For a continuous random variable with probability density function , its mean is
provided . Its variance is the expectation of the squared deviation from its mean:
This is finite when ; otherwise it may be if the finite mean exists.
Assume a finite second moment, so the following expectations are finite. Expand the square and use linearity of expectation:
Thus
The same equality holds in the extended sense when the mean is finite and the second moment is infinite.
The normalization of a probability distribution requires
The prescribed mean then gives
As , these equations have the unique solution
The second moment is
Subtracting the squared mean gives the variance
The random variable is nonnegative, so its cumulative distribution function is zero for . If , then
The last step uses the symmetry of the standard normal distribution, . Hence
Differentiate the cumulative distribution function for using the chain rule and the standard normal density:
Thus a probability density function is
The value assigned at zero is immaterial. The probability density function is unbounded near zero, but and continuity of the cumulative distribution function there show that there is no point mass. This is a gamma distribution with shape and rate .
The source prints bare differentials in both integrals; the integration variable is , so we write them as .[ref]
The normalization of a probability distribution for gives
Therefore
Next, the standard normal distribution has and , so the second moment is . Since , linearity of expectation yields . Expressing this expectation using the probability density function gives
Consequently
These evaluate the gamma function at and , respectively, directly from a transformed normal random variable.

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