= Solution
{c}
An <expected value> of a signed <random variable> is finite precisely when its absolute value is <integrable>. With the <standard normal distribution> <probability density function>,
$$
\mathbb E|Ze^{\alpha Z^2}|
=\frac2{\sqrt{2\pi}}\int_0^\infty z\exp\bigl[-(1/2-\alpha)z^2\bigr]\,dz.
$$
For $\alpha<1/2$, this <integral> equals $1/[\sqrt{2\pi}(1/2-\alpha)]<\infty$. For $\alpha\ge1/2$, its <integrand> does not decay sufficiently and the <integral> diverges. Therefore
$$
\boxed{\mathbb E[Ze^{\alpha Z^2}]\text{ exists exactly for }\alpha<\frac12,
\quad\text{and then equals }0.}
$$
The value is zero by the <odd function> and the <symmetric distribution>. For the other $\alpha$, both positive and negative parts have infinite <integrals>; formal cancellation would give only a <Cauchy principal value>, not an <expected value>. This distinction is the <integrability threshold for exponential powers of a Gaussian>.
Back to article page
