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I need to find the expectation of $\ln (x-\epsilon) $ with respect to a probability distribution $\mathbb{P}(x)$. A direct evaluation seems very difficult as the expression for $ \mathbb{P}(x)$ is very complicated. What I have however, is a closed form expression for the Stieltjes transform of $\mathbb{P}(x)$, $\mathbf{m}(\theta)$. I have derived an expression for this expectation. My questions are as follows

  1. Is the solution correct ?
  2. Any arguments that need further reasoning ?
  3. Is there a simpler way to do it ?
  4. Is there a way I can get rid of the $\epsilon$ ? using this technique I can't seem to get shake it off.

Here is the solution:

For any probability measure $ \\mathbb{P}(x) $ and a constant $ 0 < \epsilon < x $ the following identity is true \begin{equation} \int \ln (x- \epsilon)~ d\\mathbb{P}(x)= \ln \epsilon +\dfrac{1}{c} \ln (1+c\mathbf{m}(\epsilon))+\ln(-(1+\underline{\mathbf{m}}(\epsilon))) +\epsilon\mathbf{m}(\epsilon)\underline{\mathbf{m}}(\epsilon) \end{equation}

Define the Stieltjes transform of the probability measure $ \\mathbb{P} (x) $ as follows \begin{equation} \mathbf{m}(\theta)=\int \dfrac{1}{x-\theta} d\\mathbb{P}(x); ~ \theta \in \mathbb{C}^+ \end{equation} For $ 0 < c <1 $ define the function $ \underline{\mathbf{m}}(\theta)= c\mathbf{m}(\theta)-\dfrac{1-c}{\theta} $. Observe that \begin{equation} \mathbf{m}(\theta)=-\dfrac{1}{\theta(1+\mathbf{\underline{m}} (\theta)) } \end{equation} \begin{equation} \mathbf{\underline{m}} (\theta)=-\dfrac{1}{\theta(1+c\mathbf{m(\theta)})} \end{equation}

First observe that $ \ln(x-\epsilon) $ can be written as follows \begin{equation} \ln(x-\epsilon)= \ln(\epsilon) -i\pi + \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta} + \dfrac{1}{x-\theta} d\theta \end{equation} Integrating with respect to the probability measure $ \\mathbb{P}(x) $, we get \begin{equation} \int \ln(x-\epsilon)d\\mathbb{P}(x)= \ln(\epsilon) -i\pi + \int \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta} + \dfrac{1}{x-\theta} d\theta d\\mathbb{P}(x) \end{equation} On the R.H.S, applying Fubini's theorem and using the definition of $ \mathbf{m}(\theta)=\int \dfrac{1}{x-\theta}d\\mathbb{P}(x) $, we get \begin{equation} =\ln \epsilon -i\pi + \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta}+\mathbf{m}(\theta) d\theta \end{equation} Using the relation $ \underline{\mathbf{m}}(\theta)= c\mathbf{m}(\theta)-\dfrac{1-c}{\theta} $ we can write $ \dfrac{1+\theta\underline{\mathbf{m}}(\theta)}{\theta} $ as $ -\mathbf{m}(\theta)\underline{\mathbf{m}}(\theta) $. Hence we get \begin{equation} =\ln \epsilon -i\pi - \int\limits_{\epsilon}^{\infty} \mathbf{m}(\theta)\underline{\mathbf{m}}(\theta) d\theta \end{equation} Consider $ \Lambda(\theta,\mathbf{m}(\theta),\underline{\mathbf{m}}(\theta))=\dfrac{1}{c}\ln(1+c\mathbf{m}(\theta))+\ln(1+\underline{\mathbf{m}}(\theta))+\mathbf{m}(\theta)\underline{\mathbf{m}}(\theta)\theta $. It is not hard to see that the derivative of $ \Lambda $ w.r.t $ \theta $ is $ \mathbf{m}\underline{\mathbf{m}} $. Using this and plugging in the limits $ \epsilon, ~\infty $ and simplifying we get. \begin{equation} \int\ln(x-\epsilon)d\\mathbb{P}(x)= \ln(\epsilon) +\dfrac{1}{c} \ln(1+c\mathbf{m}(\epsilon))+\ln(-(1+\underline{\mathbf{m}}(\epsilon)))+\epsilon \mathbf{m}(\epsilon)\underline{\mathbf{m}}(\epsilon) \end{equation}

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