I have a probability distribution function as following, in my simulations, when I plot f(x) versus x, it is symmetric around zero, but I can't prove it mathematically, can anybody help?
\begin{equation} f(x)= (1/2\pi) \int _{-\infty}^{\infty} \frac{e^{-\frac{(x-\theta)^2}{2(\theta^2+1)}} e^{-\frac{\theta^2}{2}} }{\sqrt{\theta^2+1}} d\theta \end{equation}
actually assume that $x=h\theta+n$ where $h,\theta,n$ all have Gaussian distribution, i.e $h \sim N(1,1), ~~\theta \sim N(0,1), ~~n \sim N(0,1)$, so
\begin{equation} f(x)= \int _{-\infty}^{\infty} f(x|\theta) f(\theta) d\theta \end{equation}
If you substitute $t=-θ$ $$\begin{equation} f(x)= (1/2\pi) \int _{-\infty}^{\infty} \frac{e^{-\frac{(x-\theta)^2}{2(\theta^2+1)}} e^{\frac{\theta^2}{2}} }{\sqrt{\theta^2+1}} d\theta \\= (1/2\pi) \int _{-\infty}^{\infty} \frac{e^{-\frac{(x+t)^2}{2(t^2+1)}} e^{\frac{t^2}{2}} }{\sqrt{t^2+1}} dt= \\ (1/2\pi) \int _{-\infty}^{\infty} \frac{e^{-\frac{(-x-t)^2}{2(t^2+1)}} e^{\frac{t^2}{2}} }{\sqrt{t^2+1}} dt=f(-x) \end{equation}$$