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I want the expectation of the following random variable: $\log\left(\frac{X}{k-X}+\alpha \right)$ with $X \sim Bin_{(k-1),p}$ and $\alpha > 0$, Therefore I derived the Taylor Series: \begin{equation} T_{x_0=(k-1)p}\left[\log\left(\frac{X}{k-X}\right)\right] = \log\left(\frac{x_0}{k-x_0}\right)+ \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(-1)^{n-1}}{x_0^n}+\frac{1}{(k-x_0)^n}\right)\left(x-x_0\right)^n \end{equation} Plugging in $x_0=(k-1)p$ and calculating the expecation leads me to: \begin{equation} E\left[\log\left(\frac{X}{k-X}\right)\right]= \log\frac{(k-1)p}{k-(k-1)p}+\sum_{n=1}^{\infty}\frac{1}{nk^n}\left(\frac{(-1)^{n-1}}{(p-\frac{p}{k})^n}+\frac{1}{(1-p+\frac{p}{k})^n}\right)E\left[\left(x-(k-1)p\right)^n\right] \end{equation}

I know that the Taylor series of $\log(x+1)$ only converges within the open ball $(-1,1)$. Does that apply for the random variable? That is, that for $f(X) = \log\left(\frac{X}{k-X}\right)$ it holds that: \begin{equation} E\left[f(X)\right]=E\left[T_{x_0=(k-1)p}f(X)\right] \Leftrightarrow f(X) \in (-1,1) \end{equation} Clearly, that would depend on the values of $k$ and $p$. However, for me $k$ can take values up to a few hundred, so $f(X) \in (-1,1)$ does not hold $\forall X$. When I wasn't aware that this convergence could go wrong, I showed that in the parameter settings $p=0.5$ and $k>>1$, so that $k \approx (k-1)$, the expectation is $0$, given the Taylor approximation is true. Now, can I still use this result from the Taylor series?

I want the expectation of the following random variable: $\log\left(\frac{X}{k-X}+\alpha \right)$ with $X \sim Bin_{(k-1),p}$ and $\alpha > 0$, Therefore I derived the Taylor Series: \begin{equation} T_{x_0=(k-1)p}\left[\log\left(\frac{X+\alpha}{k-X}\right)\right] = \log\left(\frac{x_0+\alpha}{k-x_0}\right)+ \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(-1)^{n-1}}{(x_0+\alpha)^n}+\frac{1}{(k-x_0)^n}\right)\left(x-x_0\right)^n \end{equation} Plugging in $x_0=(k-1)p$ and calculating the expectation leads me to: \begin{equation} E\left[\log\left(\frac{X+\alpha}{k-X}\right)\right]= \log\left(\frac{(k-1)p+\alpha}{k-(k-1)p}\right)+\sum_{n=1}^{\infty}\frac{1}{nk^n}\left(\frac{(-1)^{n-1}}{\left(p+\frac{\alpha - p}{k}\right)^n}+\frac{1}{\left(1-p+\frac{p}{k}\right)^n}\right)E\left[\left(x-(k-1)p\right)^n\right] \end{equation}

I know that the Taylor series of $\log(x+1)$ only converges within the open ball $(-1,1)$. Does that apply for the random variable? That is, that for $f(X) = \log\left(\frac{X+\alpha}{k-X}\right)$ it holds that: \begin{equation} E\left[f(X)\right]=E\left[T_{x_0=(k-1)p}f(X)\right] \Leftrightarrow f(X) \in (-1,1) \end{equation} Clearly, that would depend on the values of $k$ and $p$. However, for me $k$ can take values up to a few hundred, so $f(X) \in (-1,1)$ does not hold $\forall X$. When I wasn't aware that this convergence could go wrong, I showed that in the parameter settings $p=0.5$ and $k>>1$, so that $k \approx (k-1)$, the expectation is $0$, given the Taylor approximation is true. Now, can I still use this result from the Taylor series?

I want the expectation of the following random variable: $\log\left(\frac{X}{k-X}\right)$ with $X \sim Bin_{(k-1),p}$, Therefore I derived the Taylor Series: \begin{equation} T_{x_0=(k-1)p}\left[\log\left(\frac{X}{k-X}\right)\right] = \log\left(\frac{x_0}{k-x_0}\right)+ \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(-1)^{n-1}}{x_0^n}+\frac{1}{(k-x_0)^n}\right)\left(x-x_0\right)^n \end{equation} Plugging in $x_0=(k-1)p$ and calculating the expecation leads me to: \begin{equation} E\left[\log\left(\frac{X}{k-X}\right)\right]= \log\frac{(k-1)p}{k-(k-1)p}+\sum_{n=1}^{\infty}\frac{1}{nk^n}\left(\frac{(-1)^{n-1}}{(p-\frac{p}{k})^n}+\frac{1}{(1-p+\frac{p}{k})^n}\right)E\left[\left(x-(k-1)p\right)^n\right] \end{equation}

I know that the Taylor series of $\log(x+1)$ only converges within the open ball $(-1,1)$. Does that apply for the random variable? That is, that for $f(X) = \log\left(\frac{X}{k-X}\right)$ it holds that: \begin{equation} E\left[f(X)\right]=E\left[T_{x_0=(k-1)p}f(X)\right] \Leftrightarrow f(X) \in (-1,1) \end{equation} Clearly, that would depend on the values of $k$ and $p$. However, for me $k$ can take values up to a few hundred, so $f(X) \in (-1,1)$ does not hold $\forall X$. When I wasn't aware that this convergence could go wrong, I showed that in the parameter settings $p=0.5$ and $k>>1$, so that $k \approx (k-1)$, the expectation is $0$, given the Taylor approximation is true. Now, can I still use this result from the Taylor series?

I want the expectation of the following random variable: $\log\left(\frac{X}{k-X}+\alpha \right)$ with $X \sim Bin_{(k-1),p}$ and $\alpha > 0$, Therefore I derived the Taylor Series: \begin{equation} T_{x_0=(k-1)p}\left[\log\left(\frac{X}{k-X}\right)\right] = \log\left(\frac{x_0}{k-x_0}\right)+ \sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{(-1)^{n-1}}{x_0^n}+\frac{1}{(k-x_0)^n}\right)\left(x-x_0\right)^n \end{equation} Plugging in $x_0=(k-1)p$ and calculating the expecation leads me to: \begin{equation} E\left[\log\left(\frac{X}{k-X}\right)\right]= \log\frac{(k-1)p}{k-(k-1)p}+\sum_{n=1}^{\infty}\frac{1}{nk^n}\left(\frac{(-1)^{n-1}}{(p-\frac{p}{k})^n}+\frac{1}{(1-p+\frac{p}{k})^n}\right)E\left[\left(x-(k-1)p\right)^n\right] \end{equation}

I know that the Taylor series of $\log(x+1)$ only converges within the open ball $(-1,1)$. Does that apply for the random variable? That is, that for $f(X) = \log\left(\frac{X}{k-X}\right)$ it holds that: \begin{equation} E\left[f(X)\right]=E\left[T_{x_0=(k-1)p}f(X)\right] \Leftrightarrow f(X) \in (-1,1) \end{equation} Clearly, that would depend on the values of $k$ and $p$. However, for me $k$ can take values up to a few hundred, so $f(X) \in (-1,1)$ does not hold $\forall X$. When I wasn't aware that this convergence could go wrong, I showed that in the parameter settings $p=0.5$ and $k>>1$, so that $k \approx (k-1)$, the expectation is $0$, given the Taylor approximation is true. Now, can I still use this result from the Taylor series?