Understanding a proof

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iocal

Junior Member
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Jun 30, 2013
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Hi all,

I am looking for some help to understand a very important proof.

The theorem states that if plim (Xn) = α and the real function g is continuous at α. Then plim (g(Xn)) = g(α)
The proof is as following:

Let ε>0. Then since g is continuous at α, there exists a δ>0 such that if |x-α|< δ, then |g(x)-g(α)|< ε. Thus

|g(x)-g(α)|≥ ε ⇒ |x-α| ≥δ

Substituting Xn for x in the above implication, we obtain:

P[|g(Xn)-g(α)|≥ε] ≤ P[|Xn-α|≥δ]

By hypothesis the last term goes to zero as n goes to infinity which gives us the result

I understand that the first part is from the definition of a continuous function, what I do not understand is how we get inequality sign in the middle after we substitute Xn for x. I am very puzzled and therefore any help is greatly appreciated.
Thanks in advance.


 
Last edited:
iocal said:
Hi all,

I am looking for some help to understand a very important proof.

The theorem states that if plim (Xn) = α and the real function g is continuous at α. Then plim (g(Xn)) = g(α)
The proof is as following:

Let ε>0. Then since g is continuous at α, there exists a δ>0 such that if |x-α|< δ, then |g(x)-g(α)|< ε. Thus

|g(x)-g(α)|≥ ε ⇒ |x-α| ≥δ

Substituting Xn for x in the above implication, we obtain:

P[|g(Xn)-g(α)|≥ε] ≤ P[|Xn-α|≥δ]

By hypothesis the last term goes to zero as n goes to infinity which gives us the result

I understand that the first part is from the definition of a continuous function, what I do not understand is how we get inequality sign in the middle after we substitute Xn for x. I am very puzzled and therefore any help is greatly appreciated.
Thanks in advance.
Click to expand...
It isn't the substitution that made the difference, it is interpreting the \(\displaystyle \Rightarrow\) operation.

If \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is true, then it is also necessarily true that \(\displaystyle |x-\alpha|\geq \delta\).

But, it is also possible for \(\displaystyle |x-\alpha|\geq \delta\) to be true even if \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is false.

Thus the probability of \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is \(\displaystyle \leq\) the probability of \(\displaystyle |x-α|\geq \delta\).
 
DrPhil said:
It isn't the substitution that made the difference, it is interpreting the \(\displaystyle \Rightarrow\) operation.

If \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is true, then it is also necessarily true that \(\displaystyle |x-\alpha|\geq \delta\).

But, it is also possible for \(\displaystyle |x-\alpha|\geq \delta\) to be true even if \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is false.

Thus the probability of \(\displaystyle |g(x)-g(\alpha)|\geq \epsilon\) is \(\displaystyle \leq\) the probability of \(\displaystyle |x-α|\geq \delta\).
Click to expand...

That makes sense, thanks a lot.
 
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