Stein's lemma condition

Question

(Apologies if I break some conventions, this is my first time posting!)

I am working on proving Stein's characterization of the Normal distribution: for Z $\sim N(0,1)$ and some differentiable function $f$ with $E[|f'(Z)|] < \infty$, $$E[Zf(Z)] = E[f'(Z)]$$ Writing the LHS expression in integral form and integrating by parts, I eventually obtain: $$E[Zf(Z)] = \frac{1}{\sqrt{2\pi}} \left[ -f(z) \cdot \exp \left\{ \frac{-z^2}{2} \right\} \right] \Bigg|_{-\infty}^{\infty} + E[f'(Z)]$$ Now I need to show that the first expression on the right hand size is zero. Intuitively, this seems clear because of the exponential term, but I am having trouble explicitly applying the condition on $f'$ to prove this rigorously. Any ideas?

Answer 1

Given $E\left[ \lvert f'(z) \rvert \right] < \infty$, we want to show that $$\frac{1}{\sqrt{2\pi}} \left[ -f(z) \cdot \exp \left\{ \frac{-z^2}{2} \right\} \right] \Bigg|_{-\infty}^{\infty} = 0,$$ or alternatively that $$ \lim_{z \to \infty} \bigg\{ f(-z) \exp\left(-z^2 / 2\right) - f(z) \exp\left(-z^2 / 2\right) \bigg\} = \lim_{z \to \infty} \bigg\{ \left( f(-z) - f(z) \right) \exp\left(-z^2 / 2\right) \bigg\} = 0. $$

Since $f$ is differentiable everywhere, we have that $f(z) - f(-z) = \int_{-z}^z f'(x) dx.$ Then $$ \begin{align} \bigg\lvert \frac{1}{\sqrt{2 \pi}} \left( f(-z) - f(z) \right) \exp\left(-z^2 / 2\right) \bigg\rvert &= \bigg\lvert \frac{1}{\sqrt{2 \pi}} \exp\left(-z^2 / 2\right) \int_{-z}^z f'(x) dx \bigg\rvert \\ &= \bigg\lvert \int_{-z}^z \frac{1}{\sqrt{2 \pi}} \exp\left(-z^2 / 2\right) f'(x) dx \bigg\rvert \\ &= \bigg\lvert \int_{-z}^z f'(x) \, \phi(z) \, dx \bigg\rvert \\ &\le \int_{-z}^z \big\lvert f'(x) \, \phi(z) \big\rvert \, dx \\ &= \int_{-z}^z \big\lvert f'(x) \big\rvert \, \phi(z) \, dx \\ &= \int_{-z}^z \big\lvert f'(x) \big\rvert \, \phi(x) \;\times\; \frac{\phi(z)}{\phi(x)} \, dx \\ &\le \int_{-z}^z \big\lvert f'(x) \big\rvert \, \phi(x) \;\times\; \frac{\phi(z)}{\phi(0)} \, dx \\ &= \frac{\phi(z)}{\phi(0)} \int_{-z}^z \big\lvert f'(x) \big\rvert \, \phi(x) \, dx. \end{align} $$

The first factor, $\phi(z) / \phi(0)$, has limit 0. The second factor has limit $E\Big[\big\lvert f'(x) \big\rvert\Big]$, which is finite. So their product has limit 0, as desired.

Answer 2

Let $Z\sim \mathcal{N}(0,1)$ and $f$ a differentiable function with $E[|f'(Z)|]<\infty$. Then $$ \begin{align} E[Zf(Z)]&=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}zf(z)\exp\left(-\frac{z^2}{2}\right)\,\mathrm dz=\int_{-\infty}^\infty \frac{1}{\sqrt{2\pi}}zf(z)\exp\left(-\frac{z^2}{2}\right)\,\mathrm dz-f(0)E[Z]\\ &=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}z\left[f(z)-f(0)\right]\exp\left(-\frac{z^2}{2}\right)\,\mathrm dz\\ &=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}z\left[\int_0^zf'(u)\,\mathrm du\right]\exp\left(-\frac{z^2}{2}\right)\,\mathrm dz. \end{align} $$ On the other hand $$ \begin{align} E[f'(Z)]=&\int_{-\infty}^\infty f'(z)\frac{1}{\sqrt{2\pi}}\exp\left(-\frac{z^2}{2}\right)\,\mathrm dz\\ =&\frac{1}{\sqrt{2\pi}}\int_{-\infty}^0f'(z)\left[\int_{-\infty}^z-u\exp\left(-\frac{u^2}{2}\right)\mathrm du\right]\,\mathrm dz\\ +&\frac{1}{\sqrt{2\pi}}\int_0^\infty f'(z)\left[\int_z^\infty u\exp\left(-\frac{u^2}{2}\right)\mathrm du\right]\,\mathrm dz. \end{align} $$ So let us treat these two integrals seperately and use Fubini's theorem (justified by the assumption): $$ \int_0^\infty f'(z)\left[\int_z^\infty u\exp\left(-\frac{u^2}{2}\right)\mathrm du\right]\,\mathrm dz=\int_0^\infty\int_z^\infty f'(z)u\exp\left(-\frac{u^2}{2}\right)\mathrm du\,\mathrm dz\\ =\int_0^\infty \int_0^u f'(z)u\exp\left(-\frac{u^2}{2}\right)\mathrm dz\,\mathrm du $$ and similarly $$ \begin{align} &\int_{-\infty}^0f'(z)\left[\int_{-\infty}^z-u\exp\left(-\frac{u^2}{2}\right)\mathrm du\right]\,\mathrm dz=\int_{-\infty}^0\int_{u}^0f'(z)(-u)\exp\left(-\frac{u^2}{2}\right)\mathrm dz\,\mathrm du\\ &=\int_{-\infty}^0\int_{0}^u f'(z)u\exp\left(-\frac{u^2}{2}\right)\mathrm dz\,\mathrm du \end{align} $$ and thus $$ E[f'(Z)]=\int_{-\infty}^\infty\frac{1}{\sqrt{2\pi}}u\left[\int_{0}^u f'(z)\,\mathrm dz\right]\exp\left(-\frac{u^2}{2}\right)\,\mathrm du=E[Zf(Z)] $$

Answer 3

The condition $$\mathbb{E}[f'(x)] < \infty$$ tells you that $f$ is Lipschitz continuous $\mathbb{P}$-a.s. and thus a.s. locally bounded. You might then consider sequences of of functions $$f_n(x)= \left\{\begin{array}{lr} f(x): x \in [-n,n] \\ 0: \text{otherwise}\end{array}\right..$$ This is useful because you know that $f_n$ is a.s. bounded on $[-n,n]$ or $|f_n(x)| < M \cdot (2n)$ for $x \in [-n,n]$ and $M \ge 0$. If you then split $f_n$ into its positive and negative parts, you have that $f_n \le f_{n+1} \uparrow f$, allowing you to apply monotone convergence theorem so that you can pass to the limit. You might apply this to Dougal's $$\bigg\lvert \frac{1}{\sqrt{2 \pi}} \exp\left(-z^2 / 2\right) \int_{-z}^z f'(x) dx \bigg\rvert,$$ allowing you to avoid the subsequent computation.