Question:

One million random numbers are generated from a statistically stationary process with a Gaussian distribution with mean zero and standard deviation $\sigma_{0$. The $\sigma_{0}$ is estimated by randomly drawing out 10,000 samples $(x_{n})$. The estimates $\hat{\sigma}_{1}, \hat{\sigma}_{2}$ are computed in the following two ways:} \[ \hat{\sigma}_{1}^{2} = \frac{1}{10000}\sum_{n=1}^{10000} x_{n}^{2}, \hat{\sigma}_{2}^{2} = \frac{1}{9999}\sum_{n=1}^{10000} x_{n}^{2} \] Which of the following statements is true?

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When estimating variance, dividing by $N$ gives an unbiased estimator for zero-mean Gaussian signals. Dividing by $N-1$ is used in unbiased sample variance estimation when the mean is unknown and also estimated from data.
Updated On: Aug 28, 2025
  • $E(\hat{\sigma}_{2}^{2}) = \sigma_{0}^{2}$
  • $E(\hat{\sigma}_{2}) = \sigma_{0}$
  • $E(\hat{\sigma}_{1}^{2}) = \sigma_{0}^{2}$
  • $E(\hat{\sigma}_{1}) = E(\hat{\sigma}_{2})$
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The Correct Option is C

Solution and Explanation

Step 1: Recall definition of variance for Gaussian random variables.
For a Gaussian random variable with mean zero and standard deviation $\sigma_{0}$, \[ E[x_{n}^{2}] = \sigma_{0}^{2}. \]

Step 2: Expectation of $\hat{\sigma_{1}^{2}$.}
\[ \hat{\sigma}_{1}^{2} = \frac{1}{10000}\sum_{n=1}^{10000} x_{n}^{2}. \] Taking expectation: \[ E[\hat{\sigma}_{1}^{2}] = \frac{1}{10000}\sum_{n=1}^{10000} E[x_{n}^{2}] = \frac{1}{10000} \times 10000 \times \sigma_{0}^{2} = \sigma_{0}^{2}. \] Thus, $\hat{\sigma}_{1}^{2}$ is an unbiased estimator of variance.

Step 3: Expectation of $\hat{\sigma_{2}^{2}$.}
\[ \hat{\sigma}_{2}^{2} = \frac{1}{9999}\sum_{n=1}^{10000} x_{n}^{2}. \] Taking expectation: \[ E[\hat{\sigma}_{2}^{2}] = \frac{1}{9999} \times 10000 \times \sigma_{0}^{2} = \frac{10000}{9999}\sigma_{0}^{2}. \] This is slightly larger than $\sigma_{0}^{2}$, so $\hat{\sigma}_{2}^{2}$ is a biased estimator.

Step 4: Check the given options.
- (A) False, since $E[\hat{\sigma}_{2}^{2}] \neq \sigma_{0}^{2}$. - (B) False, since expectation of $\hat{\sigma}_{2}$ (square root form) is not equal to $\sigma_{0}$. - (C) True, because $E[\hat{\sigma}_{1}^{2}] = \sigma_{0}^{2}$. - (D) False, because $\hat{\sigma}_{1}$ and $\hat{\sigma}_{2}$ are computed differently. % Final Answer \[ \boxed{\text{Option (C): } E(\hat{\sigma}_{1}^{2}) = \sigma_{0}^{2}} \]

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