Question

Let \(x_1,x_2,\ldots,x_n\) (\(n\ge2\)) be the observed values of a random sample from the following probability density function
\[f(x)=\begin{cases}\dfrac{\lambda^{\alpha}}{\Gamma(\alpha)}x^{\alpha-1}e^{-\lambda x} & \text{if } x>0\\0 & \text{otherwise,}\end{cases}\]
where \(\alpha\in(0,\infty)\) and \(\lambda\in(0,\infty)\) are unknown parameters. If
\[\frac{x_1+x_2+\cdots+x_n}{n}=2 \quad\text{and}\quad \frac{x_1^2+x_2^2+\cdots+x_n^2}{n}=5,\]
then the method of moments estimate of \(\alpha\) equals ______ (answer in integer).

Hint:

Hint:
Equate the population mean \(\alpha/\lambda\) and second moment \(\alpha/\lambda^2+(\alpha/\lambda)^2\) to the given sample moments \(2\) and \(5\), then solve the two equations together.

Answer 1

Correct Answer: 4

Step 1: Use the sample mean and sample variance instead of raw moments.
The sample mean is $\bar{x}=2$. The sample variance about the mean is
\[ s^2=m_2'-\bar{x}^2=5-4=1 \]

Step 2: Match these to the Gamma parameters.
For the Gamma density, $\bar{x}$ estimates $\alpha/\lambda$ and $s^2$ estimates $\alpha/\lambda^2$. So
\[ \frac{\alpha}{\lambda}=2, \qquad \frac{\alpha}{\lambda^2}=1 \]

Step 3: Take the ratio of the two equations.
Dividing the variance equation by the mean equation cancels $\alpha$:
\[ \frac{\alpha/\lambda^2}{\alpha/\lambda}=\frac{1}{2} \quad\Rightarrow\quad \frac{1}{\lambda}=\frac{1}{2} \quad\Rightarrow\quad \lambda=2 \]

Step 4: Recover alpha from the mean equation.
\[ \alpha=2\lambda=2\times2=4 \]

Step 5: Confirm with the variance.
With $\alpha=4,\lambda=2$: $\alpha/\lambda^2=4/4=1$, matching $s^2=1$ exactly.
\[ \boxed{\alpha=4} \]