# Expectation and Stieltjes transformation

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 $$\int \ln (x- \epsilon)~ d\\mathbb{P}(x)= \ln \epsilon +\dfrac{1}{c} \ln (1+c\mathbf{m}(\epsilon))+\ln(-(1+\underline{\mathbf{m}}(\epsilon))) +\epsilon\mathbf{m}(\epsilon)\underline{\mathbf{m}}(\epsilon)$$

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

First observe that $\ln(x-\epsilon)$ can be written as follows $$\ln(x-\epsilon)= \ln(\epsilon) -i\pi + \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta} + \dfrac{1}{x-\theta} d\theta$$ Integrating with respect to the probability measure $\\mathbb{P}(x)$, we get $$\int \ln(x-\epsilon)d\\mathbb{P}(x)= \ln(\epsilon) -i\pi + \int \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta} + \dfrac{1}{x-\theta} d\theta d\\mathbb{P}(x)$$ 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 $$=\ln \epsilon -i\pi + \int\limits_{\epsilon}^{\infty} \dfrac{1}{\theta}+\mathbf{m}(\theta) d\theta$$ 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 $$=\ln \epsilon -i\pi - \int\limits_{\epsilon}^{\infty} \mathbf{m}(\theta)\underline{\mathbf{m}}(\theta) d\theta$$ 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. $$\int\ln(x-\epsilon)d\\mathbb{P}(x)= \ln(\epsilon) +\dfrac{1}{c} \ln(1+c\mathbf{m}(\epsilon))+\ln(-(1+\underline{\mathbf{m}}(\epsilon)))+\epsilon \mathbf{m}(\epsilon)\underline{\mathbf{m}}(\epsilon)$$

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