Independence

Finally, we shall define the notion of independence between two continu­ous random variables.

DEFINITION 3.4.6 Continuous random variables X and Y are said to be independent if f{x, y) = fix)fiy) for all x and y.

This definition can be shown to be equivalent to stating

Fix і ^ X ^ X2, yi ^ Y — у2) = P(xi < X ^ X2)Piyi — Y ^ yfj

for all X], x%, yi, jo such that x < x%, yi ^ y^- Thus stated, its connection to Definition 3.2.3, which defined independence for a pair of discrete random variables, is more apparent.

Definition 3.4.6 implies that in order to check the independence be­tween a pair of continuous random variables, we should obtain the mar­ginal densities and check whether their product equals the joint density. This may be a time-consuming process. The following theorem, stated without proof, provides a quicker method for determining independence.

THEOREM 3.4.4 Let 5 be a subset of the plane such that/(x, у) > 0 over S and f (x, у) = 0 outside S. Then X and Y are independent if and only if S is a rectangle (allowing — со or to be an end point) with sides parallel

to the axes and/(x, y) = g(x)h(y) over S, where g(x) and h(y) are some functions of x and y, respectively. Note that if g(x) = e/(x) for some c,

Hy) = c_1/(y).

As examples of using Theorem 3.4.4, consider Examples 3.4.1 and 3.4.4. In Example 3.4.1, X and Y are independent because S is a rectangle and xye~(x+y> = xe~x • ye~y over S. In Example 3.4.4, X and Y are not indepen­dent because S is a triangle, as shown in Figure 3.5, even though the joint density 24xy factors out as the product of a function of x alone and that of у alone over S. One can ascertain this fact by noting that/(1/2» %) = 0 since the point (V2, %) is outside of the triangle whereas both f(x = У2) and f(y = %) are clearly nonzero.

The next definition generalizes Definition 3.4.6 in the same way that Definition 3.2.5 generalizes 3.2.3.

DEFINITION 3.4.7 A finite set of continuous random variables X, Y, Z,

. . . are said to be mutually independent if

/(x, y, Z, . . .) = f(x)f(y)f{z) ….

(We have never defined a. multivariate joint density f{x, y,z, . . .), but the reader should be able to generalize Definition 3.4.1 to a multivariate case.)

1.3.3 Examples

We shall give examples involving marginal density, conditional density, and independence.

example 3.4.5 Suppose f(x, y) = (%)(x2 + у2) for 0 < x < 1, 0 < 3? < 1 and = 0 otherwise. Calculate P(0 < X < 0.5 | 0 < Y < 0.5) and P(0 < X < 0.5 I Y = 0.5) and determine if X and Y are independent.

Подпись: (3.4.31)

image077

We have

By a simple double integration it is easy to determine that the numerator is equal to Уі6. To obtain the denominator, we must first obtain the marginal density f(y). By Theorem 3.4.1 we have (3.4.32) f(y) = ||o (x + /)dx= і + I/.

Подпись: (3.4.33) P(0 < Y < 0.5) image079 Подпись: _5_ 16 ‘

Therefore

Therefore P(0 < X < 0.5 | 0 < Y < 0.5) = У5.

To calculate P(0 < X < 0.5 I Y = 0.5), we must first obtain the condi­tional density f(x I y). By Theorem 3.4.3,

Подпись: Я*. У) f(y) Подпись:f <** + />

1 , 3 2

2 + 2^

Putting у = 0.5 in (3.4.34), we have

3 12

(3.4.35) f(x I Y = 0.5) = – + — x2.

Подпись: (3.4.36) Подпись: P(0 < X < 0.5 I Y =
image085 image086
image087

Therefore

That X and Y are not independent is immediately known because /(x, y) cannot be expressed as the product of a function of x alone and a function of у alone. We can ascertain this fact by showing that

(l

3

9

(1

, 3

2

– + –

X

H—

У

2

J

І2

2

(3.4.37) I (x2 + у2) Ф

EXAMPLE 3.4.6 Let fix, y) be the same as in Example 3.4.4. That is, fix, y) = 24x)> for 0 < x < 1 and 0 < у < 1 — x and = 0 otherwise. Calculate P(0 < Г < % | X = У2).

We have

(3.4.38) fix) = 24 [ xydy= 12x(l — x)2, 0 < x < 1.

Jo

Therefore

(3.4.39) fiy I x) = ^X’ ^ = ———9 ’ for 0<х<1, 0 < 31 < 1 — x,

f{x) (1-х)

= 0 otherwise.

Подпись: (3.4.40) Подпись: 0 < Y < - 4 image090

Therefore

Note that in (3.4.40) the range of the first integral is from 0 to %, whereas the range of the second integral is from 0 to У2. This is because f(y | x — У2) = 83 only for 0 < у < У2 and f(y x = У2) =0 for У2 < у, as can be seen either from (3.4.39) or from Figure 3.5. Such an observation is very important in solving this kind of problem, and diagrams such as Figure

3.5 are very useful in this regard.

EXAMPLE 3.4.7 Suppose f(x, у) = У2 over the rectangle determined by the four corner points (1, 0), (0, 1), ( — 1, 0), and (0, —1) and = 0 other­wise. Calculate marginal density fiy)-

We should calculate /(3) separately for 3 > 0 and 3 < 0 because the range of integration with respect to x differs in two cases. We have

(3.4.41) fiy) = [ fix, y)dx = [ У dx — 1 — 3 if0^3^1

J – a. J 31-1 2

and

f°° fl+y 1

(3.4.42) fiy) = fix, y)dx = — dx = 1 + 3 if —1 < 3 < 0.

J – CO J -1-у 2

Note that in (3.4.41), for example, fix, y) is integrated with respect to x from —00 to °° but У2 is integrated from 3 — 1 to 1 — 3, since/(x, 3) = У2

figure 3.7 Marginal density

if x is within the interval (y — 1, 1 — y) and = 0 if л: is outside the interval. Figure 3.7 describes (3.4.41) as the area of the shaded region.

example 3.4.8 Suppose f(x, у) = 1 for 0 ^ x < 1 and 0 < у < 1 and = 0 otherwise. Obtain/(x | X < F).

This example is an application of Definition 3.4.3. Put a = 0, b = 1, h(x) = x, and g(x) = 1 in Figure 3.3. Then from (3.4.22) we have

[ldy

(3.4.43) image091f(x I X < Y) = —— = 2(1 – x) for 0 < x < 1,

Подпись: = 0

Подпись: x
image094

otherwise.

EXAMPLE 3.4.9 Assume f(x, у) = 1 for 0 < x < 1, 0<y<l and = 0 otherwise. Obtain the conditional density f(x Y = 0.5 + X)..

This example is an application of Theorem 3.4.2. The answer is imme­diately obtained by putting yi = y/% с = 1 in (3.4.26) and noting that the range of X given Y = У2 + X is the interval (0, У2)- Thus

Подпись: (3.5.2) F(x) =Подпись: = 2, for 0 < x <

image097

= 0 otherwise.

Leave a reply

You may use these HTML tags and attributes: <a href="" title=""> <abbr title=""> <acronym title=""> <b> <blockquote cite=""> <cite> <code> <del datetime=""> <em> <i> <q cite=""> <s> <strike> <strong>