# Mutual Information of the summation of a Chi-square random variable and a Gaussian variable

As the title, $$X$$ is an random variable subject to $$N(0,1)$$, $$N$$ is an random variable subject to $$N(0,\sigma^2)$$, and $$X$$ and $$N$$ are independent. I want to calculate the mutual information $$I(X,Y)$$ between $$X$$ and $$Y=X^2+N$$. The problem can be reduced to calculate the differential entropy $$h(Y)$$ of $$Y$$. However, $$h(Y)=-\int_{-\infty}^{+\infty}p_Y(y)log(p_Y(y))\mathrm{d}y$$, and the $$p_Y(y)$$ can not get the explicit form. How can I calculate $$h(Y)$$ or $$I(X,Y)$$ can be calculated in other ways?

It is not quite true that $$p_Y(y)$$ has no explicit form, it is given by $$p_Y(y)=\int_0^\infty \frac{e^{-(w-y)^2/2 \sigma^2-w/2}}{2 \pi \sigma \sqrt{w}}\,dw=f_1(y)\theta(2y-\sigma^2)+f_2(y)\theta(\sigma^2-2y),$$ with $$\theta(y)$$ the unit step function and $$f_1(y)=\frac{1}{4\sigma\sqrt 2}e^{\frac{\sigma^4-4 \sigma^2 y-4 y^2}{16 \sigma^2}} \sqrt{2y-\sigma^2} \left[I_{1/4}\left(\frac{\left(\sigma^2-2 y\right)^2}{16 \sigma^2}\right)+I_{-1/4}\left(\frac{\left(\sigma^2-2 y\right)^2}{16 \sigma^2}\right)\right],$$ $$f_2(y)=\frac{1}{4\pi\sigma}e^{\frac{\sigma^4-4 \sigma^2 y-4 y^2}{16 \sigma^2}} \sqrt{\sigma^2-2 y}\; K_{1/4}\left(\frac{\left(\sigma^2-2 y\right)^2}{16 \sigma^2}\right).$$ The differential entropy $$h(Y)$$ can be readily evaluated numerically, here is a plot of $$h(Y)$$ versus $$\sigma$$.