Why $\Bbb E\{ \log_2(1+x)\} \approx \log_2(1+\Bbb E(x))$?

Question

$\Bbb E\{ \log_2(1+x)\} \approx \log_2(1+\Bbb E(x))$ ,where $\Bbb E$ is the notation of expectation. We think when $x$ is small or big enough, so the function $\log_2(1+x)$ seems to be linear, so the expectation and the function $log$ would change the order of calculation? I 'm confused about that. Any comments would appreciated!

Answer 1

As you say, this is just an implication of Taylor's expansion. If $x$ is small

$$\ln\left(1+x\right)\approx x$$

and therefore

$$\mathbb{E}\left[\ln\left(1+x\right)\right]\approx\mathbb{E}\left[x\right]\approx\ln\left(1+\mathbb{E}\left[x\right]\right)$$

Answer 2

$$\begin{align*} \Bbb E\{ \log_2(1+x)\} &= \int p(x)\log_2(1+x)\,dx \text{ where } p(x) \text{ is some probability density function}\\ &= \log_2(\int (1+x)^{p(x)}\,dx)\\ &\approx \log_2(1+\int p(x)\cdot x\,dx) \text{ with the binomial expansion}\\ &=\log_2\big(1+\mathbb{E}(x)\big) \end{align*}$$

Answer 3

Do a Taylor series around $E(x)$, that is $\mathbb{E}[\log(x)]\approx\log({E}[x])-\frac{{V}[x]}{2{E}[x]^2} \>$ according to this paper. Do the same thing for $\log(1+x)$ that is $${E}[\log(1+x)]\approx\log(1+{E}[x])-\frac{\mathbb{V}[x]}{2(1+{E}[x])^2} \>.$$