I need some help with this exercise. I'm not sure how to deal with it:
Let $f(x)=e^{-x^2}$, $\mu$ the Lebesgue measure in $[0,1]$ and $\nu$ the Lebesgue measure in $[2,\infty)$.
I have to find the convolutions $f\ast f$, $\mu\ast f$ and $\mu\ast\nu$.
Using the definition I don't get much, so I was wondering if there is a best way to compute them.
Thanks for any help.
By definition, we have
$$\begin{align*} (f \ast f)(x) &= \int_{\mathbb{R}} f(x-y) \cdot f(y) \, dy = \int_{\mathbb{R}} \exp \left(-(x-y)^2 - y^2 \right) \, dy \\ &= \sqrt{2\pi \sigma^2} e^{-x^2} \int_{\mathbb{R}} \exp(2xy) \cdot \underbrace{\frac{1}{\sqrt{2\pi \sigma^2}} \exp \left( - \frac{y^2}{2\sigma^2} \right)}_{=:p(y)} \, dy. \end{align*}$$
for $\sigma^2 := 1/4$. As $y \mapsto p(y)$ is the density of the normal distribution with mean $0$ and variance $\sigma^2=1/4$, the integral can be computed using the widely known formula for exponential moments of the normal distribution. Thus,
$$(f \ast f)(x) = \sqrt{\frac{\pi}{2}} e^{-x^2} \cdot e^{\frac{1}{2} x^2} = \sqrt{\frac{\pi}{2}} \cdot e^{-\frac{1}{2} x^2}$$
Alternatively, a more probabilistic reasoning can be used; it is based on the fact that the convolution of two distributions $\mu$ and $\nu$ equals the distribution of the random variable $X+Y$ where $X \sim \mu$, $Y \sim \nu$ are independent random variables. As $f$ is basically the density of a Gaussian random variable, the calculation of the convolution is pretty easy using some known properties of the Gaussian distribution.
Similarly, we find by the definition of the convolution that the measure $(\mu \ast \nu)$ has the density
$$\begin{align*} h(x) &= \int 1_{[0,1]}(y) \cdot 1_{[2,\infty)}(x-y) \, dy \\ &= \int_0^1 1_{(-\infty,x-2]}(y) \, dy = \begin{cases} 0 & x \leq 2 \\ x-2 & x \in [2,3] \\ 1 & x \geq 3 \end{cases} \end{align*}$$