# Proof of asymptotic variance

How do you prove that

$X_n - E[X_n] = O_p(\sqrt{Var(X_n)})$

It's used in my textbook and I don't know where they get it from.

• Are there any assumptions stated regarding the sequence of random variables $\{X_n\}$? Are they identically distributed? Independently? – Alecos Papadopoulos Aug 24 '14 at 11:26
• The joint law of the $X_n$ is not important here. The result is a straightforward application of Chebyshev's inequality (and the definition of big O). – Stéphane Laurent Aug 24 '14 at 11:58
• @StéphaneLaurent My question to the OP intended to find out the framework into which the textbook in question presents the equality in question -not the actual conditions needed (or needed not) for the equality to hold. – Alecos Papadopoulos Aug 24 '14 at 12:12
• @AlecosPapadopoulos My first sentence was not intended to contradict you, but only to set the framework of my second sentence. – Stéphane Laurent Aug 24 '14 at 12:21

For completion, I will provide an answer I found from Theorem 14.4-1 in Bishop et al. Understanding this helped me, and I wish to share it with others on this forum.

It goes as follows:

### Step 1: Definition of $$O_p(1)$$

$$X_n = O_p(1)$$ if for every $$\eta >0$$ there exist a constant $$K(\eta)$$ and an integer $$n(\eta)$$ such that if $$n\geq n(\eta)$$ then $$P(|X_n|\leq K(\eta)) \geq 1-\eta$$.

Also $$X_n = O_p(b_n)$$ if $$X_n / b_n = O_p(1)$$, or equivalently $$X_n = b_nO_p(1)$$. It is sometimes useful to say that $$X_n$$ is bounded in probability.

### Step 2: Tchebychev's inequality

If $$X$$ is a random variable with mean $$\mu$$ and variance $$\sigma^2<\infty$$ then $$P(|X-\mu|\leq h\sigma)\geq 1-h^{-2}$$

### Step 3: the proof

In step 2, set $$h=\eta^{-1/2}$$ for any $$0<\eta<1$$ and apply it to $$X_n$$ we get $$P(\dfrac{|X_n - E(X_n)|}{\sqrt{Var(X_n)}} < \eta^{-1/2}) \geq 1-\eta$$.

This holds for $$n=1,2,3...$$. Setting $$K(\eta) =\eta^{-1/2}$$, we apply the definition in Step 1 and conclude that $$(X_n - E(X_n))/\sqrt{Var(X_n)} = O_p(1)$$.

Hence $$X_n - E(X_n) = O_p(\sqrt{Var(X_n)})$$ as required.