Question

If G is a geometric mean of observations $x_{1}, x_{2}, \dots, x_{n}$ then the geometric mean of $y_{i} = e^{-\alpha \log_{e} x_{i}}$, $i = 1, 2, \dots, n$ is

• A. $\sqrt[\alpha]{G}$
• B. $G^{\alpha}$
• C. G
• D. $G^{-\alpha}$

Hint:

A useful property of the Geometric Mean is that it is "power-preserving." If every data point is raised to a power $k$, the G.M. of the new data is the original G.M. raised to that same power $k$. Here, the complex $e^{-\alpha \ln x}$ is just $x^{-\alpha}$, so the result is $G^{-\alpha}$ immediately.

Answer 1

The Correct Option is D

We are tasked with finding the Geometric Mean (G.M.) of a transformed set of observations $y_i$ based on the G.M. of the original set $x_i$.

Step 1: \color{redDefine the Geometric Mean of x
The geometric mean $G$ of observations $x_{1}, x_{2}, \dots, x_{n}$ is defined by the formula:
$G = (x_{1} \cdot x_{2} \cdot \dots \cdot x_{n})^{1/n}$.
Alternatively, in logarithmic form:
$\log_{e} G = \frac{1}{n} \sum_{i=1}^{n} \log_{e} x_{i}$.

Step 2: \color{redSimplify the Transformation for y
The transformation given is $y_{i} = e^{-\alpha \log_{e} x_{i}}$.
Using the logarithmic power property, $-\alpha \log_{e} x_{i} = \log_{e} (x_{i}^{-\alpha})$.
Substituting this into the exponent:
$y_{i} = e^{\log_{e} (x_{i}^{-\alpha})}$.
Since the exponential and natural log functions are inverses, $e^{\log_{e} w} = w$.
Thus, $y_{i} = x_{i}^{-\alpha}$.

Step 3: \color{redSetup the Geometric Mean Formula for y
Let $G_{y}$ be the geometric mean of the observations $y_{i}$.
$G_{y} = (y_{1} \cdot y_{2} \cdot \dots \cdot y_{n})^{1/n}$.
Substituting $y_{i} = x_{i}^{-\alpha}$ into this formula:
$G_{y} = (x_{1}^{-\alpha} \cdot x_{2}^{-\alpha} \cdot \dots \cdot x_{n}^{-\alpha})^{1/n}$.

Step 4: \color{redApply Laws of Exponents to Finalize Result
Using the property that $(a \cdot b \cdot \dots)^{k} = a^{k} \cdot b^{k} \cdot \dots$:
$G_{y} = [(x_{1} \cdot x_{2} \cdot \dots \cdot x_{n})^{-\alpha}]^{1/n}$.
By the power of a power rule, $(A^{m})^{n} = A^{m \cdot n}$, so we can swap the exponents:
$G_{y} = [(x_{1} \cdot x_{2} \cdot \dots \cdot x_{n})^{1/n}]^{-\alpha}$.
The term inside the square brackets is exactly the definition of $G$.
Therefore, $G_{y} = G^{-\alpha}$.
This corresponds to Option (4).