Question about an exponential Bernoulli distribution

Question

Mohie El-Din and Amein (2011) define a distribution in formula (1.2) which they call the exponential Bernoulli distribution (EBD). The distribution has the following form: $$\displaystyle f \left(t \right) = \left(1-p \right)~\alpha ~e^{-\alpha ~t }+p~\left(\alpha +\beta \right)~e^{-\left(\alpha +\beta \right)~t }$$ or with $\lambda = \alpha+\beta$ the distribution can also be written as : $$\displaystyle f \left(t \right) = \left(1-p \right)~\alpha ~e^{-\alpha ~t }+p~\lambda ~e^{-\lambda ~t }$$ In Mohie El-Din and Amein (2011) these densities are written in $x$. The logic behind this distribution is as follows. The Bernoulli variable $X$ with $X = 0, 1$ has a probability distribution with $P (X = 1) = p$ and $P (X = 0) = (1-p)$. Furthermore, there is a random variable $T$ with $0 \leq T$. If $X = 1$, then $T $ has an exponential distribution with rate parameter $\lambda$. If $X = 0$, then $T $ has an exponential distribution with rate parameter $\alpha$.

Now I have another case. In my case the following applies. If $X = 1$ then $T$ also has an exponential distribution: $$\displaystyle f \left(t \right) = p ~\lambda ~e^{-\lambda ~t }$$ But if $X = 0$, then $T = 0$. The final distribution is a proper probability distribution because: $$\displaystyle 1-p+\int_{0}^{\infty }\!p \, \lambda\,{{\rm e}^{-\lambda\,t}}\,{\rm d}t=1-p+p\int_{0}^{\infty }\!\lambda\,{{\rm e}^{-\lambda\,t}}\,{\rm d}t \, = \, 1$$ The expectation of $T$ can be written as: $$\displaystyle \mu_{{1}}\, = 0 \, (1-p) \, + \, \int_{0}^{\infty }\!p \, t \, \lambda\,{{\rm e}^{-\lambda\,t}}\,{\rm d}t \,= \, \frac {p}{\lambda}$$ The variance of $T$ can be written as: $$\displaystyle \mu_{{2}}\, = \, (1-p) \, \left(0-\frac{p}{\lambda}\right) ^2 +\int_{0}^{\infty }\!p \, \left( t-{\frac {p}{\lambda}} \right) ^{2}\lambda\,{{\rm e}^{-\lambda\,t}}\,{\rm d}t \, = \,-{\frac {p \left( p-2 \right) }{{\lambda}^{2}}}$$ The third central moment can be written as: $$\displaystyle \mu_{{3}}\, = \, (1-p) \, \left(0-\frac{p}{\lambda}\right) ^3 +\int_{0}^{\infty }\!p \, \left( t-{\frac {p}{\lambda}} \right) ^{3}\lambda\,{{\rm e}^{-\lambda\,t}}\,{\rm d}t \, = \,2\,{\frac {p \left( {p}^{2}-3\,p+3 \right) }{{\lambda}^{3}}}$$ I now have three questions. The first question is: what is the best way to name this distribution? The second question is: how can I best write down this distribution in a formula? The third question is: are the derivations of the three given central moments correct?

Reference

Mohie El-Din, M. M. and Amein, M. M. (2011). Estimation of Parameters of the Exponential Bernoulli Distribution Based on Progressively Censored Data. Applied Mathematical Sciences, Vol. 5, no. 58, 2883 - 2890.

Answer 1

I looked into the paper and they define the EBD density function for $\alpha, \beta \geq 0$, which would include your case.

But for $\alpha \rightarrow 0$ the right tail of the distribution gets fatter. The mean of the EBD is $\frac{p}{\alpha+\beta} + \frac{1-p}{\alpha}$, which obviously doesn't exist for $\alpha=0$.

Notice that you can always give the cumulative distribution function for the EBD (see the paper you mentioned), but you cannot define a distribution density function for the case $\alpha=0$. Even though $\lim_{\alpha\rightarrow0}\alpha e^{-\alpha t}=0$, the resulting $\lim_{\alpha \rightarrow 0} f(t) = 0 + p (0+\beta)e^{-(0+\beta)t} = p\beta e^{-\beta t}$ is not a density function (integral is not equal to $1$).

You can still calculate the moments with the cdf (see here), though! Which gives the same central moments like your calculations.

Reagrding the EBD, see also a similar question here.