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I read the following result based on a second order Taylor expansion (here: Expected value of a natural logarithm)

$$\mathbb{E}(\log(x)) = \log(\mathbb{E}(x))-\frac{1}{2}\frac{\mathbb{V}\text{ar}(x)}{\mathbb{E}(x)^2}$$

If we assume that $x\sim N(\mu,\sigma^2)$ then $$\mathbb{E}(\log(x)) = \log(\mu) - \frac{\sigma^2}{2\mu}.$$ What I would like to know is what happens if the random variable is negative and we want the expectation of its log, i.e., what is $\mathbb{E}(\log(-x))$.

Is the following correct?

$$\mathbb{E}(\log(x)) = \log(-\mu) + \frac{\sigma^2}{2\mu}.$$ Or should we still be subtracting $\sigma^2/2\mu$?

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    $\begingroup$ Because logarithms of negative numbers are undefined, the Normal example in the question makes no sense. $\endgroup$
    – whuber
    Jul 28, 2022 at 15:18
  • $\begingroup$ I just want to confirm that you understand that the log of a negative number is multi-valued and complex. $\endgroup$ Jul 28, 2022 at 15:18
  • $\begingroup$ @John That would take us off on an irrelevant tangent. In statistical applications, it suffices to remember that logarithms can be meaningfully applied only to positive numbers. $\endgroup$
    – whuber
    Jul 28, 2022 at 15:19
  • $\begingroup$ @whuber Our posts were simultaneous; I was addressing the OP :) $\endgroup$ Jul 28, 2022 at 15:20
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    $\begingroup$ @JohnSmith it's just that, even for the case E[log(x)] with x Normal, since the probability that x < 0 is positive for any mu\inR and sigma2>0, the expectation is never a real number; rather its some infinite collection of complex numbers. Thence whuber's suspicion that this isn't what you're really looking for. $\endgroup$ Jul 28, 2022 at 15:52

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As discussed in the comments, this expectation is ill defined.

But even just mechanically, no, you don't need to add, it's still subtraction, as you can verify by checking that $\frac{\partial^2}{\partial x^2} log(x) = \frac{\partial^2}{\partial x^2} log(-x)$.

We can check the series expansions of both in mathematica:

enter image description here

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