Let $X$ and $Y$ be two independent random variables, each with the standard normal density. Compute $f_{X+Y} (a)$.
The solution uses a convolution of the form
$$f_{X+Y}(a)= {1\over 2\pi} \int_{-\infty}^{\infty} e^{-{(a-y)^2}\over2}e^{-{y^2}\over2} \ dy$$
I am wondering why the bounds are $\infty$ and not $a$ and $0$ such as I found in another problem- Let $X$ and $Y$ be two independent exponential random variables with common parameter $λ$. Compute $f_{X+Y} (a)$, where the solution uses $f_{X+Y}(a)= \int_{0}^{a} λe^{-λ(a-y)}λe^{-λy}\ dy$
Why are the bounds $\infty$ when the sum of $X$ and $Y$ is no more than $a$? Is this something special about the standard normal variable?