= Mixture of a point mass and an exponential distribution
{title2=$\mathcal L(Y)=p\delta_{-c}+(1-p)\operatorname{Exp}(\lambda)$}
Let $c>0$, $0\leq p\leq1$ and $\lambda>0$. With probability $p$, take $Y=-c$; otherwise draw an <exponential distribution> of rate $\lambda$. This <mixed probability distribution> has a negative atom and a continuous nonnegative component. Its moments are $\mathbb E[Y]=-pc+(1-p)/\lambda$ and $\mathbb E[Y^2]=pc^2+2(1-p)/\lambda^2$. The <law of total variance> equivalently gives $\operatorname{Var}(Y)=(1-p)/\lambda^2+p(1-p)(c+1/\lambda)^2$.
When $p<1$, conditioning on $Y\geq0$ selects exactly the exponential component, so $P(Y\geq t\mid Y\geq0)=e^{-\lambda t}$ for $t\geq0$. An ordinary <probability density function> for the continuous component integrates to $1-p$, not one; the missing mass is the atom, not a normalization error.
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