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Use integration by substitution with $x=2u$ and the Gaussian being an even function:The factor two from the differential exactly cancels the half-line factor in the given full-line Gaussian integral.

The numerator event is contained in the conditioning event. The definition of conditional probability therefore givesThe denominator is positive for all $\alpha >0$, and endpoint strictness is irrelevant for a continuous probability distribution.

Integrating the probability density function from its lower endpoint to $x$ gives the cumulative distribution functionIt is continuous and strictly increasing on the support. Hence the unique median $m$ satisfies $F(m)=1/2$, or $m^{-\alpha }=1/2$. Therefore

The second moment is $E[X^2]=\alpha {\int }_1^{\infty }{x}^{1-\alpha }dx$, finite exactly for $\alpha >2$, where it equals $\alpha /(\alpha -2)$. Subtracting the square of the mean givesFor $1<\alpha \le 2$, the mean exists but the variance is infinite. For $0<\alpha \le 1$, the ordinary centered variance is not defined because the mean is not finite. Thus there is no finite variance for any $\alpha \le 2$.

The expected value is the improper integral $\alpha {\int }_1^{\infty }{x}^{-\alpha }dx$. It converges exactly when $\alpha >1$, yieldingAt $\alpha =1$ this improper integral diverges logarithmically, and for $0<\alpha <1$ it diverges as a power of its upper endpoint. In both cases the nonnegative variable has infinite mean. This is the first Pareto moment threshold.

The probability density function is nonnegative for $\alpha >0$. Its integral over the support isso it is normalized for every $\alpha >0$. This is the unit-threshold Pareto distribution.

The complement of an event here is that all three dice avoid six, with probability $(5/6{)}^3$. Therefore

For independent events from fair dice, the probability of a six on each roll is $1/6$. Multiplication gives