Variance of an Exponentiated Gamma Random Variable
Let $X \sim \text{Gamma}(a, b)$ with shape parameter $a$ and scale parameter $b$. Compute $\text{Var}(e^X)$ when $a = 4$ and $b = 1/4$.
Round your answer to the nearest thousandth.
Hints
- When you need moments of $e^X$, think about what function is defined as $E[e^{\theta X}]$.
- Write $\text{Var}(e^X) = M_X(2) - (M_X(1))^2$ using the Gamma MGF $M_X(\theta) = (1 - b\theta)^{-a}$. Check that the MGF exists at $\theta = 2$.
- Plug in $a = 4$, $b = 1/4$: compute $(1 - 1/2)^{-4}$ and $((1 - 1/4)^{-4})^2$ separately, then subtract.
Worked Solution
How to Think About It: Whenever you see $e^X$ for some random variable $X$, your first reflex should be moment generating functions. The MGF is literally $M_X(\theta) = E[e^{\theta X}]$, so $E[e^X] = M_X(1)$ and $E[e^{2X}] = M_X(2)$. That is all the machinery you need. The only subtlety is checking that the MGF is actually finite at the values you need -- for Gamma this constrains $\theta < 1/b$.
Quick Estimate: The Gamma$(4, 1/4)$ distribution has mean $ab = 1$ and variance $ab^2 = 1/4$, so $X$ is concentrated around 1 with moderate spread. A rough lower bound: if $X$ were exactly 1, then $e^X = e \approx 2.72$ and the variance would be 0. The spread in $X$ inflates $E[e^{2X}]$ much more than $(E[e^X])^2$ because the exponential amplifies the right tail. The MGF of Gamma$(4,1/4)$ at $\theta = 1$ is $(3/4)^{-4} \approx 3.16$, so $E[e^X] \approx 3.16$. At $\theta = 2$, it is $(1/2)^{-4} = 16$. So $\text{Var}(e^X) \approx 16 - 3.16^2 \approx 16 - 9.99 \approx 6.0$. We should land close to 6.
Approach: Use the variance identity $\text{Var}(e^X) = E[e^{2X}] - (E[e^X])^2$ and express each term via the Gamma MGF.
Formal Solution:
The variance identity gives
$$\text{Var}(e^X) = E[(e^X)^2] - (E[e^X])^2 = E[e^{2X}] - (E[e^X])^2 = M_X(2) - (M_X(1))^2$$
The MGF of a $\text{Gamma}(a, b)$ random variable (with scale parameterization) is
$$M_X(\theta) = (1 - b\theta)^{-a}, \quad \theta < \frac{1}{b}$$
We need $M_X(2)$, which requires $2 < 1/b = 4$. This holds, so we proceed.
With $a = 4$ and $b = 1/4$:
$$M_X(1) = \left(1 - \frac{1}{4}\right)^{-4} = \left(\frac{3}{4}\right)^{-4} = \left(\frac{4}{3}\right)^4 = \frac{256}{81}$$
$$M_X(2) = \left(1 - \frac{2}{4}\right)^{-4} = \left(\frac{1}{2}\right)^{-4} = 2^4 = 16$$
Therefore
$$\text{Var}(e^X) = 16 - \left(\frac{256}{81}\right)^2 = 16 - \frac{65536}{6561} = \frac{104976 - 65536}{6561} = \frac{39440}{6561} \approx 6.011$$
Answer: $\text{Var}(e^X) = 16 - (4/3)^8 = 39440/6561 \approx 6.011$.
Intuition
The moment generating function is one of the most underused tools in probability interviews. People memorize it as an abstract transform, but its power is concrete: any time you need $E[e^{\theta X}]$ for specific $\theta$, you are literally just evaluating the MGF at a point. The variance of $e^X$ decomposes into $M_X(2) - (M_X(1))^2$, which turns a potentially ugly integral into plugging two numbers into a closed-form expression. This trick generalizes -- the $k$-th moment of $e^X$ is $M_X(k)$, so if you know the MGF in closed form, you have all moments of $e^X$ for free.
The domain constraint is the subtle part that separates good answers from great ones. The Gamma MGF $(1 - b\theta)^{-a}$ blows up at $\theta = 1/b$, so if $b \geq 1/2$ then $M_X(2)$ does not exist and $\text{Var}(e^X)$ is infinite. This is not a technicality -- it reflects the heavy right tail of the Gamma distribution. When the scale parameter is large enough, $e^X$ has such extreme upside that its second moment diverges. In practice, this is exactly the kind of thing that matters when modeling portfolio returns or insurance losses: the exponential of a heavy-tailed variable can have infinite variance even when the underlying variable is perfectly well-behaved.