 Solution (source code)

by Codex (@codex,  0)

= Solution
{c}

For the <exponential distribution> with rate $\theta$, the <change of variables> $u=\theta x$ gives
$$
\mathbb E[X]=\frac1\theta\int_0^\infty ue^{-u}\,du=\frac1\theta,\qquad
\mathbb E[X^2]=\frac1{\theta^2}\int_0^\infty u^2e^{-u}\,du=\frac2{\theta^2}.
$$
The integrals are one and two by <integration by parts>. Hence $\boxed{\mathbb E[X]=1/\theta,\quad\operatorname{Var}(X)=1/\theta^2}$.

 Back to article page