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## New answers tagged t-distribution

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Indeed, because $X_2+X_3\sim N(0,2)\implies\left(\frac{X_2+X_3}{\sqrt 2}\right)^2\sim \chi^2_1$, independently of $X_1\sim N(0,1)$, $$\frac{\sqrt 2X_1}{|X_2+X_3|}=\frac{X_1}{\sqrt{\left(\frac{X_2+X_3}{\sqrt 2}\right)^2}}\sim t_1$$ Alternatively, since $\frac{X_2+X_3}{\sqrt 2}\sim N(0,1)$, independently of $X_1\sim N(0,1)$, $$\frac{X_1}{\left|\frac{X_2+X_3}{... 1 While this is not as elementary as Stirling's approximation, the pointwise convergence of the density can be shown using dominated convergence theorem. The density of a t-distribution with n degrees of freedom is of the form$$f_n(x)=c_n\cdot\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}\quad,\,x\in\mathbb R$$Let g_n(x)=\left(1+\frac{x^2}{n}\right)^{-(n+1)/2}... 1 A generalization uncovers a fundamental idea. One nice thing about it is how it circumvents calculation altogether: the Gamma functions don't play any role and, in fact, neither do the specific expressions for the Normal and Chi-squared pdfs. Recall that the Student t distribution with \nu degrees of freedom originates (both historically, pedagogically,... 5 Stirling's approximation gives$$\Gamma(z) = \sqrt{\frac{2\pi}{z}}\,{\left(\frac{z}{e}\right)}^z \left(1 + O\left(\tfrac{1}{z}\right)\right)$$so$$\frac{\Gamma(\frac{n+1}{2})}{\Gamma(\frac{n}{2})} = \dfrac{\sqrt{\frac{2\pi}{\frac{n+1}{2}}}\,{\left(\frac{\frac{n+1}{2}}{e}\right)}^{\frac{n+1}{2}}}{\sqrt{\frac{2\pi}{\frac{n}{2}}}\,{\left(\frac{\frac{n}{2}}{...

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