Find the Expectation of $2\sum\limits_{i=1}^n \frac {(y_i-\alpha)^2-\beta ^2}{(\beta^2+(y_i-\alpha)^2)^2}$ for $y_i$ i.i.d. Cauchy$(\alpha,\beta)$

Question

Find the Expectation of $$2\sum_{i=1}^n \frac {(y_i-\alpha)^2-\beta ^2}{(\beta^2+(y_i-\alpha)^2)^2}$$ given $y_1,y_2...$ iid ~Cauchy$(\alpha,\beta)$ with pdf $(-\infty < y<\infty , \beta>0)$:

$$f(y;\alpha,\beta)=\frac{\beta}{\pi\{\beta^2+(y-\alpha)^2\}}$$

P.S. the answer is $-n/2\beta^2$.

I can find $$\bar y=\sum_{-\infty}^{\infty}\frac{y\beta}{\pi\{\beta^2+(y-\alpha)^2\}}$$

I tried putting the expectation inside the sum, but the $E(y_i)=y_i$ if I understand correctly, and it will simply remain the same.

Could you give me a hint?

Answer 1

\begin{align} E\frac {(Y_1-\alpha)^2-\beta ^2}{(\beta^2+(Y_1-\alpha)^2)^2}&=\int_{-\infty}^{\infty}\frac {(y_1-\alpha)^2-\beta ^2}{(\beta^2+(y_1-\alpha)^2)^2}f(y_1)dy_1\\ &=\int_{-\infty}^{\infty}\frac {(y_1-\alpha)^2-\beta ^2}{(\beta^2+(y_1-\alpha)^2)^2}\frac{\beta/\pi}{\beta^2+(y-\alpha)^2}dy_1\\ &=\frac{1}{\pi\beta^2}\int_{-\infty}^{\infty}\frac {x^2-1}{(x^2+1)^3}dx\\ &=-\frac{1}{\pi\beta^2}\int_{-\infty}^{\infty}\frac {4-4x^2}{4(x^2+1)^3}dx\\ &=-\frac{1}{\pi\beta^2}\int_{-\infty}^{\infty}\frac {x^4+2x^4+1-x^4-6x^2+3}{4(x^2+1)^3}dx\\ &=-\frac{1}{\pi\beta^2}\int_{-\infty}^{\infty}\frac {(x^2+1)^2-x^4-6x^2+3}{4(x^2+1)^3}dx\\ ...&=-\frac{1}{\pi\beta^2}\times \frac{\pi}{4} \end{align}