For $K>0$, the <Cauchy-Schwarz inequality> and part a give
$$
\mathbb E\left[Y_n^-\mathbf1_{\{Y_n^->K\}}\right]
\leq\sqrt{\mathbb E[(Y_n^-)^2]}
\sqrt{\mathbb P(Y_n^->K)}
\leq\frac1K.
$$
Thus $(Y_n^-)$ is <uniform integrability>[uniformly integrable]. Combining this with the <weak convergence of random variables> from part b yields convergence of the first moments:
$$
\mathbb E[Y_n^-]\longrightarrow\mathbb E[Y^-].
$$
By symmetry of the standard normal density,
$$
\mathbb E[Y^-]
=\int_0^\infty x\frac{e^{-x^2/2}}{\sqrt{2\pi}}\,dx
=\frac1{\sqrt{2\pi}}.
$$

Solution

Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 201 / 3 / c / Solution

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2023/iii/paper-201/3/c/solution