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How can we relate the notions of the order of some term with his expectation and variance? I was reading a paper which aims to find the order of some random sequence $X_n$ and it says (it doesn't matter what the terms mean)

Let's prove that $X_n = O_p(1) \left(\frac{1}{n} + \frac{C(p,1)}{n \sigma^T \sigma} \right).$ Since $X_n$ is positive we need only to compute the expectation

and then proves that $$E(X_n) = O_p(1) \left(\frac{1}{n} + \frac{C(p,1)}{n \sigma^T \sigma} \right).$$

In other cases, the paper find the expected value and the variance, for example

$$E(X_n-1) = \frac{n(n+2)}{n^2} - 1 \quad \text{and} \quad \text{var}(X_n - 1) = O_p(n^{-1}) \Longrightarrow X_n - 1=O_p(n^{-1/2})$$

My question is, how does it work the benefit of being positive?, why the variance is not needed in the first case? Thanks

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The nonnegativity of $X_n$ means Markov's inequality can be used:

$$P(|X_n| > M) < \frac{E(X_n)}{M}$$

If $X_n$ weren't nonnegative, you could have a situation like:

$$ P(X_n = n) = 0.5 \\ P(X_n = -n) = 0.5 $$

Then

$$ E(X_n) = 0 $$

But it would not be true that

$$ X_n = O_p(1) $$

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  • $\begingroup$ Thanks. So, if I don't have the positivity of the random variables I need the expected value and the variance to use Chebyshev insted of Markov? $\endgroup$ – Ejrionm Jan 29 at 12:26
  • $\begingroup$ Yes Chebyshev's inequality would be helpful for the non nonnegative case. $\endgroup$ – fblundun Jan 31 at 0:43

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