Discussion
Let \(X\) be a random variable having the moment generating function \(M_{X}(t) = e^{2(e^{t}-1)},\ t \in \mathbb{R}\). Then, \(Var(2X + 3)\) equals
Answer: D
The variance is 8.
The moment generating function \(e^{\lambda(e^{t} - 1)}\) belongs to the Poisson distribution with mean \(\lambda\). Here \(\lambda = 2\), so \(X \sim\) Poisson(2) and \(Var(X) = 2\).
Adding a constant does not change variance, and multiplying by 2 multiplies it by \(2^{2}\):
\[ Var(2X + 3) = 4\,Var(X) = 4 \times 2 = 8 \]
16 squares the variance as well as the coefficient. 4 is \(2^{2}\) with \(Var(X)\) forgotten, and 6 adds the constant 3 to \(2\,Var(X)\), but a shift has no effect on spread.
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