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qwert
qwert
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So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$$p\geq\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p\geq\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

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qwert
qwert
  • 89
  • 8

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[X+\alpha\right]-E\left[Y+1\right]$$E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[X+\alpha\right]-E\left[Y+1\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[\log\left(X+\alpha\right)\right]-E\left[\log\left(Y+1\right)\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

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qwert
qwert
  • 89
  • 8

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[X+\alpha\right]-E\left[Y+1\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p>\frac12$$p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[X+\alpha\right]-E\left[Y+1\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p>\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

So I use hint from Clement with $E\left[\log\frac{X+\alpha}{k - X}\right]$, where $X\sim Bin_{(k-1),p}$ and $\alpha > 0$ is identical to calculating $E\left[X+\alpha\right]-E\left[Y+1\right]$ with $Y\sim Bin_{(k-1),(1-p)}$. To first check the convergence disc for the log-Taylor series I apply the ratio test. \begin{equation} T_{x_0}\left(\log(x+\alpha)\right)=\log\left(x_0+\alpha\right)+\sum_{i=1}^{\infty}\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}(x-x_0)^i \end{equation} My coefficient of the power series is $c_i=\frac{(-1)^{i-1}}{i(x_0+\alpha)^i}$. Thus, the radius around the center $x_0$ is: \begin{equation} r=\lim_{i\rightarrow \infty}\left|\frac{c_i}{c_{i+1}}\right|=\lim_{i\rightarrow \infty}\left|\frac{(i-1)(x_0+\alpha)^{i+1}}{i(x_0+\alpha)^i}\right|=x_0+\alpha \end{equation} Thus, plugging in $x_0 = (k-1)p$ causes no trouble if $p=\frac12$, otherwise, the realization can be outside of the convergent disc centered at $x_0$. As in my setting, if $p\neq 0.5$ I could be either for $X$ or for $Y$ outside of the convergent disc.

So in the case of $p=0.5$, the Taylor series is convergent (sum of two convergent series) and therefore indeed it holds that \begin{equation} \lim_{k\rightarrow \infty}E\left[\log\frac{X+\alpha}{k-X}\right]=0 \end{equation} as the odd terms disappear because of symmetry and the even terms disappear because the big sum in the coeffient of the joint Taylor series above.

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qwert
qwert
  • 89
  • 8