91Ó°ÊÓ

Let X have the normal distribution whose p.d.f. is given by (5.6.6) that is,

\(f\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\exp \left( { - \frac{1}{2}{x^2}} \right)\)

The p.d.f. of X is,

\(f\left( x \right) = \frac{1}{{\sqrt {2\pi } }}\exp \left( { - \frac{1}{2}{x^2}} \right)\)

The mean of X is,

\(E\left( X \right) = \int_{ - \infty }^\infty {\frac{x}{{\sqrt {2\pi } }}} \exp \left[ { - \frac{{{x^2}}}{2}} \right]dx\)

The integrand is a function f with the property that\(f\left( { - x} \right) = - f\left( x \right)\).

Since the range of integration is symmetric around 0, the integral is 0.

The integral that defines the variance is then,

\(\begin{array}{c}Var\left( X \right) = E\left( {{X^2}} \right)\\ = \int_{ - \infty }^\infty {\frac{{{x^2}}}{{\sqrt {2\pi } }}} \exp \left[ { - \frac{{{x^2}}}{2}} \right]dx\end{array}\)

In this integral, let\(u = x\)and\(dv = \frac{1}{{\sqrt {2\pi } }}\exp \left[ { - \frac{{{x^2}}}{2}} \right]dx\)

It is easy to see that\(du = dx\)and\(v = - \frac{1}{{\sqrt {2\pi } }}\exp \left[ { - \frac{{{x^2}}}{2}} \right]\)

Integration by parts yields

\(Var\left( X \right) = - \frac{x}{{\sqrt {2\pi } }}\exp \left[ { - \frac{{{x^2}}}{2}} \right]_{x = - \infty }^\infty + \int {\frac{1}{{\sqrt {2\pi } }}\exp \left[ { - \frac{{{x^2}}}{2}} \right]dx} \)

This term on the right above o at both\(\infty \,\,and\,\, - \infty \)

The remaining integral is 1 because it is the integral of the standard normal p.d.f.

So, \(Var\left( X \right) = 1\).