Convergence in Probability
One of the handiest tools in regression is the asymptotic analysis of estimators as the
number of observations becomes large. This is handy for the following reason. When you have
a nonlinear function of a random variable g(X), when you take an expectation E[g(X)], this is not
the same as g(E[X]). For example, for a mean centered X, E[X2] is the variance and this is not
the same as (E[X])2=(0)2=0. However, as we will soon show, plim[g(Xn)] is equal to
g(plim[Xn]), so asymptotic properties can be passed through nonlinear operators. What is this
“plim” in the above?
Suppose that we have a sequence of random variables X1, X2,...,Xn,…each with
cumulative distribution functions F1(X), F2(X),...,Fn(X),… We say that this sequence converges
in probability to a constant c if lim Pr ob X n  c    0 for any small >0. In terms of the
n 
CDFs, this says that Fn gets very close to a staircase step function at the value c, as seen in the
figure below.
F1
Fn
1.0
1.0
c
X
c
X
That is, the limiting random variable is very unlikely to be either below or above the value of c.
We use the notation plim[Xn]=c to denote the above lim Pr ob X n  c    0 , and we say that
n 
the probability limit of Xn is c.
If g(X) is a continuous function then values of X close to c produce values of g(X) that
are close to g(c), since that is what “continuous” means. If the plim[Xn]=c, then almost certainly
Xn is near c, so almost certainly g(Xn) is near g(c). That is, plim[g(Xn)]=g(plim[Xn]). For
example, if g(X)=X2 and plim[ x n ]=, then the plim[ x n 2]= (plim[ x n ])2=2. It is not true, of
course, that E[ x n 2]=( E[ x n ])2=2. In fact, since var( x n )=E[( x n -)2]=E[ x n 2]-2, we know that
E[ x n 2]=2+var( x n )=2+2/n ≠2.
1
1
Chebychev’s Inequality: Pr ob X    k  2 .
k
Chebychev’s inequality (sometimes written Chebysheff) says that only an insignificant portion
of the probability falls beyond a given number of standard deviations of the mean, regardless of
the probability distribution. For example, less than 25% of the probability can be more than 2
standard deviations of the mean; of course, for a normal distribution, we can be more specific –
less than 5% of the probability is more than 2 standard deviations from the mean.
Proof: 2  


( x  )2 f ( x )dx  
  k

( x  )2 f ( x )dx  

  k
( x  )2 f ( x )dx because we have
left out the middle piece of the sum of positive numbers.
a. Look at the first integral on the right of the inequality. Notice that in this region of integration
x  -k or x--k, but x< so x-<0. Multiple both sides of x--k by -1 to get |x-|k
Thus,
  k

  k 2 2
k  f (x )dx  k22 F(  k) .

(x  )2 f (x )dx  
b. By similar reasoning for the second integral on the right of the integral

 k (x  )
2
f ( x )dx  k 22 (1  F(  k))
Putting a and b together with the first integral, we have
2  k 22 F(  k)  k 22 (1  F(  k))  k 22 Pr ob( X    k) . Dividing both sides by
k22 gives Chebychev’s inequality. Q.E.D.
Note: This implies that Pr ob X     
2
2
.
Law of Large Numbers
Let us combine Chebychev’s inequality and “plim” as follows. Suppose that we have a
sequence of independent and identically distributed (iid) random variables and we calculate the
x  x 2  x n
average of them: x n  1
. We know that the average has a mean  and a variance
n
n
  1

2/n. Apply Chebychev’s inequality to get Pr ob x n    k
, so
  2 . Let k  

n k

Pr ob x n      
2
2


lim
Pr
ob
x





lim
 0 , for
.
Take
the
limit
as
n,
n
n
n   n2
n 2
any >0. Hence,
plim[ x n ]=.
This is the famous Law of Large Numbers that states that the empirical mean of a growing list of
iid random variables will almost certainly equal the population mean of the random variable. If
you flip enough fair coins, then the average number of heads for all those flips will almost
certainly be 0.5.
2
Application of “plim” in regression.
Suppose Y=X+ and the OLS estimator is b=(X’X)-1X’Y=+(X’X/n)-1X’/n. The plim
b=+(X’X/n)-1plim(X’/n). For the kth variable, plim(Xiki/n)=0, since E[Xiki/n]=0 and
2

 X ik  1

var(Xiki/n)=2Xik2/n2. By Chebychev’s inequality Pr ob X ik  i / n  k
. Set

 k2
n


 X ik
 2 X ik 2
. Taking the limit with respect
n
n 2
to n, plim(Xiki/n)=0. That is, the plim b =  and we say that the OLS estimator is consistent.
k
2
  and we have Pr ob X ik  i / n   
3