), the probability that (X, Y) falls into a more general region (for example, the inside of a circle) can be also evaluated as a double integral of f (x, y). We write this statement mathematically as

(3.4.9)

where S is a subset of the plane. The double integral on the right-hand side of (3.4.9) may be intuitively interpreted as the volume of the space under f(x, y) over S. If /(x, y) is a simple function, this intuitive interpre­tation will enable us to calculate the double integral without actually

performing the mathematical operation of integration, as in the following example.

EXAMPLE 3.4.2 Assuming f(x, у) = 1 for 0 < ас < 1, 0 < 3? < 1 and = 0 otherwise, calculate P(X > Y) and P(X2 + Y2 < 1).

The event (X > Y) means that (X, Y) falls into the shaded triangle in Figure 3.1. Since P(X > Y) is the volume under the density over the triangle, it must equal the area of the triangle times the height of the density, which is 1. Therefore, P(X > Y) = x/%. The event (X2 + Y2 < 1) means that (X, Y) falls into the shaded quarter of the unit circle in Figure 3.2. Therefore, P(X2 + Y2 < 1) = тг/4. Note that the square in each figure indicates the total range of (X, Y).

This geometric approach of evaluating the double integral (3.4.9) may not work if f (x, y) is a complicated function. We shall show the algebraic approach to evaluating the double integral (3.4.9), which will have a much more general usage than the geometric approach. We shall consider only the case where S is a region surrounded by two vertical line segments and two functions of x on the (x, y) plane, as shown in Figure 3.3.

A region surrounded by two horizontal line segments and two functions of у may be similarly treated. Once we know how to evaluate the double integral over such a region, we can treat any general region of practical interest since it can be expressed as a union of such regions.

Let S be as in Figure 3.3. Then we have

[b Г fg(x) j

f(x, y)dxdy = f{x, y)dy dx.

J a |_J Kx) J

We shall show graphically why the right-hand side of (3.4.10) indeed gives the volume under f(x, y) over S. In Figure 3.4 the volume to be evaluated is drawn as a loaf-like figure.

The first slice of the loaf is also described in the figure. We have approximately

Г £(*«)

(3.4.11) Volume of the zth slice = f(xv y)dy • (x{ — Xi-i),

J h(xj

і = 1, 2, . . . , n.

Summing both sides of (3.4.11) over i, we get

(3.4.12) Volume = 2^ f(xi, y)dyixi-xi–l).
i= 1 J AOO

But the limit of the right-hand side of (3.4.12) as n goes to °° is clearly equal to the right-hand side of (3.4.10).

The following examples use (3.4.10) to evaluate the joint probability.

EXAMPLE 3.4.3 We shall calculate the two probabilities asked for in Example 3.4.2 using formula (3.4.10). Note that the shaded regions in

Figures 3.1 and 3.2 are special cases of region 5 depicted in Figure 3.3. To evaluate P(X > Y), we put f(x, y) = 1, a = 0, b = 1, g(x) = x, and h(x) = 0 in (3.4.10) so that we have

( dy I dx = [ xdx = 4 •

oU<> J Jo 2

To evaluate P(X2 + Y2 < 1), we put f(x, y) = 1, a = 0, b = 1, g(x) = VT — x2, and h(x) = 0 so that we have

To evaluate the last integral above, we need the following formula for integration by change of variables:

г/[ф(0]Ф'(<Ж

J ty if ф is a monotonic function such that ф(Ь) = xi and ф^) = x2- Here,

ф'(£) denotes the derivative of ф with respect to t. Next, we shall put x =

2 2

cos 0. Then, since dx/dQ = —sin 0 and sin 0 + cos 0 = 1, we have

 Vl — x2dx = fir/2 2 sim0d0 = —^ sin 0 cos 0 + 4 0 Jo Jo L 2 2 j
 (3.4.16)

example 3.4.4 Suppose f{x, y) = 24 xy for 0 < x < 1, 0 < 3» < 1 — x and = 0 otherwise. Calculate P(Y < У2).

Event (Y < У2) means that (X, T) falls into the shaded region of Figure 3.5. In order to apply (3.4.10) to this problem, we must reverse the role of x and у and put a = 0, b = У2, g(y) = 1 — y, h(y) = 0 in (3.4.10) so that we have