GATE 2023 EE Q.25: solution and answer

Official answer: CSource: Official answer key, GATE 2023 EE (IIT Kanpur)

1. Every value drawn comes from the Gaussian with mean $0$ and variance ${\sigma }_o^2$. So for each sample $$E(x_n^2)=\text{Var}(x_n)+{\left(E[x_n]\right)}^2={\sigma }_o^2+0={\sigma }_o^2$$

2. (C). By linearity of expectation (this needs no independence between the draws): $$E({\hat{\sigma }}_1^2)=\frac{1}{10000}\sum _{n=1}^{10000}E(x_n^2)=\frac{10000{\sigma }_o^2}{10000}={\sigma }_o^2$$ So (C) is true, and ${\hat{\sigma }}_1^2$ is an unbiased estimator of ${\sigma }_o^2$.

3. (A). The same step for ${\hat{\sigma }}_2^2$ gives $$E({\hat{\sigma }}_2^2)=\frac{10000{\sigma }_o^2}{9999}\ne {\sigma }_o^2$$ so (A) is false. Dividing by $N-1$ (Bessel's correction) removes the bias only when the sample mean $\bar{x}$ is subtracted, as in $\sum (x_n-\bar{x}{)}^2$. Here the mean is known to be zero and nothing is subtracted, so dividing by $9999$ adds bias.

4. (D). For every set of samples, ${\hat{\sigma }}_2=\sqrt{\frac{10000}{9999}}{\hat{\sigma }}_1$. So $E({\hat{\sigma }}_2)=\sqrt{\frac{10000}{9999}}E({\hat{\sigma }}_1)$. Since ${\hat{\sigma }}_1\ge 0$ and is not always zero, $E({\hat{\sigma }}_1)>0$, and so $E({\hat{\sigma }}_2)\ne E({\hat{\sigma }}_1)$. (D) is false.

5. (B), exam shortcut. Exactly one option is correct, and (C) has been proved true exactly, so (B) must be false. Linearity cannot settle (B) by itself, because the square root is not linear: $E({\hat{\sigma }}_2)\ne \sqrt{E({\hat{\sigma }}_2^2)}$ in general. Jensen's inequality gives $E({\hat{\sigma }}_1)<{\sigma }_o$, but the factor $\sqrt{10000/9999}>1$ pushes the other way. The next two steps settle it exactly for completeness.

6. (B), exact. Treat the $N=10000$ draws as independent. Then $Q_N=\frac{1}{{\sigma }_o^2}{\sum }_{n=1}^N{x}_n^2$ is chi-square with $N$ degrees of freedom, with density $\frac{q^{N/2-1}{e}^{-q/2}}{2^{N/2}\Gamma (N/2)}$ for $q>0$. Its mean square root is $$g(N)=E\left(\sqrt{Q_N}\right)={\int }_0^{\infty }\frac{q^{(N+1)/2-1}{e}^{-q/2}}{2^{N/2}\Gamma (N/2)}dq=\frac{\sqrt{2}\Gamma \left(\frac{N+1}{2}\right)}{\Gamma \left(\frac{N}{2}\right)}$$ and ${\hat{\sigma }}_2={\sigma }_o\sqrt{Q_N/(N-1)}$, so $E({\hat{\sigma }}_2)=\frac{{\sigma }_{o}g(N)}{\sqrt{N-1}}$.

7. Using $\Gamma (z+1)=z\Gamma (z)$ with $z=\frac{N-1}{2}$: $$g(N-1)g(N)=\frac{2\Gamma \left(\frac{N+1}{2}\right)}{\Gamma \left(\frac{N-1}{2}\right)}=2\cdot \frac{N-1}{2}=N-1$$ Let $Q_{N-1}$ be the sum of the first $N-1$ of the same standardised squares; it is chi-square with $N-1$ degrees of freedom and $Q_N$ adds one more square, which is positive with probability 1. So $\sqrt{Q_N}>\sqrt{Q_{N-1}}$ and $g(N)>g(N-1)$. Hence $g(N{)}^2>g(N-1)g(N)=N-1$, so $g(N)>\sqrt{N-1}$ and $E({\hat{\sigma }}_2)>{\sigma }_o$. Numerically $E({\hat{\sigma }}_2)\approx 1.000025{\sigma }_o$ (and $E({\hat{\sigma }}_1)\approx 0.999975{\sigma }_o$). (B) is false.

8. Only (C) holds. Answer: (C).