Question:

Let \( X \) be a random variable with probability density function \[ f(x; \lambda) = \begin{cases} \frac{1}{\lambda} e^{- \frac{x}{\lambda}} & \text{if } x > 0, \\ 0 & \text{otherwise} \end{cases} \] where \( \lambda > 0 \) is an unknown parameter. Let \( Y_1, Y_2, \dots, Y_n \) be a random sample of size \( n \) from a population having the same distribution as \( X^2 \). If \( \overline{Y} = \frac{1}{n} \sum_{i=1}^{n} Y_i \), then which one of the following statements is true?

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- The method of moments estimator is obtained by equating sample moments to population moments.
- For exponential distributions, \( E(X^2) = 2\lambda^2 \), which helps in estimating \( \lambda \).
Updated On: Aug 30, 2025
  • \( \sqrt{\overline{Y}/2} \) is a method of moments estimator of \( \lambda \)
  • \( \sqrt{\overline{Y}} \) is a method of moments estimator of \( \lambda \)
  • \( \frac{1}{2} \sqrt{\overline{Y}} \) is a method of moments estimator of \( \lambda \)
  • \( 2 \sqrt{\overline{Y}} \) is a method of moments estimator of \( \lambda \)
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The Correct Option is A

Solution and Explanation

1) Understanding the distribution of \( X \):
The given probability density function is for an exponential distribution with rate parameter \( \lambda \). The expected value \( E(X) \) is \( \lambda \). Since \( Y_i = X^2 \), the expected value of \( Y_i \) is \( E(Y_i) = E(X^2) \). For an exponential random variable \( X \), we have: \[ E(X^2) = \lambda^2 + \lambda^2 = 2\lambda^2 \] 2) Applying the method of moments:
The method of moments estimator is found by equating the sample moments to the population moments. The first moment of \( Y \) is \( 2\lambda^2 \). Hence, the method of moments estimator for \( \lambda \) is: \[ \hat{\lambda} = \sqrt{\frac{\overline{Y}}{2}} \]
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