Page 78 - Computational Statistics Handbook with MATLAB

64 Computational Statistics Handbook with MATLAB

which is the product of the individual density functions evaluated at each
or sample point.
x i
ˆ
θ
In most cases, to find the value that maximizes the likelihood function,
we take the derivative of L, set it equal to 0 and solve for θ. Thus, we solve the
following likelihood equation
d L θ() = . 0 (3.24)
d θ
It can be shown that the likelihood function, L θ() , and logarithm of the
likelihood function, ln L θ() , have their maxima at the same value of θ. It is
sometimes easier to find the maximum of ln L θ() , especially when working
with an exponential function. However, keep in mind that a solution to the
above equation does not imply that it is a maximum; it could be a minimum.
It is important to ensure this is the case before using the result as a maximum
likelihood estimator.
When a distribution has more than one parameter, then the likelihood func-
tion is a function of all parameters that pertain to the distribution. In these sit-
uations, the maximum likelihood estimates are obtained by taking the partial
derivatives of the likelihood function (or ln L θ() ), setting them all equal to
zero, and solving the system of equations. The resulting estimators are called
the joint maximum likelihood estimators. We see an example of this below,
where we derive the maximum likelihood estimators for µ and σ 2 for the
normal distribution.

Example 3.3
In this example, we derive the maximum likelihood estimators for the
parameters of the normal distribution. We start off with the likelihood func-
tion for a random sample of size n given by

⁄
n  ( µ)   1  n 2  n 
2
1
x i –
1
L θ() = ∏ -------------- exp  – --------------------  =  ------------ 2 exp –  --------- 2∑ ( x i – µ) 2 . 
2πσ
σ 2π  2σ 2     2σ 
i = 1 i = 1
Since this has the exponential function in it, we will take the logarithm to
obtain
n --- n
 1  2  1 
2
ln [ L θ()] = ln  ------------ 2 + ln exp –  --------- 2∑ ( x i – µ)  .
2πσ
   2σ 
i = 1
This simplifies to

© 2002 by Chapman & Hall/CRC