Variance of infinitesimal term in Taylor expansion Assume I want to compute the variance of $\log (1+X)$
In my setting I know that $X$ is "small", so I want to approximate the variance using a first order Taylor approximation 
$$Var(\log (1+X)) = Var(X + o(X)) $$
How can I quantify the error I commit by ignoring the $o(X)$ term?
 A: I don’t have a simple solution but I think I can point you in the right direction. First, the most important step here is to assume |X|<1 with probability 1. Otherwise, the Taylor expansion of the log won’t play nicely with the expectations. The easy part then is re-characterize your random variable as
$$\log(1+X)=-∑_{k=1}^∞ (-1)^k X^k/k=X-∑_{k=2}^∞ (-1)^k X^k/k$$
This allows you to decompose the variance as
$$Var⁡(\log(1+X))=Var⁡(X-∑_{k=2}^∞ (-1)^k X^k/k)=Var⁡(X)+Var⁡(∑_{k=2}^∞ (-1)^k X^k/k)-2Cov⁡(X,∑_{k=2}^∞ (-1)^k X^k/k)=Var⁡(X)+E[(∑_{k=2}^∞ (-1)^k X^k/k)^2]-E^2(∑_{k=2}^∞ (-1)^k X^k/k)-2E(∑_{k=2}^∞ (-1)^k X^(k+1)/k)+2E(X)E(∑_{k=2}^∞ (-1)^k X^k/k)$$
Now, a thorny issue arises when you start to consider the interchange of Big-O’s and Expectations. For example, consider that
$$∑_{k=2}^∞ ((-1)^k x^k)/k=x^2+O(x^3)$$
But contract that with what happens when we take the expectation of this (we end up losing our Big-O expression):
$$E(∑_{k=2}^∞ (-1)^k X^k/k)=∫(∑_{k=2}^∞ (-1)^k x^k/k)f(x)dx=E(X^2)+∑_{k=3}^∞ m_k E(X^k)$$
Here, $\{m_k\}$ is just some sequence of constants that can be arbitrarily large. The fact that this happens makes it impossible to derive a convenient result like $Var⁡(\log⁡(1+X))=Var⁡(X)+a(X)$, where $a(X)$ is some simple function of the random variable. As an alternative – and perhaps the only way around this issue – is to make a more restrictive assumption on $X$.
The easiest approach would be to do something like this: Choose an $ϵ$ and a $δ$ such that $P(|X|≤ϵ) \ge 1-δ$. Then, the worst-case scenario (bound-wise) is that $P(|X|=ϵ)=1-δ$ and $P(|X|=1)=δ$. Now technically, if you assume $|X|<1$, then $P(|X|=1)=δ$ makes no sense. Instead, you’d have to assume $P(|X|=1-λ)=δ$ but this added technical machinery is annoying and won’t end up affecting your bound if you pretend like this isn’t an issue.  If you proceed forward with this setup, you can fairly easily bound each of the terms in the variance expression in terms of $ϵ$ and $δ$.
