Let \(X_1\) and \(X_2\) be independent and identically distributed random variables following normal distribution with mean \(\theta\in(-\infty,\infty)\) and variance \(1\). Then which of the following estimators of their expected values attains the Cramer-Rao lower bound?
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An estimator attains the Cramer-Rao bound only if it is an affine function of the score. Since the score is \((x_1+x_2)-2\theta\), only \(X_1+X_2\) matches it exactly.
Step 1: Recall what attaining the Cramer-Rao bound really means. An estimator hits the Cramer-Rao bound exactly when it is a constant times the score function of the likelihood, plus a shift. For a normal model with known variance, this happens only for statistics that are linear combinations of the data with equal weights on every observation.
Step 2: Write the score function for this specific model. With $X_1,X_2\sim N(\theta,1)$ independently, \[ \frac{\partial}{\partial\theta}\log L(\theta)=(x_1-\theta)+(x_2-\theta)=(x_1+x_2)-2\theta \] This is a straight line in $x_1+x_2$, with slope $1$ and intercept $-2\theta$.
Step 3: Match option (A) against this score. Option (A) is $T=X_1+X_2$. Subtracting its mean $\psi(\theta)=2\theta$ gives $(x_1+x_2)-2\theta$, which is identical to the score function from Step 2. Whenever an estimator minus its own mean equals a constant times the score, the Cramer-Rao inequality becomes an equality. So $X_1+X_2$ attains the bound.
Step 4: Confirm with a direct variance computation. Fisher information per observation for $N(\theta,1)$ is $1$, so for two independent observations, $I_2(\theta)=2$. Since $\psi(\theta)=2\theta$ has derivative $\psi'(\theta)=2$, the Cramer-Rao bound for $T$ is \[ \frac{(\psi'(\theta))^2}{I_2(\theta)}=\frac{4}{2}=2 \] The actual variance is $Var(X_1+X_2)=1+1=2$, an exact match.
Step 5: Check option (B), the quadratic mix $\dfrac{X_1^2+X_2^2+5}{2}$. Subtracting its own mean does not give anything proportional to $(x_1+x_2)-2\theta$, since squared terms cannot be written as a constant multiple of a linear expression. So this cannot achieve equality.
Step 6: Check option (C), the squared sum $(X_1+X_2)^2$. Even though this is built only from $X_1+X_2$, squaring it makes it a quadratic function, not a straight line, of the score. Only the straight line case gives equality.
Step 7: Check option (D), the mismatched combination $2X_1-X_2$. Here the weights on $X_1$ and $X_2$ are $2$ and $-1$, not equal, so this is not a constant multiple of $(x_1+x_2)-2\theta$. A quick variance check confirms it: $Var(2X_1-X_2)=4+1=5$, while estimating $\theta$ needs a bound of $1/2$. Since $5$ is nowhere near $1/2$, this estimator is far from efficient.
Step 8: Conclude. Only the equal weighted sum $X_1+X_2$ lines up exactly with the score function. \[ \boxed{X_1+X_2\text{ attains the Cramer-Rao lower bound}} \]
