 Solution

For rate $\lambda >0$, the exponential distribution has cumulative distribution function $F(x)=1-e^{-\lambda x}$ on $x\ge 0$. Inverse transform sampling gives $-\log (1-U)/\lambda$; since $1-U$ is also uniform, an equivalent answer isFor $x\ge 0$, $P(E>x)=P(U<e^{-\lambda x})=e^{-\lambda x}$, verifying the exponential distribution directly.

The centered Laplace distribution with density $(\lambda /2)e^{-\lambda |x|}$ can be obtained by multiplying $E$ by an independent sign taking $1$ and $-1$ with equal probabilities. Alternatively, its inverse transform sampling formula uses just one uniform variate:The first branch inverts the negative half of the cumulative distribution function and the second branch inverts the positive half. Add a location parameter if a noncentered Laplace distribution is wanted.

Correct uniformity is essential for each marginal distribution, and independence is essential for independent samples and for the sign construction. For example, setting every $U_i$ equal to a single uniform $U$ gives uniform marginals but identical transformed draws. Likewise, choosing a sign from the same uniform used for the magnitude can change the Laplace distribution unless the one-uniform inverse above is used.