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Let $\sigma>0$, $$\phi(x):=\frac1{\sqrt{2\pi\sigma^2}}e^{-\frac{x^2}{2\sigma^2}}\;\;\;\text{for }x\in\mathbb R$$ and $$\mathcal N(x,B):=\int_B\varphi(y-x)\:{\rm d}y\;\;\;\text{for }(x,B)\in\mathbb R\times\mathcal B(\mathbb R).$$ If $$\iota:\mathbb R\to[0,\infty)\;,\;\;\;x\mapsto x-\lfloor x\rfloor,$$ then the wrapped normal distribution kernel on $[0,1)$ with variance $\sigma^2$ is defined by $$\mathcal W(x,\;\cdot\;):=\mathcal N(x,\;\cdot\;)\circ\iota^{-1}\;\;\;\text{for }x\in[0,1).$$ If $$\psi(x):=\sum_{k\in\mathbb Z}\phi(x+k)\;\;\;\text{for }x\in\mathbb R,$$ we can show that $$\mathcal W(x,B)=\int_B\psi(y-x)\:{\rm d}y\;\;\;\text{for all }(x,B)\in[0,1)\times\mathcal B([0,1))\tag1.$$

Let $x\in[0,1)$. How can we compute the expectation of a random variable $Y\sim\mathcal W(x,\;\cdot\;)$?

Clearly, $$\operatorname E[Y]=\int_0^1\psi(y-x)y\:{\rm d}y\tag2.$$

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The wrapped Normal distribution can be written as an infinite mixture of truncated Normal distributions since, if $\phi_\sigma$ and $\Phi_\sigma$ denote respectively the density and cdf of the Normal $\mathcal N(0,\sigma^2)$ distribution, then $$\psi(y-x)=\sum_{k\in\mathbb Z}\phi_\sigma(y-x-k)\mathbb I_{(0,1)}(y)$$ can be rewritten as \begin{align}\psi(y-x)&=\sum_{k\in\mathbb Z}[\Phi_\sigma(1-x-k)-\Phi_\sigma(-x-k)]\times\dfrac{\phi_\sigma(y-x-k)\mathbb I_{(0,1)}(y)}{\Phi_\sigma(1-x-k)-\Phi_\sigma(-x-k)}\\ &=\sum_{k\in\mathbb Z}\underbrace{[\Phi_\sigma(x+k)-\Phi_\sigma(x+k-1)]}_{p_k(x)}\times\dfrac{\phi_\sigma(y-x-k)\mathbb I_{(0,1)}(y)}{\Phi_\sigma(x+k)-\Phi_\sigma(x+k-1)}\end{align} Therefore its mean is the weighted sum of the means of the truncated Normal distributions $\mathcal N_0^1(x+k,\sigma^2)$, given by $$x+k-\sigma^2\dfrac{\phi_\sigma(1-x-k)-\phi_\sigma(-x-k)}{\Phi_\sigma(x+k)-\Phi_\sigma(x+k-1)}$$ That is, $$\mathbb E_x[Y]=x+\sum_{k\in\mathbb Z} kp_k(x)-\sigma^2 \sum_{k\in\mathbb Z}[\phi_\sigma(1-x-k)-\phi_\sigma(-x-k)]=x+\sum_{k\in\mathbb Z} kp_k(x)$$ since the last sum is equal to zero. Except when $x=0,1/2,1$, where the expectation is $1/2$, I do not know whether or not the mean has a closed form expression. A vague simplification using the symmetry properties of the Normal cdf is $$\sum_{k\in\mathbb Z} kp_k(x)=\Phi(-x)+\sum_{k=1}^\infty[\Phi(x-k)+\Phi(-x-k)]$$

Here is the Monte Carlo approximation to the mean when $\sigma=.1$ (although there is no visible difference in the graph when moving $\sigma$):

enter image description here

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  • $\begingroup$ I see. However, the relation to the truncated normal distribution might be useful. I've got a related question, you might be able to answer as well: math.stackexchange.com/q/3594286/47771. $\endgroup$ – 0xbadf00d Mar 25 at 10:41

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