\[\begin{align*} E(X) &= \int_{-\infty}^\infty x\cdot \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})}\left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}} dx \\ &= \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})} \int_{-\infty}^\infty x \left(1+\frac{x^2}{\nu} \right)^{-\frac{\nu+1}{2}} dx\\ &= \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})} \left[\int_{-\infty}^0 x \left(1+\frac{x^2}{\nu} \right)^{-\frac{\nu+1}{2}} dx + \int_0^{\infty} x \left(1+\frac{x^2}{\nu} \right)^{-\frac{\nu+1}{2}} dx \right] \\ \text{let: } u &= 1+\frac{x^2}{\nu}, \\ \text{then: } du &= \frac{2xdx}{\nu} \Rightarrow xdx = \frac{\nu du}{2}\\ E(X) &= \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})} \left[\int_{-\infty}^1 \frac{\nu}{2} u^{-\frac{\nu+1}{2}}du + \int_1^{\infty} \frac{\nu}{2} u^{-\frac{\nu+1}{2}}du \right] \\ &= \frac{\nu\Gamma(\frac{\nu + 1}{2})}{2\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})} \left[-\int_1^{\infty} u^{-\frac{\nu+1}{2}}du + \int_1^{\infty}u^{-\frac{\nu+1}{2}}du \right] \\ &= 0 \quad \text{for } \nu > 1 \end{align*}\]
When \(\nu=1\), we have the natural log evaluated at infinity, where it is undefined. \[\begin{align*} &= \frac{\Gamma(1)}{2\sqrt{\pi} \Gamma(\frac{1}{2})} \left[-\int_1^{\infty} \frac{1}{u}du + \int_1^{\infty} \frac{1}{u}du \right] \\ &= \frac{\Gamma(1)}{2\sqrt{\pi} \Gamma(\frac{1}{2})} \left(-\left[\ln(u) \right]_1^\infty + \left[\ln(u) \right]_1^\infty\right) \end{align*}\] \(\therefore\) the expectation of \(X\) does not exist at \(\nu = 1\)
\[\begin{align*} E(X^2) &= \int_{-\infty}^\infty x^2\cdot \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})}\left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}} dx \\ &= \frac{\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})}\int_{-\infty}^\infty x^2\cdot \left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}} dx \\ &= \frac{2\Gamma(\frac{\nu + 1}{2})}{\sqrt{\pi\nu} \Gamma(\frac{\nu}{2})} \left[\int_0^\infty x^2\cdot \left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}} dx\right] \\ &= \frac{2\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu}\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left[\int_0^\infty x^2\cdot \left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu+1}{2}} dx\right]\\ &= \frac{2\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu}\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left[\int_0^\infty \left(x\left(1 + \frac{x^2}{\nu}\right)^{-\frac{1}{2}} \right) \left(\left(1 + \frac{x^2}{\nu}\right)^{-\frac{\nu-4}{2}} \right) \left(x \left(1 + \frac{x^2}{\nu}\right)^{-2} \right) dx\right] \\ \text{let: } u &= \frac{\frac{x^2}{k}}{\left(1+\frac{x^2}{k}\right)} \\ E(X^2) &= \frac{2\Gamma(\frac{\nu+1}{2})}{\sqrt{\nu}\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left[ \int_0^1 (u\nu)^{\frac{1}{2}} (1-u)^{\frac{\nu-4}{2}} \frac{\nu}{2} du\right] \\ &= \frac{2\Gamma(\frac{\nu+1}{2})\frac{\nu\sqrt{\nu}}{2}}{\sqrt{\nu}\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left[ \int_0^1 u^{\frac{1}{2}} (1-u)^{\frac{\nu-4}{2}} du\right] \\ &= \frac{\nu\Gamma(\frac{\nu+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left[ \int_0^1 u^{\frac{3}{2}-1} (1-u)^{\frac{\nu-2}{2}-1} du\right]\\ &= \frac{\nu\Gamma(\frac{\nu+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})}\left(B\left(\frac{3}{2}, \frac{\nu-2}{2}\right)\right)\\ &= \frac{\nu\Gamma(\frac{\nu+1}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \left(\frac{\Gamma(\frac{3}{2})\Gamma(\frac{\nu-2}{2})}{\Gamma(\frac{\nu+1}{2})} \right) \\ &= \frac{\nu \Gamma(\frac{3}{2}) \Gamma(\frac{\nu-2}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \\ &= \frac{\nu \Gamma(\frac{1}{2} + 1) \Gamma(\frac{\nu-2}{2})}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \\ &= \frac{\nu \frac{1}{2} \Gamma(\frac{1}{2}) \frac{\Gamma(\frac{\nu}{2})}{\frac{\nu-2}{2}}}{\Gamma(\frac{1}{2})\Gamma(\frac{\nu}{2})} \\ &= \frac{\nu}{\nu-2} \\ Var(X) &= E(X^2) - [E(X)]^2 \\ &= \frac{\nu}{\nu-2} - 0 \\ &= \frac{\nu}{\nu-2} \quad \text{for } \nu > 2 \end{align*}\]
Step 1: Derivation of \(du\): \[\begin{align*} u &= \frac{\frac{x^2}{\nu}}{\left(1+\frac{x^2}{\nu} \right)} \\ \frac{du}{dx} &= \frac{\left(1+\frac{x^2}{\nu}\right)\left(\frac{2x}{\nu}\right) - \left(\frac{2x}{\nu}\right)\left(\frac{x^2}{\nu}\right)}{\left(1+\frac{x^2}{\nu}\right)^2} \\ &= \frac{\frac{2x}{\nu}\left(1+\frac{x^2}{\nu} - \frac{x^2}{\nu} \right)}{\left(1+\frac{x^2}{\nu}\right)^2}\\ \frac{\nu}{2}du &= \frac{xdx}{\left(1+\frac{x^2}{\nu}\right)^2}\\ \end{align*}\]
Step 2: \(1-u\): \[\begin{align*} 1-u&=1-\frac{\frac{x^2}{\nu}}{1+\frac{x^2}{\nu}} \\ &= \frac{1 + \frac{x^2}{\nu}-\frac{x^2}{\nu}}{1+\frac{x^2}{\nu}} \\ &= \frac{1}{1+\frac{x^2}{\nu}} \end{align*}\]
Step 3: Solving for \(x\left(1+\frac{x^2}{\nu}\right)^{-\frac{1}{2}}\) \[\begin{align*} u &= \frac{\frac{x^2}{\nu}}{1+\frac{x^2}{\nu}} \\ u\nu &= \frac{x^2}{1 +\frac{x^2}{\nu}} \\ \sqrt{u\nu} &= \frac{x}{\left(1 +\frac{x^2}{\nu} \right)^{\frac{1}{2}}} \end{align*}\]
From the above, we observe the following: