I'm wondering how to get variance of exp. distribution from the raw variance computed using the moment generating function. Here's my line of reasoning:
PDF of Exponential distriution is
$$ p_X(x) = \lambda \cdot e^{-\lambda x} $$
for $x > 0$, and $0$ for $x \leq 0$.
Deriving the MGF:
$$ \begin{aligned} M_X(t) &= \mathbb{E}\left[e^{t X}\right] && \text{definition} \\ &= \int_{- \infty}^{\infty} x \cdot p_X(x) dx&& \text{just definition of expectation} \\ &= \int_{- \infty}^{\infty} e^{t x} \cdot \lambda e^{-\lambda x} dx&& \text{LOTUS} \\ &= \int_{0}^{\infty} e^{t x} \cdot \lambda e^{-\lambda x} dx&& \text{since } x > 0 \\ &= \lambda \int_{0}^{\infty} e^{t x} \cdot e^{-\lambda x} dx&& \text{the constant multiple rule} \\ &= \lambda \int_{0}^{\infty} e^{t x -\lambda x} dx \\ &= \lambda \int_{0}^{\infty} e^{x (t -\lambda )} dx \\ &= \lambda \cdot \frac{1}{\lambda - t} && \text{closed form solution for } t < \lambda \\ &= \frac{\lambda}{\lambda - t} \qquad \boxed{\checkmark} \text{ Wikipedia check} \end{aligned} $$
Getting moments of exponential distributions by derivating MGF$$ M_X(t) = \frac{\lambda}{\lambda - t} $$
First moment (expectation)
$$ M_X^{(1)}(t) = \frac{\partial}{\partial t} \left( \frac{\lambda}{\lambda - t} \right) = \frac{\lambda}{(\lambda - t)^2} $$
- And evaluate at $t = 0$:
$$ \frac{\lambda}{(\lambda - t)^2} \bigg\vert_{t=0} = \frac{\lambda}{\lambda^2} = \frac{1}{\lambda} \qquad \boxed{\checkmark} \text{ Wikipedia check} $$
Second moment
$$ M_X^{(2)}(t) = \frac{\partial^2}{\partial^2 t} \left( \frac{\lambda}{\lambda - t} \right) = \frac{2 \lambda}{(\lambda - t)^3} $$
$$ \frac{2 \lambda}{(\lambda - t)^3} \bigg\vert_{t=0} = \frac{2}{\lambda^2} $$
So this is raw variance but not the actual variance $\frac{1}{\lambda^2}$... how to get there?
