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Let $\Phi $ denote the standard normal CDF, and $\phi$ the standard normal PDF. Fix $\alpha > 0$.

Let $$ Z\left( r\right) =r\int_{0}^{\infty } e^{-(r+\alpha )t} \mathbb{E} \left[ \Phi \left( \frac{e^{-\alpha t}(\theta -\omega )}{\sqrt{\frac{1% }{2}(1-e^{-2\alpha t})}}\right) \frac{1-\Phi (\theta )}{\phi(\theta)} \; \right] \; dt $$

The expectation is over the random variables $\theta $ and $\omega$ and these two variables are drawn from the standard normal distribution, independently.

Question: What is the following limit? $$ \lim_{r\rightarrow 0^{+}}Z( r ) $$

Low-quality numerical evidence seems to suggest zero is the answer, but this is only a hunch, and I have not been able to prove it.

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Do you have some simple bounds on that complicated expectation? That doesn't involve $r$. In fact, you are just considering $\int_0^\infty re^{-rt} H(t) dt$ for some complicated function $H$. This should be roughly $\int_0^\infty e^{-rt} H'(t) dt$ up to some constants; so we would expect the limit to be roughly $\int_0^\infty H'(t) dt$, again up to constants; if you can show that $\int_0^\infty |H'(t)| dt < \infty$ then you can use Dominated Convergence. – Zen Harper May 6 '11 at 1:18
up vote 3 down vote accepted

Let $U$, $X$ and $Y$ denote independent random variables with $U$ uniform on $(0,1)$ and $X$ and $Y$ standard normal. For every positive $s$, introduce $$ K(s)=\mathbb{E}\left[U^{s}\Phi\left(W(X-Y)\right)\frac{1-\Phi(X)}{\phi(X)}\right],\qquad W=\frac{U}{\sqrt{\frac12(1-U^2)}}. $$ Some simple computations show that for every positive $r$ and $\alpha$, $$ Z(r)=K(r/\alpha)r/\alpha. $$ Hence, when $r\to0^+$, $Z(r)$ converges to the limit of $sK(s)$ when $s\to0^+$. We now study this limit.

First step: rewriting $K(s)$ Let $Z$ denote a standard normal random variable independent from $Y$, $w$ a positive real number and $x$ a real number. Then $$ \mathbb{E}\left[\Phi\left(w(x-Y)\right)\right]= \mathbb{P}\left[Z\le w(x-Y)\right]= \mathbb{P}\left[Y+\frac{1}{w}Z\le x\right]. $$ Since $\displaystyle Y+\frac{1}{w}Z$ is a centered normal random variable with variance $\displaystyle 1+\frac1{w^2}$, $$ \mathbb{E}\left[\Phi\left(w(x-Y)\right)\right]=\Phi\left(xv\right),\qquad \frac1{v^2}=1+\frac1{w^2}. $$ Integrating this relation over $W$ and $X$, one gets $$ K(s)=\mathbb{E}\left[U^{s}\Phi\left(XV\right)\frac{1-\Phi(X)}{\phi(X)}\right], \qquad V=\sqrt{\frac{2U^2}{1+U^2}}. $$ Let us write $K(s)$ as $K(s)=K_+(s)+K_-(s)$ where $K_+(s)$ corresponds to the expectation on $X\ge0$ and $K_-(s)$ to the expectation on $X\le0$.

Second step: behaviour of $K_+(s)$ Note that, if $X\ge0$, then $U^s\Phi(XV)\le1$, hence $$ K_+(s)\le\mathbb{E}\left[\frac{1-\Phi(X)}{\phi(X)};X\ge0\right]=\int_0^{+\infty}\mathrm{d}x\int_{x}^{+\infty}\phi(y)\mathrm{d}y=\int_0^{+\infty}y\phi(y)\mathrm{d}y=\frac1{\sqrt{2\pi}}. $$

Third step: behaviour of $K_-(s)$ Note that, if $X\le0$, then $1/2\le 1-\Phi(X)\le1$, hence $L(s)/2\le K_-(s)\le L(s)$, with $$ L(s)=\mathbb{E}\left[U^{s}\Phi\left(XV\right)\frac1{\phi(X)};X\le0\right] =\mathbb{E}\left[U^{s}(1-\Phi\left(XV\right))\frac1{\phi(X)};X\ge0\right]. $$ For every positive $v$, $$ \mathbb{E}\left[(1-\Phi\left(Xv\right))\frac1{\phi(X)};X\ge0\right] =\int_0^{+\infty}\mathrm{d}x\int_{xv}^{+\infty}\phi(y)\mathrm{d}y =\int_0^{+\infty}\frac{y}{v}\phi(y)\mathrm{d}y=\frac1{v\sqrt{2\pi}}. $$ Hence, $$ \sqrt{2\pi}L(s)=\mathbb{E}\left[U^{s}\frac1V\right]. $$ Using the upper bound $V\le\sqrt2U$, one gets $$ 2\sqrt{\pi}L(s)\ge\mathbb{E}\left[U^{s-1}\right]=\frac1s. $$

Conclusion All this proves that $$ \liminf Z(r)=\liminf sK_-(s)\ge\frac12\liminf sL(s)\ge\frac1{4\sqrt\pi}, $$ hence $Z(r)$ does not converge to zero. Finally, we mention that it is not much more difficult to show that in fact $$ \lim Z(r)=\lim sK_-(s)=\lim sL(s)=\frac1{2\sqrt\pi}. $$

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Thank you for this beautiful solution! – Ben Golub May 16 '11 at 5:39
This is beautiful! – Chee Mar 13 at 3:18

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