Sum of independent random variables is $\sim\mathrm{Exp}(\lambda)$

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Consider a parameter $\lambda > 0$ and a random variable $Y$ taking values in $\mathbb{N}$, with the support of its distribution being $\mathbb{N}$. Given that $Y=n$ for some $n\in\mathbb{N}$, is there a canonical way to construct independent random variables $(X_i^{(n)})_{i=1}^n$ such that $$\sum_{i=1}^n X_i^{(n)} \sim \mathrm{Exp}(\lambda)?$$

Note that $\mathrm{Exp}(\lambda)$ must not depend on $n$.

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The exponential distribution is a special case of the Gamma distribution with $\alpha=1,\,\beta=\lambda$. In general its characteristic function is $\left(1-\frac{it}{\beta}\right)^{-\alpha}$. We can thus take iids $X_i^{(n)}$ to also be Gamma-distributed with $\alpha=\frac{1}{n},\,\beta=\lambda$.