My Math Forum Analytic convariance constant for normal distributions

 May 22nd, 2012, 08:01 PM #1 Newbie   Joined: Apr 2012 Posts: 26 Thanks: 0 Analytic convariance constant for normal distributions Hi, I've come across a constant that I have found useful in calculations involving normal distributions with sign based peturbations (Cyhelsky skew). However, I haven't been able to calculate it's value analytically, but only experimentally. I was wondering if anyone either knew how to calculate it , or if anyone happens to know a name for it and an analytical expression for it. Given two normal distribution random variables a,b; Each has a mean of 0, a standard deviation of 1, and NO covariance between them; Adding the two variables together will give a result with a standard deviation of (2)**0.5. I am interested in a very similar problem; Take only samples of a and b which have the same sign and add them; reject all samples in a and b which have opposite signs; compute the result. The problem introduces a covariance that is positive, but of an unkown value. I know experimentally, that the result will have a mean of 0, a standard deviation of ~1.809213(2), However, I can't figure out the solution analytically. Is this constant a known value that someone has computed before? If not, can someone give me some pointers on how to calculate the constant? I know how to take the integral of z*exp(-0.5*x**2)**2, and then using the area of the bell curve (2*pi)**0.5 to compute the variance or standard deviation of the distribution. I'm just not sure how to set the problem up for two independent normal distributions that are added with the restrictions I have given...
 June 1st, 2012, 11:12 PM #2 Newbie   Joined: Apr 2012 Posts: 26 Thanks: 0 Re: Analytic convariance constant for normal distributions O.K. I attempted a solution.... Problem setup, and analogy: I am interested in finding the deviation of adding two normal distributions together that are 100\% correlated in the *Sign* of the data elements, but otherwise are uncorrelated. If I were not adding them together, but just attempting to get the deviation of a normal distribution of single data points; I would approach it as a weighted integral to obtain the average variance, and then convert that to a deviation. The probability of each data element is: $p(x)= \sqrt{2 \over \pi}e^{-{1\over2}x^2}$ and variance is: $\sigma^2(x)=x^2$, which makes the weighted variance: $\sqrt{2 \over \pi}{x^2e^{-{1\over2}x^2}}$ and the total of the weighted variances is therefore: $\overline{\sigma^2}=\sqrt{2 \over \pi}\int\limits_{-\infty}^{\infty}{x^2e^{-{1\over2}x^2}}$ From an integral table... $\int\limits_{-\infty}^{\infty} {x^2}e^{-{1\over2}x^2}\ dx= \left\{ \left. \sqrt{\pi\over2} \left(erf({x\sqrt{2}\over2})\right)-xe^{-{1 \over 2 }x^2}\right\}\right|_{-\infty}^{\infty}$ $=\sqrt{\pi\over2}$ So the final result is 1.0, as expected and $\sigma=1$ Now .......................... ! Trying to do this for an addition, by the same methodology. The relative probability is: $p(x)={1\over k}(e^{-{1\over2}x^2}e^{-{1\over2}y^2})$ But, I only need two quadrants formed by x,y since the data is sign correlated; eg: quadrants I and III. By symmetry, the problem can be reduced to solving only for quadrant I; but keeping in mind that quadrant III makes $\mu$ stay at 0. Computing total area to normalize probability for quadrant I: $k=\int\limits_{0}^{\infty}\int\limits_{0}^{\infty } e^{-{1\over2}(x^2+y^2)}dxdy={\pi\over4}$ The variance for addition is: $\sigma^2(x,y)=(x+y)^2$ Which makes the weighted variance: ${4\over\pi}(x+y)^2e^{-{1\over2}x^2}e^{-{1\over2}y^2}={4\over\pi}(x+y)^2e^{-{1\over2}(x^2+y^2)}$ So, the combined equation yields for quadrant I: $\overline{\sigma^2}=\int\limits_0^{\infty}\int\li mits_0^{\infty}{4\over\pi}(x+y)^2e^{-{1\over2}(x^2+y^2)}dxdy$ ... and this is as far as I have gotten... Any pointers on solving this last bit? Anyone have Mathematica that can do a brute force? :P
 June 4th, 2012, 07:01 PM #3 Newbie   Joined: Apr 2012 Posts: 26 Thanks: 0 Re: Analytic convariance constant for normal distributions I made a mistake about what erf(0) was... and royally messed up the normalization constant of the bell curve, it ought to uniformly be changed to : $\sqrt{1 \over 2\pi}$ in the original example. I don't think I made the same mistake in the attempted solution, but if someone sees a mistake -- please let me know.
 June 19th, 2012, 04:56 PM #4 Newbie   Joined: Apr 2012 Posts: 26 Thanks: 0 Re: Analytic convariance constant for normal distributions I was able to finish solving for it analytically... thanks to some help on another forum. BTW: Does anyone know a name for this constant ? (I'd hate to invent one if there was a standard convention...) $\overline { \sigma }= \sqrt{ 2 + { 4 \over \pi } } \approx 1.809209646$ Or, perhaps, the Pearson covariance constant equivalent (PPMCC) ? $PPMCC(auto)= { 2 \over \pi } \approx 0.636619772$

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### cyhelsky's skewness

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