Let $X \sim \mathcal{N} (0, I_n)$ be a standard gaussian in $\mathbb{R}^n$ and $Y$ be a another random variable with a smooth density $f$. I want to show that for $s>0$, the density of $Y + \sqrt{s} X$ (say $f_s$) is a solution to the heat equation, $$\frac{\partial}{\partial s} f_s (x) = \frac{1}{2} \Delta f_s(x)$$
I'm not sure how to proceed as I have difficulty finding $f_s$. Any help?
The distribution of $Z=Y+\sqrt{s}X$ is the following $$f_Z(z,s)=\int_{\mathbb{R}^n}\phi(z-y,s)f(y)dy, \ \ z \in \mathbb{R}^n,s \in [0,T]$$ where $f$ is the density of $Y$ and $\phi(x,s)$ is the density of a multivariate normal with covariance matrix $s\mathbf{I}_n$.This is the solution to the PDE $$\frac{\partial}{\partial s}f_Z(z,s)=\frac{1}{2}\sum_{k=1}^n\frac{\partial^2}{\partial z_k^2}f_Z(z,s)$$ with initial condition $f_Z(z,0)=f(z)$. This is because the fundamental solution to the PDE is a multivariate normal density with covariance matrix $s\mathbf{I}_n$, so the particular solution is given by the convolution with the given initial condition
Note: I assumed $X$ and $Y$ are independent.
It can be shown that the convolution is a solution by using the theory of Brownian motions. Consider a Brownian motion $W_t \in \mathbb{R}^n$. The evolution of its probability density is described by the $n$-dimensional Forward Kolmogorov Equation $$\frac{\partial}{\partial t}f_W(w,t)=\frac{1}{2}\sum_{k=1}^n\frac{\partial^2}{\partial w_k^2}f_W(w,t)$$ with general initial condition $f_W(w,0)=f(w)$ (which is a starting probability density). The solution is given by the convolution $$f_W(w,t)=\int_{\mathbb{R}^n}\phi(w-y,t)f(y)dy, \ \ w \in \mathbb{R}^n,t \in [0,T]$$ where $\phi(w,t)$ is the density of a multivariate normal with covariance matrix $s\mathbf{I}_n$.