= Solution
For $X\sim\operatorname{Exp}(\lambda)$, integration gives $M_X(t)=\lambda/(\lambda-t)$ for $t<\lambda$. For $0<c<1$, the candidate remainder <moment-generating function> is
$$
\frac{M_X(t)}{M_X(ct)}=\frac{\lambda-ct}{\lambda-t}=c+(1-c)\frac{\lambda}{\lambda-t}.
$$
This is the <moment-generating function> of the <mixture distribution>
$$
\boxed{\mathcal L(R_c)=c\delta_0+(1-c)\operatorname{Exp}(\lambda).}
$$
Construct 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 $X$, and
$$
M_{cX+R_c}(t)=\frac{\lambda}{\lambda-ct}\frac{\lambda-ct}{\lambda-t}=\frac{\lambda}{\lambda-t}=M_X(t)
$$
on a neighborhood of zero. By the <uniqueness theorem for moment-generating functions>, $cX+R_c\overset d=X$, so the <exponential distribution> is a <self-decomposable distribution>.
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