Question

Let \(X_1,X_2,X_3,X_4\) be a random sample of size \(4\) from a distribution with the probability density function

Hint:

In the method of moments, equate sample moments with corresponding population moments and solve for the unknown parameter.

Answer 1

Correct Answer: 0.67

Step 1: Find the sample mean.
The observed data are
\[ 0.2,\;0.1,\;0.3,\;0.4 \] So,
\[ \bar{x}=\frac{0.2+0.1+0.3+0.4}{4} \] \[ =\frac{1.0}{4} \] \[ =0.25 \]

Step 2: Find the theoretical mean.
\[ E(X)=\int_0^1 x\alpha(\alpha+1)x^{\alpha-1}(1-x)\,dx \] \[ =\alpha(\alpha+1)\int_0^1 x^\alpha(1-x)\,dx \] \[ =\alpha(\alpha+1)\left[\int_0^1 x^\alpha\,dx-\int_0^1 x^{\alpha+1}\,dx\right] \] \[ =\alpha(\alpha+1)\left[\frac{1}{\alpha+1}-\frac{1}{\alpha+2}\right] \] \[ =\alpha(\alpha+1)\cdot \frac{1}{(\alpha+1)(\alpha+2)} \] \[ =\frac{\alpha}{\alpha+2} \]

Step 3: Apply method of moments.
Equate the theoretical mean to the sample mean:
\[ \frac{\alpha}{\alpha+2}=0.25 \] \[ \frac{\alpha}{\alpha+2}=\frac14 \] \[ 4\alpha=\alpha+2 \] \[ 3\alpha=2 \] \[ \alpha=\frac23 \] \[ \alpha=0.666\ldots \] Rounded off to two decimal places,
\[ \alpha=0.67 \]

Step 4: Final conclusion.
Hence, the method of moments estimate of \(\alpha\) is
\[ \boxed{0.67} \]