Consider a r.v. X with MGF $m(t) = \frac{e^{k(e^t-1)}}{(1-bt)^a} $
$(a,b,k) ∈ \Bbb{R}$
Differentiating twice we retrieve $m''(t) = (a+1)ab^2e^{k(e^t-1)}(1-bt)^{-(a+2)}+2abke^{k(e^t-1)+t}(1-bt)^{-(a+1)}+ke^{k(e^t-1)+t}(ke^t+1)(1-bt)^{-a}$
Evaluating at t=0, we retrieve $VarX$
$VarX =m''(0)=a^2b^2+ab^2+2abk+k^2+k$
Now we observe that $m(t) =e^{k(e^t-1)}(1-bt)^{-a} = m_y(t)m_z(t) $
and our original r.v. $X = Y+ Z$
For two r.v.'s $Y$ and $Z$, where $Y$~$Pois(k)$ and $Z$~$\Gamma(a,b)$
(From the reproductive property of MGF's and comparing our equation with commonly known MGFs)
Now since $X=Y+Z$, we can again find the variance:
$Var(X)=Var(Y+Z)=Var(Y)+Var(Z)=ab^2+k$
As $Var(Y) = k$ and $Var(Z) = ab^2$
Which contradicts the first expression we found for the variance.
Can anyone please explain to me find the error I have made?
The differentiation seems like the obvious candidate for my error, but I have checked a few times and also done it using Wolfram Alpha.
https://www.wolframalpha.com/input/?i=second+derivative+of+(e%5E(11(e%5Et-1))*(1-bt)%5E(-a)
Note that I used 11 instead of k and then just put k back in where multiples of 11 appeared, because Wolfram doesn't seem to allow for 3 constants when it differentiates.
I also have assumed that Y and Z are independent implicitly, but I am unsure of how to test this.
variance of X is not equal to value of second order derivative of mgf at t= 0.In fact value of nth order derivative of mgf is nth order moment about origin not about mean. so you have calculated E(X^2) rather than variance of X.That's why contradiction took place.