Probability density function (pdf)
The derivative of the CDF
FX (x) is called the probability
density function f X (x) of continuous random variable X.
f X ( x) 
dFX ( x )
.
dx
dFX ( x)
F ( x  x)  FX ( x)
 lim X
,

x

0
dx
x
from the non decreasing nature of
FX (x),
f X (x) will be a continuous function,
f X ( x)  0
for all x
Properties of PDF
FX ( x )  
x

f x (u ) du.
FX ()  1,



f x ( x )dx  1,
f X (x)
x1 x2
x
P x1  X ( )  x2 
 FX ( x2 )  FX ( x1 )

x2
x1
Area underf X (x)
the probability .
f X ( x ) dx.
( x1 , x2 )
in the interval
represents
f X ( x)  e  x .
1
x
The probability that x is any exact particular value (such as 1.9976) is 0
We can only assign probabilities to possible ranges of x.
f X ( x)  e  x .
1
x
1
2
The probability of x falling within 1 to 2:
2

P(1  x  2)  e  x  e  x
1
2
1
 e 2  e 1  .135  .368  .23
The uniform distribution: all values are equally likely.
f X ( x )  1.
0  x 1
f X (x )
1
x
1
Probability distribution because it integrates to 1 (the area under the curve is 1):
1
1  x
0
1
0
1 0 1
What is the probability that x is between 0 and 0.5?
f X (x )
1
0
0.5
0 .5
1
1
P(0  x  )   1 dx  .5
2
0
x
Consider a continuous random variable X having the following probability
density function.
1 2
 x ; 1  x  2
f X ( x)   3
0 ; elsewhere
1) Verify that the above function is non-negative and integrates to 1
2) Find P(0<X<1)
3) Find the CDF of X
1 2
 x ; 1  x  2
f X ( x)   3
0 ; elsewhere
f X (x)  0 because it is a quadratic function.

1
2

1 2
- f X (x) dx  - 0 dx  -1 3 x dx  2 0 dx
1 3 x  2 
1 2
  x dx   x

x


1
3
9
-1


2
1
 (8  (1))  1
9
1
1
1 2
P (0  X  1)   f X ( x) dx   x dx
3
0
0
1 3 x  1 
 x

x

0
9


1
 (1  (0))
9
1

9
Computation of CDF
1 2
 x ; 1  x  2
f X ( x)   3
0 ; elsewhere
For
x  1
x
x
-
-
FX ( x)   f(t) dt   0 dt  0
For
1  x  2
x
1
x
1
FX ( x)   f X (t) dt   0 dt   t 2 dt
3
-
-
-1
x
1 2
  t dt
3
-1
1 3 t  x  1 3
1 3
 t

(
x

(

1
))

( x  1)

t  1 9
9
9
For
x2
x
1
2
x
1
FX ( x)   f X (t) dt   0 dt   t 2 dt   0 dt
3
-
-
-1
2
2
1 2
-1 3 t dt  1
 0 ; x  1
1

F ( x)  P ( X  x)   ( x 3  1) ;  1  x  2
9
 1 ; x  2
P (0  X  1)  FX (1)  FX (0)
2 1

9 9
1

9

Continuous random variables
1. Exponential:
 e  x , x  0,
f X ( x)  
 0, otherwise.
f X (x)
x
1  e  x , x  0,
FX ( x )  
 0, otherwise.
1  e  x , x  0,
FX ( x )  
 0, otherwise.
18
Used for modeling the waiting
time until arrival of a customer,
memoryless property
If you have not observed
a customer until time a, the
distribution of waiting time
(from time a)
until the next customer is the
same as when you started at
time zero.
19
P( X  x  a | X  a) 
P ({ X  x  a}  { X  a}
P ({ X  a})
P ({ X  x  a})

P ({ X  a})
1  FX ( x  a )

1  FX (a )
e  ( x  a )
  a
e
 e  x
 P( X  x)
20
2. Uniform:
 1

, a  x  b,
f X ( x)   b  a

 0, otherwise.
1
ba
f X (x)

a
b
X ~ U ( a, b)
x
Suppose a random noise voltage X across
an electronic circuit is uniformly distributed
between -4 V and 5 V. What is the
probability that the noise voltage will lie
between 2 V and3.53.5 V ?
1
dx
5

(

4
)
2
P ( 2  X  3.5)  
3 .5
1
dx
9
2
 

1 .5
9
23
3. Normal (Gaussian): X is said to be normal or Gaussian r.v, if
f X ( x) 
1
2 2
e
 ( x   ) 2 / 2 2
This is a bell shaped curve, symmetric around the parameter
 and its

distribution function is given by
x
1

2 2
FX ( x )  
.
e
 ( y   ) 2 / 2 2
dy
f X (x) depends on two parameters  and  2
X  N (  , 2 )
f X (x)

x