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: Recall the moments of the Gamma distribution.
For \(X\sim\text{Gamma}(\alpha,\lambda)\) with the given density, the mean and variance are
\[ E(X)=\frac{\alpha}{\lambda}, \qquad \text{Var}(X)=\frac{\alpha}{\lambda^2} \]
The method of moments equates population moments to sample moments and solves for the parameters.

Step 2: Write the first two sample moments.
The sample mean is
\[ \bar{x}=\frac{x_1+x_2+\cdots+x_n}{n}=2 \]
The second sample moment about the origin is
\[ m_2'=\frac{x_1^2+x_2^2+\cdots+x_n^2}{n}=5 \]

Step 3: Set up the moment equations.
Equating the first population moment to \(\bar{x}\):
\[ \frac{\alpha}{\lambda}=2 \quad\Rightarrow\quad \alpha=2\lambda \]
Since \(E(X^2)=\text{Var}(X)+[E(X)]^2\), equating it to \(m_2'\) gives
\[ \frac{\alpha}{\lambda^2}+\left(\frac{\alpha}{\lambda}\right)^2=5 \]
Using \(\alpha/\lambda=2\), this becomes
\[ \frac{\alpha}{\lambda^2}+4=5 \quad\Rightarrow\quad \frac{\alpha}{\lambda^2}=1 \]

Step 4: Solve the two equations together.
Substitute \(\alpha=2\lambda\) into \(\alpha/\lambda^2=1\):
\[ \frac{2\lambda}{\lambda^2}=1 \quad\Rightarrow\quad \frac{2}{\lambda}=1 \quad\Rightarrow\quad \lambda=2 \]
Then
\[ \alpha=2\lambda=2(2)=4 \]

Step 5: Check the answer.
With \(\alpha=4\) and \(\lambda=2\): mean \(=4/2=2\), which matches. Variance \(=4/4=1\), so second moment \(=1+4=5\), which also matches.

Final Answer:
The method of moments estimate of \(\alpha\) is \[ \boxed{\alpha=4} \]