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168 CHAPTER 5 JOINT PROBABILITY DISTRIBUTIONS
The requested probability is P(X
1000, X
1000, X
1000, X
1000), which
2
1
4
3
equals the multiple integral of f 1x , x , x , x 2 over the region x 1
1000, x
1000,
X 1 X 2 X 3 X 4 1 2 3 4 2
x
1000, x
1000. The joint probability density function can be written as a product of
4
3
exponential functions, and each integral is the simple integral of an exponential function.
Therefore,
P1X
1000, X
1000, X
1000, X
10002 e 1 2 1.5 3 0.00055
1
4
3
2
Suppose that the joint probability density function of several continuous random vari-
ables is a constant, say c over a region R (and zero elsewhere). In this special case,
f X 1 X 2 p X p 1x , x , p , x 2 dx dx p dx c 1volume of region R2 1
p
p
2
1
2
1
p
by property (2) of Equation 5-22. Therefore, c 1 volume (R). Furthermore, by property (3)
of Equation 5-22.
P31X 1 , X 2 , p , X p 2 B4
p f 1x 1 , x 2 , p , x p 2 dx 1 dx 2 p dx p c volume 1B ¨ R2
X 1 X 2 p X p
B
volume 1B ¨ R2
volume 1R2
When the joint probability density function is constant, the probability that the random vari-
ables assume a value in the region B is just the ratio of the volume of the region B ¨ R to the
volume of the region R for which the probability is positive.
EXAMPLE 5-24 Suppose the joint probability density function of the continuous random variables X and Y is
2
2
2
2
constant over the region x y 4. Determine the probability that X Y 1.
The region that receives positive probability is a circle of radius 2. Therefore, the area of
2
2
this region is 4 . The area of the region x y 1 is . Consequently, the requested prob-
ability is 14 2 1 4.
Definition
If the joint probability density function of continuous random variables X , X , p , X p
2
1
is f 1x , x p , x 2 the marginal probability density function of X i is
2
p
1
X 1 X 2 p X p
f 1x 2 p f 1x , x , p , x 2 dx dx p dx dx p dx (5-23)
X i i X 1 X 2 p X p 1 2 p 1 2 i 1 i 1 p
R x i
where R denotes the set of all points in the range of X , X , p , X p for which
1
2
x i
x .
X i i