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For $X\sim \mathrm{Exp}(\lambda )$, integration gives $M_X(t)=\lambda /(\lambda -t)$ for $t<\lambda$. For $0<c<1$, the candidate remainder moment-generating function isThis is the moment-generating function of the mixture distributionConstruct it by taking an exponential random variable $E$ of rate $\lambda$ and a Bernoulli random variable $J$ with $\mathbb{P}(J=1)=1-c$, with $E,J,X$ mutually independent, and setting $R_c=JE$. Then $R_c$ is independent of $X$, andon a neighborhood of zero. By the uniqueness theorem for moment-generating functions, $cX+R_c\stackrel{d}{=}X$, so the exponential distribution is a self-decomposable distribution.

Fix $0<c<1$ and choose $R_c$, independently of $X$, with the normal distributionThe linear combination of independent normal random variables $cX+R_c$ is normal. Its expected value is $c\mu +(1-c)\mu =\mu$ and its variance is $c^2{\sigma }^2+(1-c^2){\sigma }^2={\sigma }^2$. Hence $cX+R_c\stackrel{d}{=}X$ for every $c$, proving that every specified normal distribution is a self-decomposable distribution.

Let $X$ have a strictly stable distribution of index $\alpha$, and let $X^{\prime }$ be an independent copy. For integers $1\le k<n$, split a sum of $n$ independent copies into blocks of sizes $k$ and $n-k$. Strict stability of each block and of the full sum givesFor any $0<c<1$, choose rationals $k_m/n_m\to c^{\alpha }$. On the fixed probability space carrying $X,X^{\prime }$, the left side converges almost surely to $cX+(1-c^{\alpha }{)}^{1/\alpha }{X}^{\prime }$. Thus it converges in distribution; since its probability distribution is constant, the limit also has the law of $X$. This proves a self-decomposable distribution with the displayed remainder. The proof does not require any finite moments.