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Question about an exponential Bernoulli distribution

Cross Validated Asked by Ad van der Ven on January 11, 2021

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:
$$displaystyle 1-p+int_{0}^{infty }!p , lambda,{{rm e}^{-lambda,t}},{rm d}t=1-p+pint_{0}^{infty }!lambda,{{rm e}^{-lambda,t}},{rm d}t , = , 1$$
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.

One Answer

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 $lim_{alpharightarrow0}alpha e^{-alpha t}=0$, the resulting $lim_{alpha rightarrow 0} f(t) = 0 + p (0+beta)e^{-(0+beta)t} = pbeta e^{-beta t}$ 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.

Answered by Edgar on January 11, 2021

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