INDIAN

INSTITUTE OF SCIENCE

STOCHASTIC HYDROLOGY
Lecture -2
Course Instructor : Prof. P. P. MUJUMDAR
Department of Civil Engg., IISc.
 Summary of the previous lecture
Concept of a random variable
Discrete and conAnuous random variables
Probability mass funcAon, density funcAon
and cumulaAve distribuAon funcAons

2
 Corrections in the slides of Lecture-1
Slide Title of
No. Original Correction
slide
Continuous P [ x < X x + dx ] P [ x < X x + dx ]
23 f ( x) = lim f ( x) = lim
RVs dx dx dx 0 dx
d d

f1 ( x) + P [ x = d ] + f1 ( x)dx + P [ x = d ] +
Mixed

32
distributions

f ( x) = 1.0
d
2
f ( x)dx = 1.0
2
d

39 P[5 < x < 3] P[5 < X < 3]

3
 CorrecAons in Lecture-1
Slide No.23-Continuous RVs

f(x)
P [x < X < x+dx]

dx x
P [ x < X x + dx ]
f ( x) = lim where f ( x)dx = 1
dx dx
CORRECTED AS
P [ x < X x + dx ]
f ( x) = lim where f ( x)dx = 1
dx 0 dx 4
 Bivariate DistribuAons
In many situations, we would be interested in
simultaneous behavior of two or more random variables.
e.g., in hydrology, we may be interested in the joint
behavior of
Rainfall Runoff
Rainfall Recharge
Rainfall intensity- Peak flood discharge
Temperature Evaporation
Soil permeability GW yield
Flow rates on two streams

5
 Bi-variate DistribuAons
We denote (X,Y) as a two-dimensional random variable
(or a two dimensional random vector).
X and Y both discrete : two dimensional discrete r.v
X and Y both continuous : two dimensional continuous
r.v.
It is possible that one of the rvs of (X, Y), say, X, is
discrete while the other is continuous. In this course,
however, we deal only with cases in which both X & Y
are either discrete or continuous.

6
 Probability distribuAon of (X, Y)
We define the joint probability mass function of a two
dimensional discrete r.v., (X,Y) as,

p(xi, yj) = P[X=xi, Y=yj]

By this we imply, P[X = xi, AND Y = yj ]

p(xi, yj) > 0

p( x , y ) =1
j =1 i =1
i j

7
Probability DistribuAon of (X, Y)
Similar to the CDF for one dimensional random
variables, we define the joint CDF of the two
dimensional discrete r.v., (X,Y) as

F(x, y) = Prob [X<x, Y<y]

= p ( xi , y j )
xi x y j y

F(, ) = P[X< , Y< ] = 1.0

8
 Example : Discrete two-d RV
Probability mass function
x
0 1 2 3 4
y
0 0 0.04 0.05 0.07 0.09
1 0.03 0.04 0.06 0.07 0.08
2 0.02 0.05 0.05 0.07 0.05
3 0.01 0.03 0.05 0.07 0.07
y = 2 x =3
F(3,2) = p ( x, y )
y =0 x =0

P[X<3, Y<2] = 0+0.04+0.05+0.07+0.03+0.04+0.06+0.07+

0.02+.0.05+.0.05+0.07+0.01+0.03+0.05+0.07
= 0.55
9
 Joint pdf of (X, Y)
For a continuous r.v. (X, Y), we define the joint probability
density function, f(x, y), as
1) f(x, y) 0

2) This states that the total
f ( x, y )dxdy = 1 volume under the surface
given by f(x, y) is 1.0

The joint pdf, f(x, y) is not a probability.

For small x, y (+ve), f(x, y) x y is approximately
equal to P[x X<x+ x, yY<y+ y]

10
 Joint cdf of (X, Y)
The joint cumulative distribution function F(x, y) of the
two dimensional random vector (x, y) is defined as
F(x, y) = P[X<x, Y<y]
y x
= f ( x, y ) dxdy

It follows from the definition, that

F(, )=1.0
F(- , y) = F(x, - ) = 0

11
 Example 1

Flows in two adjacent streams are denoted as a random

vector (X, Y) with a joint pdf
f(x, y) = c if 5 < x < 10 ; 4 < y < 9
= 0, elsewhere

X Y
1.Obtain c
2.Obtain P[X > Y]

12
 Example 1 (contd)

1. To determine c ,
f ( x, y )dxdy = 1

9 10

c dxdy = 1
4 5
9 10

c [ x ] dy = 1
4 5
9
5c [ y ]4 = 1
25c = 1 c = 1
25
13
 Example 1 (contd)
y xy
2. P[X > Y] = 1-P[X < Y] 9
9 y
= 1 f ( x, y )dxdy 5
5 5 4
9 y
1 0 x
= 1 dxdy 5 9 10
25 5 5
1 9
= 1 ( y 5)dy
25 5
9
1 y 2
= 1 5 y
25 2 5 14
 Example 1 (contd)

1 92 52
= 1 5 9 + 5 5
25 2 2
= 1 0.32
= 0.68
P[X > Y] = 0.68

15
 Example 2

Consider the joint pdf

f(x, y) = c(x2+y2) 0<x<1

0<y<1

= 0, elsewhere
Obtain
1.the constant c
2.F(x, y)
3.P[X < 1/2, Y < 3/4]
4.P[X > Y]
5.P[X+Y > 1]

16
 Example 2 (contd)

1. To obtain c , f ( x, y )dxdy = 1


1 1
2 2
c ( x + y ) dxdy = 1
0 0

1 3 1
x 2
0 c 3 + xy dy = 1
0
1
1 2
0 c 3 + y dy = 1
3 1
y y 2c
c + = 1 =1 c = 3
2
3 3 0 3
17
 Example 2 (contd)
x y x y
3 2
2. F(x, y) = 2
(
f ( x, y )dydx = x + y 2 dydx )
0 0 0 0
x 3 y
3 2 y
= x y + dx
0
2 3 0
x
3 2 y 3
= x y + dx
0
2 3
3 3 x
3 x y xy x3 y + xy 3
= + =
2 3 2 0 2
x3 y + xy 3 0<x<1
F ( x, y ) =
2 0<y<1
18
 Example 2 (contd)

3 3
1
( ) 3 + 1 3 ( )
3. P[X < 1/2, Y < 3/4] = 2 4 2 4
2
39
=
256
=0.152

19
 Example 2 (contd)
x=y
1
4. P[Y > X]

Limits x 0 to y y
x>y
y 0 to 1
1 y 0 1 x
3 2
P[Y > X] =
2
(
x + y 2 dxdy )
0 0
1 3 21 1
x 3xy 1 3 y 2 3
= + dy = + 2 y dy
0
2 2 y 0
2 2
3 4 1
y y y
= + = 1
2 2 2 2 20
0
 Example problem-2
1
5. P[X+Y > 1]

Limits x 0 to 1
y
y 1-x to 1
1 1
3 2 0
P[X+Y> 1] =
2
x + y 2
(
dydx ) 1 x
0 1 x
1 2 1 3 1
3x y y 3x 3x 2 3
= + dx = + 2 x dx
0
2 2 1 x 0
2 2
2 3 4 1
3x x y 3
= + =
4
4 2 2 0
21
Marginal Probability DistribuAon
We have seen f(x, y) as a joint probability
distribution
In discrete case, p(x, y) = P[X = x, Y = y]
indicates prob [X=x AND Y=y].
Consider the following distribution as in the
previous numerical example

22
Marginal Probability DistribuAon
Marginal distribution of Y
x
y 0 1 2 3 4 Sum
0 0 0.04 0.05 0.07 0.09 0.25
1 0.03 0.04 0.06 0.07 0.08 0.28
2 0.02 0.05 0.05 0.07 0.05 0.24
3 0.01 0.03 0.05 0.07 0.07 0.23
Sum 0.06 0.16 0.21 0.28 0.29 1.00

Marginal distribution of X

e.g., P[X < 3] = 0.06+0.16+0.21+0.28 = 0.71

23
Marginal Probability DistribuAon
An element in the body of the table indicates P[X = xi,
Y = yj].
The marginal totals give P[Y = yj] and P[X = xi] resply.
For example, if we are interested in P[Y = 0], this is given
by marginal sum as 0.25.
Since the event P[Y = 0] can occur with X=0, X=1,..
X=5. we have P[Y=0, X=0 OR Y=0, X=1 OR ..]

P[Y = 0] = P[Y=0, X=0]+P[Y=0, X=1]+ P[Y=0, X=2]+

. P[Y=0, X=5]

This indicates P[Y=0] irrespective of the value of X

24
Marginal Probability DistribuAon
In general, we may write as
p(xi) = P[X=xi]
= P[X=xi, Y=y1 or X=xi, Y=y2or]

= p( xi , y j )
j =1

The function p(xi) for i=1,2,. is called the marginal

distribution of X.
Analogously we define q(yj) = p( xi , y j ) v j, as the
marginal distribution of Y. i =1

25
 Marginal Density FuncAons
In the continuous case, we proceed as follows
Let f(x, y) denote the joint pdf of (X, Y).
We define g(x) and h(y) as the marginal probability
density functions of X & Y respectively as

g ( x) = f ( x, y )dy , h( y ) = f ( x, y )dx

These marginal pdfs which are infact derived from the

joint pdf f(x, y) correspond to the original pdf s of the
one-dimensional r.v.s X and Y.

26
 Marginal Density FuncAons
This may be seen from
P[c < X < d] = P[c < X < d, - < Y < ]
d
= f ( x, y )dy dx
c
d
= g ( x)dx
c

From the definitions of pdf s, it is thus seen that g(x) is

infact the original pdf of the r.v. X

27
 Marginal Density FuncAons

Thus g ( x) = f ( x, y )dy and F ( x) = g ( x)dx

Similarly for the r.v. Y

That is, starting with the joint pdf f(x, y), we are able to get
the pdfs of X & Y respectively.
For discrete case these results may be written as

p [ X = xi ] = p ( xi , yi ) i
j =1

p Y = y j = p ( xi , yi ) j
i =1
28
 Example 3

Consider the joint pdf in the previous example

5 < x < 10
f(x, y) = 1/25 4<y<9

= 0, elsewhere

1. Obtain the marginal density g(x), h(y)

2. Obtain CDF G(x), H(y)
3. P[X > 7]
4. P[5 < Y < 8]

29
 Example 3 (contd)

1. To obtain g(x),

g ( x) = f ( x, y )dy 5 x 10

9
1
= dy
4
25
9
y
= = 1
25 4 5

1
g ( x) = 5 x 10
5
30
 Example 3 (contd)

1. To obtain h(y),

h( y ) = f ( x, y )dx 4 y9

10
1
= dx
5
25
10
x
= = 1
25 5 5

1
h( y ) = 4 y9
5
31
 Example 3 (contd)

2. To obtain G(x),
x
G ( x) = g ( x)dx 5 x 10

x
1
= dx
5
5
x
x
=
5 5
x 5
G ( x) = 5 x 10
5
32
 Example 3 (contd)

2. To obtain H(y),
y

H ( y) = h( y)dy 4 y9

y
1
= dy
4
5
y
y
=
5 4
y4
H ( y) = 4 y9
5
33
 Example 3 (contd)

3. P[X > 7] = 1 P[X < 7]

= 1 G(7)
75 3
= 1 =
5 5

4. P[5 < Y < 8] = H(8) H(5)

4 1 3
= =
5 5 5

34
 Example 4

Consider the joint pdf

x>0
f(x, y) = e-y y>x

1. Obtain the marginal density g(x)

2. P[X > 2]

35
 Example 4 (contd)

1. To obtain g(x),

g ( x) = f ( x, y )dy x>0


= e y dy
x

= e = e x
y
x
x
x x x
G ( x) = e dx = e
0
0

= 1 e x x>0
36
 Example 4 (contd)

2. P[X > 2] = 1 P[X < 2]

(
= 1 1 e2 )
= e2

37
 CondiAonal DistribuAon
A marginal distribution is the distribution of one variable
regardless of the value of the second variable
A joint distribution is the simultaneous occurrence of the
given values of the two variables
The distribution of one variable with conditions placed on
the second variable is called conditional distribution. For
example,
Distribution of X, given that Y=y0 or distribution of Y
given that c < X < d etc.

38
 CondiAonal DistribuAon
Definition: (X, Y) is a continuous two dimensional r.v. with a
joint pdf of f(x, y).
Let g(x) and h(y) be the marginal pdfs of X and Y
respectively
The conditional pdf of X given Y = y is defined as

f ( x, y )
g (x y) = h( y ) > 0
h( y )
Read as x given y

39
 CondiAonal DistribuAon
The conditional pdf of Y given X = x is defined as
f ( x, y )
h ( y x) = g ( x) > 0
g ( x)
The conditional pdf of X given Y R and conditional
pdf of Y given X R is defined as

f ( x, y ) dy
R
f ( x, y ) dx
R
g ( x y R) = , h ( y x R) =
h ( y ) dy
R
g ( x ) dx
R

40
 CondiAonal DistribuAon
The conditional pdfs g(x/y) and h(y/x) satisfy all conditions
for a pdf.
For a given y, g(x/y)>0, as both f(x,y) and h(y) are positive.

f ( x, y )
g (x y) =

h( y )
dx

1
= f ( x, y )dx
h( y )
h( y )
= =1.0
h( y )
Cumulative conditional distributions
x y

G (x y) = g ( x y ) dx, H ( y x) = h ( y x ) dy
41

 Example 5

Consider the joint pdf

f(x, y) =
(
x 1+ 3y2 ) 0<x<2
4 0<y<1

= 0, elsewhere

1. Obtain h(y/x)

2. P[1/2 < Y < 1/X=1]

3. P[Y < 3/4 / X < 1]

42
 Example 5 (contd)

1. To obtain h(y/x), g(x) is first obtained

g ( x) = f ( x, y )dy 0 x2

=
1
(
x 1+ 3y2 ) dy
0
4
1 3 1
= xy + y
4 0

( x + 1)
g ( x) = 0 x2
4
43
 Example 5 (contd)
f ( x, y )
h ( y x) =
g ( x)

=
(
x 1+ 3y2 )
x +1
1
x
2. P[1/2 < Y < 1/X=x] = ( )
1 + 3 y 2 dy
1 x +1
2

x 3 1
= y + y 1
x +1 2

11 x
=
8 x + 1 44
 Example 5 (contd)
11 1 11
2. P[1/2 < Y < 1/X=1] = = 16
8 1 + 1
3. To obtain P[Y < 3/4/X < 1],

f ( x, y ) dx
R
h( y x R) =
g ( x ) dx
R
1

f ( x, y ) dx
0
h ( y x 1) = 1

g ( x ) dx
0 45
 Example 5 (contd)
1 1
(
x 1+ 3y2 )dx
f ( x, y)dx =
0 0
4
2 2 1
1 + 3 y x
=
4 2 0
1+ 3y2
=
8
1 1
( x + 1)
0 g ( x)dx = 0 4 dx
2 1
1 x 3
= + x =
4 2 8
0 46
 Example 5 (contd)
1+ 3 y2
h ( y x 1) =
3
3
1+ 3y2
4
h ( y 3 / 4 x 1) = dy
0
3
1 3
3
= y + y 4
3 0

1 3 27 75
= + =
3 4 64 192

47
Independence of two random variables

Intuitively, X and Y are independent r.v.s if

the distribution of one r.v. does not in any
way influence distribution of the other r.v.
Independence is a very useful assumption
for hydrologic analysis in many situations.
However, physically the assumption must
have a sound basis.

48
 As an example, inflow to a reservoir (X) and the rainfall
in the command area (Y) may be taken as independent,
if the command area is far removed from the reservoir.

Rainfall

Inflow

Reservoir

Command area 49
 In water quality problems, for example, pollutant load (X)
and stream flow (Y) may be treated as independent
variables.

Non-point Source Pollution

50
 Independent R.V.
When two r.v.s are independent, g(x/y)=g(x)
Distribution of x given y is independent of y and hence
the original pdf itself gives the conditional pdf
f ( x, y )
g (x y) =
h( y )
f ( x, y )
g ( x) =
h( y )
f ( x, y ) = g ( x).h( y )

51
 Independent R.V.
The random variables X and Y are stochastically
independent if and only if their joint density is equal to
the product of their marginal densities.
For discrete case, the two r.v.s are independent if and
only if
p(xi, yj) = p(xi) . p(yj) v i,j

52