= Solution
{c}
For a <random variable> with a <continuous probability distribution> and <probability density function> $f_X$, the <mean> and <variance> are
$$
\boxed{\mu=\mathbb E[X]=\int_{-\infty}^{\infty}x f_X(x)\,dx,\qquad \operatorname{var}(X)=\mathbb E[(X-\mu)^2]=\int_{-\infty}^{\infty}(x-\mu)^2f_X(x)\,dx.}
$$
A finite <mean> requires $\int |x|f_X(x)\,dx<\infty$. A finite <variance> additionally requires the second <moment> to be finite; without that condition the nonnegative integral for the <variance> may be infinite.
Back to article page
