In statistics, the definition of $F^k$ is the k-fold convolution of $F$ with itself, where $F$ is some common distribution. I am wondering if the following holds, if:
$$ L_{F^{k}(x)} = \left(L_{F(x)}\right)^k $$
That is, the Laplace transform of the k-fold convolution is just the Laplace transform of the common distribution multiplied $k$ times. Thanks!
Short answer: yes
Long answer: For any 2 functions $f(t)$ and $g(t)$, the Laplace tranform of the convolution is the product of Laplace transforms $$\mathcal{L} \{ f(t) * g(t) \} = \mathcal{L}\{f(t)\} \mathcal{L}\{g(t)\}$$
It makes sense that this would hold no matter how many times we do the convolution