Any solution for the following definite integaral? Here $\Phi(x)$ represents the cumulative distributive function of standard normal distribution $$\int_{\frac{-d}{\sqrt2\sigma}}^{\frac{d}{\sqrt2\sigma}} \frac{1}{2\sigma^2\sqrt{2\pi}} \left( \Phi\left(\alpha+\sqrt{d^2-\frac{x^2}{2\sigma^2}}\right)-\Phi\left(\alpha-\sqrt{d^2-\frac{x^2}{2\sigma^2}}\right)\right) \exp\left(\frac{-x^2}{2}\right) \; dx$$

The above function can be represented in terms of error function as $$\frac{1}{4\sigma^2\sqrt{2\pi}}\int_{\frac{-d}{\sqrt2\sigma}}^{\frac{d}{\sqrt2\sigma}}\exp\left(\frac{-x^2}{2}\right) \left(\text{erf}\left(\frac{\alpha+\sqrt{d^2-\frac{x^2}{2\sigma^2}}}{\sqrt{2}}\right)-\text{erf}\left(\frac{\alpha-\sqrt{d^2-\frac{x^2}{2\sigma^2}}}{\sqrt{2}}\right)\right)\; dx$$

any more help???

  • $\begingroup$ Do you mean bounds to be $\pm \sqrt{2} d \sigma$ instead of $\pm \frac{d}{\sqrt{2} \sigma}$ ? $\endgroup$ – Sasha Sep 1 '11 at 4:52
  • $\begingroup$ No, I want bounds to be $\pm \frac{d}{\sqrt{2} \sigma}$ $\endgroup$ – shaikh Sep 1 '11 at 8:37

I will take $\sigma^2 = 1$, and ignore some fine details in what follows, that is, not all $i$'s will be crossed nor will all $t$'s be dotted.

Let $X$ and $Y$ denote independent standard normal random variables. Then, the result that you are trying to calculate looks very much like the probability that the random point $(X, Y)$ lies inside the circle of radius $d$ centered at $(0, \alpha)$, that is, in the disc of radius $d$ centered at $(0, \alpha)$. For, conditioned on $X = x$, where $\vert x \vert < d$, the line $x = d$ crosses the circle at $y = \alpha - \sqrt{d^2 - x^2}$ and at $y = \alpha + \sqrt{d^2 - x^2}$. Thus, $$ \begin{align*} P\{(X, Y) \in \text{disc} ~ \mid X = x\} &= P\{\alpha - \sqrt{d^2 - x^2} < Y < \alpha + \sqrt{d^2 - x^2}\}\\ &= \Phi\left (\alpha + \sqrt{d^2 - x^2}\right) - \Phi\left (\alpha - \sqrt{d^2 - x^2}\right) \end{align*} $$ and it follows that $$ P\{(X, Y) \in \text{disc} \} = \int_{-d}^d \left [ \Phi\left (\alpha + \sqrt{d^2 - x^2}\right) - \Phi\left (\alpha - \sqrt{d^2 - x^2}\right) \right ]\phi(x) \mathrm dx. $$ This looks pretty much like the integral you want to evaluate.

To the best of my knowledge, there is no closed-form expression for this integral except when $\alpha = 0$ when it should work out to $1 - \exp(-d^2)$. For $\alpha \neq 0$, I suggest bounding the desired probability from above by the probability that $(X,Y)$ is in the circumscribing square of side $2d$, viz. $$ \begin{align*} P\{\vert X \vert < d, \alpha - d < Y < \alpha + d\} &= P\{\vert X \vert < d\}P\{\alpha - d < Y < \alpha + d\}\\ &=\left [\Phi(d) - \Phi(-d)\right ] \left [\Phi(\alpha + d) - \Phi(\alpha -d)\right ] \end{align*} $$ and bounding it from below by the probability that $(X, Y)$ is in the inscribed square of side $\sqrt{2}d$. I will leave the details to you.


You said $\Phi$ is the cumulative distribution function and you didn't add "of the standard normal distribution". But that is conventional, and if it's what you meant, then the only hope I immediately see would involve using the fact that $$ \Phi'(x) = \frac{1}{\sqrt{2\pi}} \exp\left(\frac{-x^2}{2}\right), $$ so that if $u=\Phi(x)$ then $du = \varphi(x)\,dx = \Phi'(x)\,dx = \exp\left(\dfrac{-x^2}{2}\right)\, dx$.

  • 1
    $\begingroup$ Yes, $\Phi$ is cumulative distribution of standard normal distribution. Thanks for your help $\endgroup$ – shaikh Sep 2 '11 at 1:08

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