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I am interested in upper tail bounds (or bounds on deviation from the mean) for t-distribution with n degrees of freedom (http://en.wikipedia.org/wiki/Student's_t-distribution)

A bound that is of the form similar to the tail bound for standard normal distribution Proof of upper-tail inequality for standard normal distribution would be great.

Does anyone know where to find such a result?

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Since $\frac{\mathrm d}{\mathrm dt}(1+t^2/n)^{-(n-1)/2} = -\frac{(n-1)}{2}(1+t^2/n)^{-(n+1)/2}\frac{2t}{n} = -\frac{(n-1)}{n}\cdot t \cdot (1+t^2/n)^{-(n+1)/2}$, the method used by @cardinal in deriving a bound on the upper tail probabilityof the normal distribution seems to work here as well. We have $$\begin{align*} P\{X > x\} &= \int_x^{\infty} \frac{c}{\left(1 + \frac{t^2}{n}\right)^{(n+1)/2}} \mathrm dt\\ &\leq \int_x^{\infty}\frac{t}{x} \frac{c}{\left(1 + \frac{t^2}{n}\right)^{(n+1)/2}} \mathrm dt\\ &= \frac{c}{x}\cdot\frac{n}{n-1} \left[-(1+t^2/n)^{-(n-1)/2} \right|_x^{\infty}\\ &= \frac{n}{n-1}\cdot \frac{1}{x}\cdot \frac{c}{\left(1 + \frac{x^2}{n}\right)^{(n-1)/2}} \end{align*}$$

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