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An introduction to continuous random variables, the concept of probability density functions (pdf), and the normal distribution. It covers the relationship between the pdf and cumulative distribution function (cdf), mean and variance calculations, and the importance of the normal distribution. The document also includes examples and formulas for calculating probabilities using the normal distribution.
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Reminder: a rv is said to be continuous if its cdf is a continuous function.
If the function FX (x) = Pr(X ≤ x) of x is continuous, what is Pr(X = x)?
Pr(X = x) = Pr(X ≤ x) − Pr(X < x) = 0 , by continuity
A continuous random variable does not possess a probability function.
Probability cannot be assigned to individual values of x; instead, probability is assigned to intervals. [Strictly, half-open intervals]
Consider the events {X ≤ a} and {a < X ≤ b}. These events are mutually exclusive, and
{X ≤ a} ∪ {a < X ≤ b} = {X ≤ b}.
So the addition law of probability (axiom A3) gives:
Pr(X ≤ b) = Pr(X ≤ a) + Pr(a < X ≤ b) ,
or Pr(a < X ≤ b) = Pr(X ≤ b) − Pr(X ≤ a)
= FX (b) − FX (a).
So, given the cdf for any continuous random variable X, we can calculate the probability that X lies in any interval (a, b].
Note: The probability Pr(X = a) that a continuous rv X is exactly a is 0. Because of this, we often do not distinguish between open, half-open and closed intervals for continous rvs.
4.2 Probability density function
If X is continuous, then Pr(X = x) = 0.
But what is the probability that ‘X is close to some particular value x?’. Consider Pr(x < X ≤ x + h ), for small h.
Recall: d FX (x) dx
FX (x + h) − FX (x) h
So Pr(x < X ≤ x + h) = FX (x + h) − FX (x) ' h d FX (x) dx
DEFINITION: The derivative (w.r.t. x) of the cdf of a continous rv X is called the probability density function of X.
The probability density function is the limit of Pr(x < X ≤ x + h) h
as h → 0.
The probability density function
Alternative names: pdf, density function, density.
Notation for pdf: fX (x)
Recall: The cdf of X is denoted by FX (x)
Relationship: fX (x) = d FX (x) dx
Care needed: Make sure f and F cannot be confused!
Interpretation
4.3 Mean and Variance
Reminder: for a discrete rv, the formulae for mean and variance are based on the probability function Pr(X = x). We need to adapt these formulae for use with continuous random variables.
DEFINITION: For a continuous rv X with pdf fX (x), the expectation of a function g(x) is defined as
E{g(X)} =
∫ (^) ∞ −∞
g(x) fX (x) dx
Hence, for the mean :
E(X) =
∫ (^) ∞ −∞
x fX (x) dx
Compare this with the equivalent definition for a discrete random variable:
E(X) =
∑ x
x Pr(X = x) , or E(X) =
∑ x
xpX (x).
For the variance, recall the definition.
Var(X) = E[{X − E(X)}^2 ]
Hence Var(X) =
∫ (^) ∞ −∞
(x − μ)^2 fX (x) dx
As in the discrete case, the best way to calculate a variance is by using the result:
Var(X) = E(X^2 ) − {E(X)}^2.
In practice, we therefore usually calculate
E(X^2 ) =
∫ (^) ∞ −∞
x^2 fX (x) dx
as a stepping stone on the way to obtaining Var(X).
Uniform Distribution: cdf
For this distribution the cumulative distribution function (cdf) is
FX (x) =
∫ (^) x −∞
fX (y) dy
0 , x < a , x−a b−a ,^ a^ ≤^ x^ ≤^ b , 1 , x > b. 6
-
a b
FX (x) 1
Uniform Distribution: Mean and Variance
E(X) = μ =
∫ (^) b a
x 1 b − a
dx
= 12 (a + b).
Var(X) = σ^2 = E(X^2 ) − μ^2
∫ (^) b a
x^2
b − a
dx − (a + b)^2 4
= 1 12
(b − a)^2.
For example, if a random variable is uniformly distributed on the range (20,140), then a = 20 and b = 140, so the mean is 80. The variance is 1200 , so the standard deviation is 34.64.
Properties of the exponential distribution
The distribution has pdf
fX (x) =
λe−λx, x ≥ 0 , 0 , x < 0.
and its cdf is given by
FX (x) =
∫ (^) x 0
λe−λy^ dy = 1 − e−λx, x > 0. Mean and Variance
∫ (^) ∞ 0
x λe−λx^ dx =^1 λ
For the variance, we use integration by parts to obtain
E(X^2 ) =
∫ (^) ∞ 0
x^2 λe−λx^ dx =
λ^2
Hence Var(X) = E(X^2 ) − {E(X)}^2
=
λ^2
λ
λ^2
Applications
The exponential distribution is often used to model the lengths of gaps between events occurring haphazardly (that is, quite at random, and with no memory) in time.
There are close links with the Poisson
model the number of such events occurring in a fixed time interval.
Let X be the number of events occurring in an interval of length t: then X has the Poisson distribution with mean λt. Let T be the gap until the first event occurs. Then the events {X = 0} and {T > t} are identical. We note that
Pr(X = 0) = e−λt Pr(T > t) = 1 − FT (t) = 1 − (1 − e−λt) = e−λt.
Scaling of the pdf
The function fX (x) = (^) σ√^12 π e−^
(x−μ)^2 2 σ^2 must
integrate to 1 over (−∞, ∞) if it is to be a valid pdf. The proof that it does so is tricky, and beyond the scope of this course. But it can be shown that ∫ (^) ∞ −∞
e−^
(x−μ)^2 2 σ^2 dx = σ
2 π
as is required.
Cumulative distribution function
If X ∼ N(μ, σ^2 ), the cdf of X is the integral:
FX (x) =
∫ (^) x −∞
σ
2 π
e−^
(x−μ)^2 2 σ^2 dx.
This cannot be evaluated analytically. Numerical integration is necessary: extensive tables are available.
The Standardised Normal Distribution
The Normal distribution with mean 0 and variance 1 is known as the standardised Normal distribution (SND). We usually denote a random variable with this distribution by Z. Hence
Z ∼ N(0, 1).
Special notation φ(z) is used for the pdf of N(0, 1). We write
φ(z) =
2 π
e−
(^12) z 2 , −∞ < z < ∞.
∫ (^) z −∞
φ(x) dx
∫ (^) z −∞
2 π
e−^
1 2 x^2 dx
textbooks and computer programs.
Brief extract from a table of the SND
Tables in textbooks and elsewhere contain
on, up to z = 4.0 or further.
But the range of Z is (−∞, ∞), so we need
values we use the fact that the pdf of N(0, 1) is symmetrical about z = 0. This means that
negative values of z. For example,
