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Math 111
Continuous Probability
Here is a summary of the most basic parts of the continuous theory of probability in one
dimension.
Definitions.
• A sample space is a set, S, thought of as the set of possible outcomes of an experiment.
• A function p : R → R is a probability density function if
1. p(x) ≥ 0 for all x ∈ R;
R∞
2. −∞ p(x) dx = 1.
R∞
Rs
R0
[By definition, −∞ p(x) dx = lims→∞ 0 p(x) dx + limt→−∞ t p(x) dx, provided both
limits on the right exist.]
• A random variable with probability density function p is a function X : S → R, i.e., a
function with domain equal to the sample space where the probability that X is between
the real numbers s and t is
Z b
P {s ≤ X ≤ t} :=
p(x) dx.
a
• The expectation or the expected value of a random variable X with probability density
function p is
Z ∞
E(X) :=
x p(x) dx.
−∞
Thus, the expectation is the centroid of the probability density function. It is also
called the mean of the random variable.
• Let X be a random variable with probability density function p. Let f : R → R be
a function. The composition f ◦ X, often just written f (X), is called a function of a
random variable. The expected value of f (X) is
Z ∞
E(f (X)) =
f (x)p(x) dx.
−∞
As a special case, the expected value of X, itself, is the expected value of the identity
function f (x) = x.
1
• The variance of a random variable X with probability density function p is
Z ∞
2
(x − µ)2 p(x) dx
Var(X) = E((X − µ) ) =
−∞
where µ is the expected value of X, i.e., µ = E(X) =
R∞
−∞
x p(x) dx.
The standard deviation of X is
p
σ = Var(X) =
sZ
∞
(x − µ)2 p(x) dx.
−∞
While the expected value finds the “center” of the probability distribution, the variance
and standard deviation are measures of the distribution’s “spread”.
Example.
A gasoline station is supplied with gas once each week. If its weekly volume of sales in ten
thousand gallon units is a random variable with probability density function
5(1 − x)4 if 0 < x < 1
p(x) =
0
otherwise.
Thus, the probability the sales are between s and t in 10,000 gallon units is
Rt
s
p(x) dx.
Exercises:
1. Check that p is a probability density function.
2. What is the probability that the sales is exactly 5, 000 gallons? Between 5, 000 and
6, 000?
3. What size storage tank should the station have so that the probability of the supply
being exhausted in a given week is exactly 0.01? At most 0.01?
4. What is the expected volume of sales in ten thousand gallon units?
5. If the station makes $2.00 per gallon, what is the expected volume of sales in dollars?
Solution: Let X denote the random variable which is the volume of sales in 104 gallons.
1. First note that p(x) ≥ 0. Then
Z ∞
Z
p(x) dx =
−∞
1
5(1 − x)4 dx
0
= −(1 − x)5 |10 = 1.
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2. The probability that sales are exactly 5, 000 gallons, i.e., 0.5 × 10, 000 is
Z 0.5
P{0.5 ≤ X ≤ 0.5} =
p(x) dx = 0.
0.5
The probability the sales are between 5, 000 and 6, 000 is
Z 0.6
P{0.5 ≤ X ≤ 0.6} =
p(x) dx
0.5
Z 0.6
5(1 − x)4 dx
=
0.5
0.6
= −(1 − x)5 |0.5
= −(1 − 0.6)4 + (1 − 0.5)4 = 0.0369.
3. Suppose the storage tank has size s × 10, 000 gallons. If s ≥ 1, then the probability
of selling out is 0. The supply will be exhausted if the sales during a week is greater
than s. The probability the supply is exhausted is
Z ∞
P{s < X < ∞} =
p(x) dx
s
Z 1
=
5(1 − x)4 dx
s
= −(1 − x)5 |1s
= (1 − s)5 .
Thus, the probability of exhausting the supply will be exactly 0.01 if s = 1−(0.01)1/5 ≈
0.6, and it will be at most 0.01 if s ≥ 1 − (0.01)1/5 .
4. The expected volume is
∞
Z
E(X) =
xp(x) dx
−∞
Z 1
5x(1 − x)4 dx
=
0
Z
1
x − 4x2 + 6x3 − 4x4 + x5 dx
0
1 4 6 4 1
= 5
− + − +
2 3 4 5 6
5
=
= 0.16666 . . .
30
= 5
5. If x gallons are sold, the volume of sales in dollars is f (x) = 20000x. Therefore, the
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expected volume of sales in dollars is
Z ∞
f (x)p(x) dx
E(f (X)) =
−∞
Z 1
20000x(5(1 − x)4 dx
=
0
Z 1
5x(1 − x)4 dx
= 20000
0
= 20000 E(X) = 20000(5/30) ≈ 3333.333 . . .
(The nice relation between the expected value for f (X) and the expected value for X
here is due to the fact that f is a linear function.)
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