Resultant probability distribution when taking the cosine of gaussian distributed variable - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-20T11:36:22Z http://mathoverflow.net/feeds/question/35260 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed Resultant probability distribution when taking the cosine of gaussian distributed variable Shannon Edwards 2010-08-11T18:06:23Z 2013-05-27T01:33:21Z <p>I am trying to do a measurement uncertainty calculation. I have a gaussian distributed phase angle (theta) with a mean of 0 and standard deviation of 16.6666 micro radians. The variance is the square of the standard. The formula for the measurment uses cos(theta) in the calculation. I need to know the mean, the variance and the distribution function that result from taking the cosine of theta in order to do my calculations correctly. </p> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/35318#35318 Answer by Will Jagy for Resultant probability distribution when taking the cosine of gaussian distributed variable Will Jagy 2010-08-12T05:15:59Z 2010-08-12T05:15:59Z <p>I wrote out the first few terms in the power series for $\cos \theta$ and then the first few terms of the series for $\cos^2 \theta .$ I used your hypothesis of normal distribution, the mean of $\theta$ is $\mu = 0$ while the variance is some $\sigma^2 .$</p> <p>Then I looked up the expected values of $\theta^2, \; \theta^4, \; \theta^6, \; \theta^8$ at <a href="http://en.wikipedia.org/wiki/Gaussian_distribution#Moments" rel="nofollow">http://en.wikipedia.org/wiki/Gaussian_distribution#Moments</a> and used that to find good approximations for your new mean $\mu_1$ and variance $\sigma_1^2$ in $$\mu_1 = E[ \cos \theta ] = 1 - \frac{\sigma^2}{2} + \frac{\sigma^4}{8} - \frac{\sigma^6}{48} + \cdots$$ and $$\mu_1^2 + \sigma_1^2 = E[ \cos^2 \theta ] = 1 - \sigma^2 + \sigma^4 - \frac{2 \sigma^6}{3} + \cdots$$ So when you subtract you get $\sigma_1^2 \approx \frac{\sigma^4}{2}$</p> <p>I will think about it some more, there is a large theory for calculating moments. But I do not see much to be done in the way of an explicit pdf or cdf. </p> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/37522#37522 Answer by Cody Martin for Resultant probability distribution when taking the cosine of gaussian distributed variable Cody Martin 2010-09-02T18:28:18Z 2010-09-02T19:21:51Z <p>Given a normal distribution with mean $\mu$ and variance $\sigma^2$, $X = \mathcal{N}(\mu,\sigma^2)$, if you pass it through trigonometric functions, you can approximate the result with the new normal distributions below</p> <p>1) normal distribution passed through Cosine function:</p> <p>$X_{\cos} = \mathcal{N}(\cos(\mu),\sigma^2\sin^2(\mu))$</p> <p>so the new average is $\cos(\mu)$ and the new standard deviation is $|\sigma\sin(\mu)|$.</p> <p>2) normal distribution passed through a Sine function:</p> <p>$X_{\sin} = \mathcal{N}(\sin(\mu),\sigma^2\cos^2(\mu))$</p> <p>so the new average is $\sin(\mu)$ and the new standard deviation is $|\sigma\cos(\mu)|$.</p> <p>The Matlab script that I used to find these relations is below.</p> <pre><code>%% Cody Martin % 9/2/2010 % m-file used to discover the mean and variance of a normal distribution % passed through cosine and sine functions...results: % - N(mu,sig^2) -&gt; cos(N(mu,sig^2)) = N(cos(mu),sig^2*sin^2(mu)) % - N(mu,sig^2) -&gt; sin(N(mu,sig^2)) = N(sin(mu),sig^2*cos^2(mu)) %% distribution of cosine and sine of a normal distribution? cresults = zeros(0,5); sresults = zeros(0,5); % loop from an average angle -90 degrees to +90 degrees for theta = -pi/2:pi/36:pi/2 theta1sig = pi/36; % standard deviation of orinigal normal distribution vtheta = theta + theta1sig*randn(9999,1); % create 9999 points using this avg and std vctheta = cos(vtheta); % take the cosine of those points vstheta = sin(vtheta); % take the sine of those points theta_ = min(vtheta):0.01:max(vtheta); % for plotting ideal distributions ctheta_ = min(vctheta):0.01:max(vctheta); % for plotting stheta_ = min(vstheta):0.01:max(vstheta); % for plotting figure(1); clf; subplot(211); hold on; plot(theta_,cdf('normal',theta_,theta,theta1sig),':'); % plot cdf of normal distribution with avg and std plot(sort(vtheta),[1:length(vtheta)]/length(vtheta)); % plot cdf of 9999 points plot(sort(vctheta),[1:length(vctheta)]/length(vctheta),'k','LineWidth',2); % plot cdf of cos(9999 points) plot(ctheta_,cdf('normal',ctheta_,cos(theta),... % plot cdf of norm dist with new avg and std after being passed through cos() sqrt(theta1sig^2*sin(theta)^2)),'r:'); plot(cos(theta)*[1 1],[0 1],'k:'); % vertical line @ cos(theta) - shows new average matches cos(old avg) title('Cosine of a Normal Distribution (for Different Initial Averages)'); legend('Norm CDF Theory','Norm CDF 9999','Cos(Norm CDF 9999)','Cos(Norm CDF) Theory'); axis([-pi/2 pi/2 0 1]) subplot(212); hold on; plot(theta_,cdf('normal',theta_,theta,theta1sig),':'); plot(sort(vtheta),[1:length(vtheta)]/length(vtheta)); plot(sort(vstheta),[1:length(vstheta)]/length(vstheta),'k','LineWidth',2); plot(stheta_,cdf('normal',stheta_,sin(theta),... sqrt(theta1sig^2*cos(theta)^2)),'r:'); plot(sin(theta)*[1 1],[0 1],'k:'); title('Sine of a Normal Distribution (for Different Initial Averages)'); legend('Norm CDF Theory','Norm CDF 9999','Sin(Norm CDF 9999)','Sin(Norm CDF) Theory'); axis([-pi/2 pi/2 0 1]) % fprintf('theta: %3.0f\tstd: %5.3f\tsin(theta): %5.3f\tavg: %5.3f\tstd: %5.3f\n',theta*180/pi,theta1sig,sin(theta),mean(vstheta),std(vstheta)); cresults = [cresults; theta theta1sig cos(theta) mean(vctheta) std(vctheta)]; sresults = [sresults; theta theta1sig sin(theta) mean(vstheta) std(vstheta)]; end figure(2); clf; subplot(211); hold on; plot(cresults(:,1),cresults(:,end)); plot(cresults(:,1),abs(theta1sig*sresults(:,3)),'r:'); title('Standard Deviation of Cosine of a Normal Distribution as a Function of the Original Average'); legend('From 9999 Points','Fit: std = |\sigmasin(\mu)|'); ylabel('std(cos(\theta_{vector})) [rad]'); xlabel('\theta [rad]'); subplot(212); hold on; plot(sresults(:,1),sresults(:,end)); plot(sresults(:,1),abs(theta1sig*cresults(:,3)),'r:'); title('Standard Deviation of Sine of a Normal Distribution as a Function of the Original Average'); legend('From 9999 Points','Fit: std = |\sigmacos(\mu)|'); ylabel('std(sin(\theta_{vector})) [rad]'); xlabel('\theta [rad]'); </code></pre> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/98194#98194 Answer by unknown (yahoo) for Resultant probability distribution when taking the cosine of gaussian distributed variable unknown (yahoo) 2012-05-28T14:21:01Z 2012-05-28T14:21:01Z <p>Cody, is it wright what you say? Sigma is varying with the mean? If I measure an angle of 90 degrees, then $N_{\cos}(0,{\sigma}^2)$ and $N_{\sin}(1,0)$? And if I measure an angle of 0 degrees, then $N_{\cos}(1,0)$ and $N_{\sin}(0,{\sigma}^2)$ ? Where do I find the theory of that?</p> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/100339#100339 Answer by Gabriel for Resultant probability distribution when taking the cosine of gaussian distributed variable Gabriel 2012-06-22T10:17:02Z 2012-06-22T10:24:39Z <p>Hi, I know this was asked a long time ago but I have just discovered it because I require a similar solution. It is possible to generate an expression, albeit as an infinite summation. For practical purposes, the first few terms of the summation should suffice.</p> <p>Let $X$ denote a random variable with pdf $f_X(x)$. Let $Y=g(X)$ be a function of $X$. We can specify the cdf of $Y$, denoted $F_Y(y)$ as follows:</p> <p>$F_Y(y)=\mathbb{P}(g(X)\leq y)=\int\limits_{\Omega}f_X(x)\text{d}x$,</p> <p>where the domain of integration $\Omega$ is defined as</p> <p>$\Omega=\left\lbrace x:g(x)\leq y \right\rbrace$</p> <p>In our case, $g(x)=\cos x$, so we need an expression for the domain of $x\in\mathbb{R}$ such that $\cos x\leq y$. This is given by</p> <p>$2k\pi+\arccos(y) \leq x &lt; 2(k+1)\pi-\arccos(y)\, k\in\mathbb{Z}$</p> <p>So integrating over this domain, we obtain</p> <p>$F_Y(y)=\sum\limits_{k=\infty}^{\infty} \int\limits_{2k\pi+\arccos(y)}^{2(k+1)\pi-\arccos(y)} f_X(x)\text{d}x$</p> <p>Now in our case $X\sim\mathcal{N}(0,\sigma)$, so</p> <p>$f_X(x)=\dfrac{1}{\sigma\sqrt{2\pi}}\exp\left(\dfrac{-x^2}{2\sigma^2}\right)$</p> <p>and the integral of this pdf between limits is given by the cdf of the normal distribution, which we denote $\Phi$:</p> <p>$\int\limits_{a}^{b}f_X(x)\text{d}x = \Phi(b/\sigma)-\Phi(a/\sigma)$</p> <p>The cdf of $Y$ is therefore</p> <p>$F_Y(y)=\sum\limits_{k=-\infty}^{\infty} \Phi\left(\dfrac{2(k+1)\pi-\arccos(y)}{\sigma}\right) - \Phi\left(\dfrac{2k\pi-\arccos(y)}{\sigma}\right)$</p> <p>To compute the pdf, take the derivative with respect to $y$:</p> <p>$f_Y(y)=\dfrac{dF_Y(y)}{dy} = \sum\limits_{k=-\infty}^{\infty} \dfrac{1}{\sqrt{1-y^2}}\left( f_{X}(2(k+1)\pi-\arccos(y) ) + f_{X}(2k\pi+\arccos(y)) \right)$</p> <p>There are probably better ways to do this. It's possible the final summation can be rewritten or simplified. But this seems to match with a numerical check.</p> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/129375#129375 Answer by Ran for Resultant probability distribution when taking the cosine of gaussian distributed variable Ran 2013-05-02T05:04:56Z 2013-05-02T05:04:56Z <p>Cody, I'm afraid your answer is incomplete. The problem I see lays with the variance. If $X \sim N(\mu,\sigma^2)$ indeed results with $cos(X) \sim N(cos(\mu), \sigma^2 sin^2(\mu))$ then for, e.g., $\mu = \frac{\pi}{2}$ the approximation is $N(1,0)$ regardless of $\sigma$. This seems to be a poor approximation because an increase in the variance of $X$ should always result in an increase of the variance of $cos (X)$. </p> http://mathoverflow.net/questions/35260/resultant-probability-distribution-when-taking-the-cosine-of-gaussian-distributed/131008#131008 Answer by Parastoo Q for Resultant probability distribution when taking the cosine of gaussian distributed variable Parastoo Q 2013-05-17T22:20:07Z 2013-05-27T01:33:21Z <p>A quick way to find the mean of $\cos(\theta)$, where $\theta\sim \mathcal{N}(0, \sigma^2)$, is through calculating the mean of a complex variable $e^{j\theta}=\cos(\theta)+j\sin(\theta)$. We have</p> <p>$E [e^{j\theta}]=e^{0+(j\sigma)^2/2}=e^{-\sigma^2/2}$</p> <p>which implies that the mean of the imaginary part $E [\sin(\theta)]$ equals zero and the mean of the real part $E[\cos(\theta)]$ equals $e^{-\sigma^2/2}$.</p> <p>The answer $\mu_1$ derived by Will Jagy is in fact the Taylor series expansion of $e^{-\sigma^2/2}$. </p> <p>The variance of $\cos(\theta)$ can be obtained as:</p> <p>$E[\cos^2(\theta)]-E[\cos(\theta)]^2= E[\frac{1}{2}+\frac{\cos(2\theta)}{2}]- E[\cos(\theta)]^2= \frac{1}{2}[1-e^{-\sigma^2}]^2$</p>