Mixture of a point mass and an exponential distribution ($\mathcal{L}(Y)=p{\delta }_{-c}+(1-p)\mathrm{Exp}(\lambda )$)

Let $c>0$, $0\le p\le 1$ 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 $\mathbb{E}[Y^2]=p{c}^2+2(1-p)/{\lambda }^2$. The law of total variance equivalently gives $\mathrm{Var}(Y)=(1-p)/{\lambda }^2+p(1-p)(c+1/\lambda {)}^2$.

When $p<1$, conditioning on $Y\ge 0$ selects exactly the exponential component, so $P(Y\ge t\mid Y\ge 0)=e^{-\lambda t}$ for $t\ge 0$. 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.