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CONTINUOUS RANDOM VARIABLES
• These are used to define probability models for continuous
scale measurements, e.g. distance, weight, time
• For a large data set we summarise the distribution using a
relative frequency histogram
the relative frequency of observations between a and b is
proportional to the areas of the rectangles above [a,b].
Relative Frequency Histogram
• As sample size increases :
Approximation by normal distribution
• Histogram of 1000 obs. Normal curve overlaid
Using frequency curves
• Frequency curves are drawn so that the area under the
curve is one. So, the area to the left of any value on the xaxis is merely the proportion of the population which falls
below that value.
• “What proportion of the ____ are less than 39? The
distribution reveals it’s 30%
Frequencies for the normal distribution
Probability Density Function (Freq. Curve)
• For a continuous random variable X, we describe the
probability distribution by some function f(x) e.g.
such that
(i) f(x) >= 0 for all x
(ii) area under the curve between a and b is
which is = P( a < X < b)
(iii) Total area under curve = 1.

b
a
f ( x)dx
Probability density function and c.d.f
• f(x) is called the probability density function (p.d.f.) of X.
• For a continuous random variable the probability of it taking a
particular value exactly, e.g. X = length of a bolt = 1.999965722 cms,
is zero. That is P[X = x] = 0
• Instead for continuous random variables probabilities are associated
with a range of values.e.g. 1.95  X  2.00 cms.
• The cumulative distribution function (c.d.f.) F(x) is defined as the
x
probability upto x, i.e. F(x) = P(X <x)
F ( x) 
 f ( x)dx

Example - Uniform Continuous Distribution
X can take any real value
between a and b with
probability uniform over this
interval.
• Total area = 1 = length x height
• Thus the probability density function is :
1
f ( x) 
if a  x  b
ba
0
o.w.
Generating 10 uniform random variables in S-plus
unifrv10_runif(n=10, min=a, max=b)
Uniform Continuous Distribution
• For any values c and d between a and b :
d
  f ( x)dx
c
C.d.f. F(x) =
Expectation and variance

  E ( X )   xf ( x)dx

For uniform[a, b]
E( X )  
b
a
2 b
1
x
x
dx 
ba
2
a
ba

2

  V ( X )   ( x   ) 2 f ( x)dx
2

For uniform[a, b]
ba 1
(b  a)

V (X )    x 
dx 

a
2  ba
12

b
2
2
Normal distributions
•Normal distributions are one type of continuous p.d.f.
•If X has the Normal distribution with mean µ and variance 2,
this is denoted by X~N(µ,2) (Splus uses s.d. instead of var)
•Z ~ N(µ=0,2=1) is called the standard normal distribution
•Since normal probabilities are hard to compute, tables were
made for the standard normal distribution only
•Most textbooks give areas under the curve of the N(0,1) p.d.f
Calculating standard normal probabilities
Find the probability of getting a value of Z greater than 1.05
P(Z>1.05)= 1 - P(Z<1.05)
look up P(Z<1.05) in tables
P(Z<1.05) = 0.8531
P(Z>1.05)=
Find the prob of Z between -1.05 and 1.05
P(-1.05<Z<1.05) =
P(Z<1.05) - P(Z<-1.05)
= 0.8531 - P(Z>1.05)
=
•In order to obtain probabilities for other Normal
distributions (i.e. areas under the curve), it is necessary to
express any value of X in terms of the number of standard
deviation units it is away from µ. z  x  

 X  x 
P ( X  x )  P


 
 
x 

 P Z 
  P( Z  z )
 

Example of normal distribution
• A filling machine is used to fill soft drink bottles. The
bottles are supposed to contain 300 mls. In fact the
quantities vary according to the Normal distribution with
expected value of µ = 302 ml and standard deviation
s = 3ml. What is the probability that an individual bottle
contains less than 295 mls?
Let the r.v. X denote the quantity in an individual bottle.
We are told X ~ N(302, 32), and we want Pr{X < 295}.
If X = 295 then Z = (295 - 302)/3 = -2.33
so P(X < 295) = P(Z < -2.33) = 1 - P(Z < +2.33)
= 1 - .990 = 0.01
i.e. about 1 bottle in 100 would have less than 295 ml.
1 and 2 sigma bands of Normal distribution
Normal probabilities from R
• e.g. If X ~ N(5,9)
(i) find P(X < 7)
(ii) find k such that P(X < k) = 0.05
• p7_pnorm(q=7, mean=5, sd=3) (0.7475)
• q0.05_qnorm(p=0.05, mean=5, sd=3)
(0.0654)
• Check using tables
Exponential distribution
• T = time to first arrival
• P(T > t) = P(N(t) = 0) = (lt)0 e-lt / 0! = e-lt
• c.d.f of T FT(t) = P(T< t) = 1 - e-lt
• p.d.f. of T fT(t)= d/dt FT(t) = le-lt
E(T) = 1/l
V(T) = 1/l2
Residual time distribution
T
t
P(R>r) =P(T>r+t |T>t)
=P(T>r+t)/P(T>t)
= e-l(r+t) /e-lt = = e-lr
= P (T > r)
R
time up to next arrival
is independent of
when the previous
arrival occurred
•Distr of additional lifetime is same as the original distr.
•Memoryless property of exponential
•Other distributions such as gamma, Weibull etc. are not
memoryless (exponential is the only one).
Quantiles
•
•
•
•
•
•
•
•
•
•
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Cumulative distribution function F(x):
F(x) = P( X < x)
e.g. Z ~ N(0,1) P(Z < 1.96) = ?
need to look at the c.d.f. curve to answer this question
F(1.96) = 0.975  P(Z < 1.96) = 0.975
i.e. 1.96 is the 97.5 th percentile of N(0,1). 1.96 = F-1(0.975)
What is the 50th percentile of N(0,1) ?
What is the 60th percentile of N(0,1) ?
Ans: Look at quantile plot.
Quantile plot: x-axis - cumulative probability (0,1) y-axis: F-1
Quantile plot is inverse of c.d.f. plot.
-4 -4
-4
-2 -2
-2
0 0
0
X X
X
2 2
2
CDF
CDF
of of
ZZ
CDF of Z
F(x)
F(x)F(x)
0.0 0.2 0.4 0.6 0.8 1.0
0.0 0.00.2 0.20.4 0.40.6 0.60.8 0.81.0 1.0
Density
Density
of of
ZZ
Density of Z
4 4
4
X
X X
-2
-1
0
1
2
-2 -2 -1 -1 0 0 1 1 2 2
Quantiles
Quantiles
of of
ZZ
Quantiles of Z
0.00.0
0.0
0.20.2
0.2
0.40.4
0.60.6
0.4
0.6
F(X)
F(X)
F(X)
0.80.8
0.8
1.01.0
1.0
X
X X
-2
0
2
4
6
8
-2 -2 0 0 2 2 4 4 6 6 8 8
f(x)
f(x) f(x)
0.0
0.1
0.2
0.3
0.4
0.0 0.0 0.1 0.1 0.2 0.2 0.3 0.3 0.4 0.4
Normal quantiles
0.00.0
0.0
-4 -4
-4
-2 -2
-2
0 0
0
X X
X
2 2
2
4 4
4
Quantiles
N(3,4)
Quantiles
of of
N(3,4)
Quantiles of N(3,4)
0.20.2
0.2
0.40.4
0.60.6
0.4
0.6
F(X)
F(X)
F(X)
0.80.8
0.8
1.01.0
1.0
4
2
0
-2
quantiles of N(3,4)
6
8
q-q plot
-2
-1
0
quantiles of Z
1
2
10
5
0
chisqsam
15
Q-Q plot of Chisq distribution (vs. normal)
-2
0
Quantiles of Standard Normal
2