Question

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?

• A. \(X_1+X_2\)
• B. \(\dfrac{X_1^2+X_2^2+5}{2}\)
• C. \((X_1+X_2)^2\)
• D. \(2X_1-X_2\)

Hint:

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.

Answer 1

The Correct Option is A

Step 1: Recall the Cramer-Rao lower bound for a possibly biased estimator.
For an estimator \(T\) of a parameter \(\theta\), with \(\psi(\theta)=E_\theta[T]\), the general Cramer-Rao inequality states \[ Var_\theta(T)\ \ge\ \frac{\left[\psi'(\theta)\right]^2}{I_n(\theta)}, \] where \(I_n(\theta)\) is the Fisher information in the whole sample. Equality holds exactly when \(T\) is an affine function of the score, that is, \[ T-\psi(\theta)=k(\theta)\cdot\frac{\partial}{\partial\theta}\log L(\theta), \] for some function \(k(\theta)\) not depending on the data.

Step 2: Write the likelihood and the score for this problem.
Here \(X_1,X_2\) are iid \(N(\theta,1)\), so the log likelihood is \[ \log L(\theta)=-\log(2\pi)-\frac{1}{2}\left[(x_1-\theta)^2+(x_2-\theta)^2\right], \] and the score function is \[ \frac{\partial}{\partial\theta}\log L(\theta)=(x_1-\theta)+(x_2-\theta)=(x_1+x_2)-2\theta. \]

Step 3: Find the Fisher information.
For a single \(N(\theta,1)\) observation the Fisher information is \(1\). With \(n=2\) independent observations, \[ I_2(\theta)=2. \]

Step 4: Test option (A), \(T=X_1+X_2\).
We can see directly that \[ T-\psi(\theta)=(x_1+x_2)-2\theta=\frac{\partial}{\partial\theta}\log L(\theta), \] where \(\psi(\theta)=E[T]=2\theta\). This is exactly the equality condition from Step 1, with \(k(\theta)=1\). So \(T=X_1+X_2\) attains the Cramer-Rao lower bound for estimating its own mean, \(2\theta\). As a direct check: \[ Var(X_1+X_2)=Var(X_1)+Var(X_2)=1+1=2, \] and the bound is \[ \frac{[\psi'(\theta)]^2}{I_2(\theta)}=\frac{2^2}{2}=2. \] The variance equals the bound exactly.

Step 5: Test option (B), \(T=\dfrac{X_1^2+X_2^2+5}{2}\).
This involves \(X_1^2\) and \(X_2^2\), which are not affine functions of the sufficient statistic \(X_1+X_2\). Since \(T-\psi(\theta)\) cannot be written as a constant multiple of the score, the equality condition fails and this estimator's variance exceeds its Cramer-Rao bound. So (B) does not attain the bound.

Step 6: Test option (C), \(T=(X_1+X_2)^2\).
Although built from the sufficient statistic \(X_1+X_2\), it is a quadratic function of it, not an affine one. Only affine functions of the score satisfy the equality condition, so a squared term breaks the equality. So (C) does not attain the bound.

Step 7: Test option (D), \(T=2X_1-X_2\).
This is linear in \(x_1\) and \(x_2\), but with unequal coefficients \(2\) and \(-1\), so it is not proportional to the score, which needs equal coefficients on \(x_1\) and \(x_2\). Checking directly: \[ E(2X_1-X_2)=2\theta-\theta=\theta, \qquad Var(2X_1-X_2)=4(1)+1(1)=5. \] The Cramer-Rao bound for estimating \(\theta\) itself is \[ \frac{1^2}{2}=\frac{1}{2}. \] Since \(5\) is far above \(1/2\), this estimator does not come close to attaining the bound.

Final Answer:
Only the statistic proportional to the score function, \(X_1+X_2\), achieves equality in the Cramer-Rao inequality. \[ \boxed{X_1+X_2\text{ attains the Cramer-Rao lower bound}} \]