Handout Ch 4 實習
微積分複習第二波(1)
f ( x)  h( x) g ( x), 則f ' ( x)  h' ( x) g ( x)  g ' ( x)h( x)
h( x )
h' ( x ) g ( x )  h( x ) g ' ( x )
f ( x) 
, 且g ( x)  0, 則f ' ( x) 
g ( x)
g ( x) 2
chain rule
dy
y  f ( g ( x)),則  f ' ( g ( x))  g ' ( x)
dx

Example
1. f ( x)  (2 x  3)(3x 2 )
f ' ( x)  2  (3x 2 )  6 x  (2 x  3)  18x 2  18x
3. f ( x)  ln(x 2  1)
2x
f ' ( x)  2
x 1
2
2x  3
x 1
2  ( x  1)  1 (2 x  3)
5
f ' ( x) 

( x  1) 2
( x  1) 2
2. f ( x) 
4. f ( x)  ( x 2  3x  2)17
f ' ( x)  17( x 2  3x  2)16 (2 x  3)
微積分複習第二波(2)

變數變換
Let u  f(x), f'(x)dx  du
  f(x)f'(x)dx   udu

Example

x
0 1  x2 dx  ?
令u  1  x 2 , 則du  2 xdx
原式= 

1
3

11
1
du  ln(u )  
1
2u
2
這是什麼鬼
微積分複習第二波(3)

Ch 4積分補充
d tan 1 ( x)
1

dx
1  x2


1
1

dx  tan ( x)
 1  x 2



 (  ( ))  
2
2
4
歸去來析 ( 乾脆去死,台語)
5
Expectation of a Random Variable

Discrete distribution
E( X ) 
 xf ( x)
All x

Continuous distribution

E ( X )   xf ( x)d ( x)

E(X) is called expected value, mean or expectation of X.

E(X) can be regarded as being the center of gravity of that distribution.

 x f ( x)  
E(X) exists if and only if All
x


6
E(X) exists if and only if  x f ( x)dx  . Whenever X is a bounded
random variable, then E(X) must exist.
The Expectation of a Function

Let Y  r (X ) , then Er ( X )  E (Y )   yg ( y )   r ( x) f ( x)
y


x
Let Y  r (X ), then Er ( X )  E (Y )   yg ( y)dy   r ( x) f ( x)dx
Suppose X has p.d.f as follows:


2 x for 0  x  1
f ( x)  
0 otherwise.
4
1
1
, then E (Y )  0 x1 2 (2 x)dx  2 0 x 3 2 dx  .
5

Let Y  X

Let Y  r ( X1 , X 2 ,, X n ), it can be shown that
12
E (Y )    r ( x1 ,, xn ) f ( x1 , xn )dx1  dxn .
Rn
7
Example 1 (4.1.3)

In a class of 50 students, the number of students ni
of each age i is shown in the following table:
Agei
ni

8
18
20
19
22
20
4
21
3
25
1
If a student is to be selected at random from the
class, what is the expected value of his age
Solution

9
E[X]=18*0.4+19*0.44+20*0.08+21*0.06+
25*0.02=18.92
Agei
18
19
20
21
25
ni
20
22
4
3
1
Pi
0.4
0.44
0.08
0.06
0.02
Properties of Expectations

If there exists a constant such that Pr(X  a)  1 , then E ( X )  a .
If there exists a constant b such that Pr(X  b)  1, then E ( X )  b .

If X1 ,, X n are n random variables such that each E ( X i )
exists, then E( X 1    X n )  E( X 1 )    E( X n ) .

For all constants a1,, an and b
E (a1 X1    an X n  b)  a1E ( X1 )    an E( X n )  b.

Usually E ( g (X))  g ( E (X)).
Only linear functions g satisfy E ( g (X))  g ( E (X)).

If X 1 ,, X n are n independent random variable such that
each E ( X i ) exists, then
n
n
E ( X i )   E ( X i )
i 1
10
i 1
Example 2 (4.2.7)

11
Suppose that on each play of a certain game a
gambler is equally likely to win or to lose.
Suppose that when he wins, his fortune is doubled;
and when he loses, his fortune is cut in half. If he
begins playing with a given fortune c, what is the
expected value of his fortune after n independent
plays of the game?
Solution
12
Properties of the Variance


Var(X ) = 0 if and only if there exists a constant c such that Pr(X = c) = 1.
For constant a and b, Var(aX  b)  a 2Var ( X ) .
Proof : E ( X )   , then E (aX  b)  a  b
Var (aX  b)  E (aX  b  a  b) 2  E (aX  a ) 2



 
 a 2 E ( X   ) 2  a 2 Var ( X ).

Var ( X )  E ( X 2 )  E ( X )

2

Proof : Var ( X )  E ( X   ) 2  E ( X 2  2X   2 )
 E ( X 2 )  2E ( X )   2
 E( X 2 )   2
13

Properties of the Variance

If X1 , …, Xn are independent random variables, then
Var( X 1 
 X n )  Var( X 1 ) 
 Var( X n )
Proof: Suppose n  2 first. E ( X 1  X 2 )  1  2
Var(X 1  X 2 )  E ( X 1  X 2  1  2 ) 2 
 E ( X 1  1 ) 2  ( X 2  2 ) 2  2( X 1  1 )( X 2  2 ) 
 Var( X 1 )  Var( X 2 )  2 E ( X 1  1 )( X 2  2 )
Since X 1 and X 2 are independent,
E  ( X 1  1 )( X 2  2 )  E ( X 1  1 ) E ( X 2  2 )  ( 1  1 )( 2  2 )  0
 Var (X 1  X 2 )  Var( X 1 )  Var( X 2 )

14
If X1,…, Xn are independent random variables, then
2
2
Var (a1 X 1    an X n  b)  a1 Var ( X 1 )    an Var ( X n )
Example 3 (4.3.4)


Suppose that X is a random variable for which
E(X)=μ and Var(X)=σ2.
Show that
E[X(X-1)]= ( -1)+
15
2
Solution
E[ X ( X  1)]
 E[ X  X ]
2
 E( X )  
2
 Var ( X )  [ E ( X )]  
2
          (   1)
2
16
2
2
Moment Generating Functions

Consider a given random variable X and for each real number t, we
shall let  (t )  E (etX ). The function  is called the moment
generating function (m.g.f.) of X.

Suppose that the m.g.f. of X exists for all values of t in some open
interval around t = 0. Then,
 d

d
 (0)   E (e tX )  E  e tX    E ( X )
 dt
 t 0
 dt  t  0 

More generally,

(n)
 d n tX  
 dn
tX 
(0)   n E (e )  E  n e    E ( X n etX )t 0  E ( X n )
 dt
 t 0
 t 0 
 dt


Thus,  (0)  E ( X ),  (0)  E ( X 2 ),  (0)  E ( X 3 ), and so on.
17
Properties of Moment Generating Functions

Let X has m.g.f.  1 ; let Y = aX+b has m.g.f.  2 . Then for every value of t
such that  1 (at ) exists,  2 (t )  ebt 1 (at )


tY
t ( aX b )
 ebt E(e atX )  ebt 1 (at )
Proof:  2 (t )  E(e )  E e

Suppose that X1,…, Xn are n independent random variables; and for i =
1,…, n, let  i denote the m.g.f. of Xi. Let Y  X 1    X n , and let the
m.g.f. of Y be denoted by  . Then for every value of t such that  i (t )
exists, we have
n
 (t )   i (t )
i 1

Proof:  (t )  E (e )  E e
tY
18
t ( X1  X n )

n
n
n

tX i 
tX i
 E  e    E (e )   i (t )
 i 1  i 1
i 1
The m.g.f. for the Binomial Distribution

Suppose that a random variable X has a binomial distribution with
parameters n and p. We can represent X as the sum of n independent random
variables X1,…, Xn.
Pr(X i  1)  p and Pr(X i  0)  q  1  p

Determine the m.g.f. of
X  X1   X n
 i (t )  E (etX )  (et )Pr(X i  1)  et (0) Pr(X i  0)  pet  q
i
 (t )  ( pet  q) n
19
Uniqueness of Moment Generating Functions

If the m.g.f. of two random variables X1 and X2 are identical for all
values of t in an open interval around t = 0, then the probability
distributions of X1 and X2 must be identical.

The additive property of the binomial distribution
Suppose X1 and X2 are independent random variables. They have
binomial distributions with parameters n1 and p and n2 and p. Let the
m.g.f. of X1 + X2 be denoted by  .
 (t )  ( pet  q) n  ( pet  q) n  ( pet  q) n  n
1
2
1
2
The distribution of X1 + X2 must be binomial distribution with
parameters n1 + n2 and p.
20
Example 4 (4.4.8)

Suppose that X is a random variable for which the m.g.f. is
as follows:
 (t )  e
t 2 3t

21
for -   t  
Find the mean and the variance of X
Solution
 (t )  (2t  3)e
'
t 2  3t
and (t )  (2t  3) e
''
2
t 2  3t
 2e
t 2  3t
T herefore,   (0)  3;    (0)    11 9  2
'
22
2
''
2
Properties of Variance and Covariance


If X and Y are random variables such that Var ( X )   and Var (Y )   , then
Var ( X  Y )  Var ( X )  Var (Y )  2Cov( X , Y )
Var (aX  bY  c)  a 2Var ( X )  b2Var (Y )  2abCov( X ,Y )
n
n


Var   X i    Var ( X i )  2
 i 1  i 1

i j
Correlation only measures linear relationship.


23
 Cov( X i , X j )
Two random variables can be dependent, but uncorrelated.
Example: Suppose that X can take only three values –1, 0, and 1, and that each of
these three values has the same probability. Let Y=X 2. So X and Y are dependent.
E(XY)=E(X 3)=E(X)=0, so Cov(X,Y) = E(XY) – E(X)E(Y)=0 (uncorrelated).
Example 5 (4.6.11)

24
Suppose that two random variables X and Y
cannot possibly have the following properties:
E(X)=3, E(Y)=2, E(X2)=10. E(Y2)=29, and
E(XY)=0
Solution
25