Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Given values of d, p and $\sigma$, is it possible to calculate the value of $\mu$?

$$1-\frac{1}{2\pi\sigma^2}\int_{-\infty}^{\infty}\int_{y-d}^{y+d}\exp\big(-{x^2}/{2\sigma^2}\big) \exp\big(-{(y-\mu)^2}/{2\sigma^2}\big) \,\mathrm{d}x\,\mathrm{d}y < p$$

share|improve this question
    
Presumably there is more than one value of $\mu$ since you have an inequality as opposed to an equation... –  J. M. Jul 17 '11 at 3:44
    
Yes, I am looking for exactly one value of $\mu$ for the above inequality. Since $\mu$ is constant with respect to integration variables, if it is possible to separate $\mu$ from the integral function then the above inequality can be easily solved. –  shaikh Jul 17 '11 at 3:52

3 Answers 3

up vote 2 down vote accepted

It can be easily shown (using the law of total probability)* that $$ \frac{1}{{2\pi \sigma ^2 }}\int_{ - \infty }^\infty {\int_{ y-d }^{y+d} {\exp \bigg( - \frac{{x^2 }}{{2\sigma ^2 }}\bigg)\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg) {\rm d}x} \,{\rm d}y} = \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg), $$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. Noting that the right-hand side is maximized when $\mu = 0$ (indeed, consider the integral of the ${\rm N}(0,1)$ pdf over the fixed length interval $[\frac{{\mu - d}}{{\sqrt {2\sigma ^2 } }},\frac{{\mu + d}}{{\sqrt {2\sigma ^2 } }}]$), it follows that a necessary condition for your inequality to hold is $$ \Phi \bigg(\frac{{d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{-d}}{{\sqrt {2\sigma^ 2} }}\bigg) > 1 - p. $$ On the other hand, if this condition is satisfied, then your inequality holds with $\mu=0$.

To summarize: The inequality holds for some $\mu \in \mathbb{R}$ if and only if it holds for $\mu=0$; the inequality for $\mu = 0$ is equivalent to $$ \Phi \bigg(\frac{{d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{-d}}{{\sqrt {2\sigma^ 2} }}\bigg) > 1 - p. $$

EDIT (in view of your comment below Sasha's answer): Assume that the necessary condition above is satisfied. The function $f$ defined by $$ f(\mu ) = \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg) $$ is decreasing in $\mu \in [0,\infty)$, with $f(\mu) \to 0$ as $\mu \to \infty$. By our assumption, $f(0) > 1-p$. So if you are interested in a $\mu > 0$ such that $f(\mu) \approx 1-p$, you need to find $\mu_1,\mu_2 > 0$ such that $f(\mu_1) > 1- p$ and $f(\mu_2) < 1-p$, and $f(\mu_1) - f(\mu_2) \approx 0$. Then, for any $\mu \in (\mu_1,\mu_2)$, $f(\mu) \approx 1-p$.

* EDIT: Derivation of the first equation above. Denote the left-hand side of that equation by $I$. First write $I$ as $$ I = \int_{ - \infty }^\infty {\bigg[\int_{y - d}^{y + d} {\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{x^2 }}{{2\sigma ^2 }}\bigg){\rm d}x} \bigg]\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg){\rm d}y} . $$ Then $$ I = \int_{ - \infty }^\infty {{\rm P}( - d \le X - y \le d)\frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg){\rm d}y} , $$ where $X$ is a ${\rm N}(0,\sigma^2)$ random variable. If $Y$ is a ${\rm N}(\mu,\sigma^2)$ random variable independent of $X$, then, by the law of total probability, $$ {\rm P}( - d \le X - Y \le d) = \int_{ - \infty }^\infty {{\rm P}( - d \le X - Y \le d|Y = y)f_Y (y)\,{\rm d}y} = I, $$ where $f_Y$ is the pdf of $Y$, given by $$ f_Y (y) = \frac{1}{{\sqrt {2\pi \sigma ^2 } }}\exp \bigg( - \frac{{(y - \mu )^2 }}{{2\sigma ^2 }}\bigg), $$ and where for the last equality ($\int_{ - \infty }^\infty \cdot =I$) we also used the independence of $X$ and $Y$. Now, $X-Y \sim {\rm N}(-\mu,2\sigma^2)$; hence $$ \frac{{(X - Y) - ( - \mu )}}{{\sqrt {2\sigma ^2 } }} \sim {\rm N}(0,1), $$ and, in turn, $$ I = {\rm P}\bigg(\frac{{ - d - ( - \mu )}}{{\sqrt {2\sigma ^2 } }} \le Z \le \frac{{d - ( - \mu )}}{{\sqrt {2\sigma ^2 } }}\bigg) = {\rm P}\bigg(\frac{{\mu - d}}{{\sqrt {2\sigma ^2 } }} \le Z \le \frac{{\mu + d}}{{\sqrt {2\sigma ^2 } }}\bigg), $$ where $Z \sim {\rm N}(0,1)$. Thus, finally, $$ I = \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg). $$

share|improve this answer
    
@Covo: But I am interested in the value of $\mu$ for which the inequality holds –  shaikh Jul 17 '11 at 5:57
    
shaikh: See the edit (beginning from "To summarize:"); it suffices to consider the case $\mu=0$. –  Shai Covo Jul 17 '11 at 6:06
    
@Covo: How to find $\mu_1$ and $\mu_2$? –  shaikh Jul 17 '11 at 7:57
    
shaikh: From a practical point of view, you can use a Normal distribution function calculator, for example davidmlane.com/hyperstat/z_table.html –  Shai Covo Jul 17 '11 at 8:06
1  
shaikh: You can find extremly accurate approximations for the solution $\mu$ of $f(\mu)=1-p$, using certain numerical tools available online. I'll give details later on (but not very soon). –  Shai Covo Jul 17 '11 at 8:32

Your double integral can be evaluated in closed form. This is done by evaluating $x$ integral first, differentiating with respect to $d$, and remembering that for $d=0$ the integral vanishes. Then carrying out $y$ integration, and then integration with respect to $d$ from zero to $d$. Using Mathematica:

In[19]:= 1 - 
 Integrate[
  Integrate[
   D[1/(2 Pi si^2)
       Integrate[
       Exp[-x^2/(2 si^2)] Exp[-(y - mu)^2/(2 si^2)], {x, y - dd, 
        y + dd}], dd] // FullSimplify, {y, -Infinity, Infinity}, 
   Assumptions -> si > 0], {dd, 0, d}]

Out[19]= 1 + 1/2 (-Erf[(d - mu)/(2 si)] - Erf[(d + mu)/(2 si)])

Hence your problem becomes $$1-\frac{1}{2} \left( \text{erf}\left( \frac{\mu+d}{2\sigma} \right) + \text{erf}\left( \frac{d-\mu}{2\sigma} \right) \right) < p$$.

From this inequality one may deduce the implies inequality for $\mu$. Here is an example:

enter image description here

share|improve this answer
    
Thanks a lot for your reply. I am interested in such a positive value of $\mu$ such that the above inequality becomes true. In your graph, this is the point where p and error function intersects i.e. approx $\mu$ = 1.8. Is it possible to find this value by solving inequality algebrically? –  shaikh Jul 17 '11 at 5:00
3  
Since you are interested in the intersection point, then you really mean to solve equality, rather then inequality. The solution can not be solved algebraically, I am afraid. –  Sasha Jul 17 '11 at 5:06

Let $I$, as in my first answer, denote the iterated integral (including the factor $\frac{1}{{2\pi \sigma ^2 }}$). Using probabilistic arguments, I obtained $$ I = \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg), $$ where $\Phi$ is the distribution function of the ${\rm N}(0,1)$ distribution. On the other hand, using Mathematica, Sasha obtained $$ I = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu + d}}{{2\sigma }}\bigg) + {\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg)\bigg). $$ Here ${\rm erf}$ is the error function, defined by $$ {\rm erf}(x) = \frac{2}{{\sqrt \pi }}\int_0^x {e^{ - t^2 } \,dt} , \;\; x \in \mathbb{R}. $$ (Note that ${\rm erf}(-x)=-{\rm erf}(x)$.) So, let's show that indeed $$ \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu + d}}{{2\sigma }}\bigg) + {\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg)\bigg). $$ From the standard relation $$ \Phi (x) = \frac{1}{2}\bigg[1 + {\rm erf}\bigg(\frac{x}{{\sqrt 2 }}\bigg)\bigg], $$ we get $$ \Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu + d}}{{2\sigma }}\bigg) - {\rm erf}\bigg(\frac{{\mu - d}}{{2\sigma }}\bigg)\bigg), $$ and hence the desired equality follows from $$ {\rm erf}\bigg(\frac{{d - \mu }}{{2\sigma }}\bigg) = - {\rm erf}\bigg(\frac{{\mu - d}}{{2\sigma }}\bigg). $$ Now, as in my first answer, define a function $f$ by $$ f(\mu):=\Phi \bigg(\frac{{\mu + d}}{{\sqrt {2\sigma^ 2} }}\bigg) - \Phi \bigg(\frac{{\mu - d}}{{\sqrt {2\sigma^ 2} }}\bigg) = \frac{1}{2}\bigg({\rm erf}\bigg(\frac{{\mu + d}}{{2\sigma }}\bigg) - {\rm erf}\bigg(\frac{{\mu - d}}{{2\sigma }}\bigg)\bigg). $$ Recall that $f$ is decreasing in $\mu \in [0,\infty)$, with $f(\mu) \to 0$ as $\mu \to \infty$. So if $f(0) > 1-p$, there exists a solution $\mu > 0$ to $f(\mu)=1-p$. You can find an extremely accurate approximation to $\mu$ using, for example, Wolfram Alpha (based on the representation using the error function).

share|improve this answer
    
@Covo: That's great, but Wolfram Alpha is again a graphical solution and I think that there is no direct algebraic solution for some approximate value of $\mu$. In order to use it in my program I need to test several values of $\mu$ before arriving at final approximate value. Thanks a lot for your help :) –  shaikh Jul 18 '11 at 2:10
1  
shaikh: Indeed, this is a problem for numerical analysis. The Bisection method might be relevant: en.wikipedia.org/wiki/Bisection_method –  Shai Covo Jul 18 '11 at 2:24

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.