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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.

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Are there any assumptions stated regarding the sequence of random variables $\{X_n\}$? Are they identically distributed? Independently? –  Alecos Papadopoulos Aug 24 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 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 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 at 12:21

1 Answer 1

up vote 5 down vote accepted

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.

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+1 for posting the answer. –  Alecos Papadopoulos Aug 24 at 21:35

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