Consider a random sample $(X_1, Y_1), \dots, (X_n, Y_n)$ from the bivariate normal distribution with $E(X_i) = \mu = E(Y_i), Var(X_i) = \sigma^2 = Var(Y_i)$ and $Cov(X_i, Y_i) = \rho\sigma^2$ for all $i$. Let $\hat{\mu}, \hat{\sigma}^2$ and $\hat{\rho}$ denote the maximum likelihood estimators of $\mu, \sigma^2$ and $\rho$ respectively. Also,
$S_X^2 = \sum_{i=1}^n (X_i - \bar{X})^2, S_Y^2 = \sum_{i=1}^n (Y_i - \bar{Y})^2$
and
$S_{XY} = \sum_{i=1}^n (X_i - \bar{X})(Y_i - \bar{Y})$.
Then
Analysis of Bivariate Normal Distribution Properties
The question asks to verify several statements related to a bivariate normal distribution with specific parameter constraints: $E(X_i) = E(Y_i) = \mu$, $Var(X_i) = Var(Y_i) = \sigma^2$, and $Cov(X_i, Y_i) = \rho\sigma^2$. We need to check the correctness of options A, B, C, and D.
Option A: Independence of Linear Combinations
We examine the independence of $U_1 = X_1 - Y_1$ and $V_1 = X_1 + Y_1$. For variables following a bivariate normal distribution, independence is equivalent to zero covariance.
- Calculate the covariance: $Cov(U_1, V_1) = Cov(X_1 - Y_1, X_1 + Y_1)$ $= Cov(X_1, X_1) + Cov(X_1, Y_1) - Cov(Y_1, X_1) - Cov(Y_1, Y_1)$ $= Var(X_1) + Cov(X_1, Y_1) - Cov(X_1, Y_1) - Var(Y_1)$ $= Var(X_1) - Var(Y_1)$
- Substitute the given variances: $Cov(U_1, V_1) = \sigma^2 - \sigma^2 = 0$
- Conclusion: Since the covariance is zero and the variables are from a bivariate normal distribution, $X_1 - Y_1$ and $X_1 + Y_1$ are independent. Option A is correct.
Option B: Maximum Likelihood Estimators (MLEs)
This option provides formulas for the MLEs: $\hat{\mu} = \frac{1}{2}(\bar{X} + \bar{Y})$, $\hat{\sigma}^2 = \frac{1}{2n}(S_X^2 + S_Y^2)$, and $\hat{\rho} = \frac{2S_{XY}}{S_X^2 + S_Y^2}$. These are standard MLEs for the specific symmetric bivariate normal distribution described.
- The MLE for the common mean $\mu$ is indeed $\hat{\mu} = \frac{\bar{X} + \bar{Y}}{2}$.
- The derivation for $\hat{\sigma}^2$ and $\hat{\rho}$ involves maximizing the likelihood function under the constraints $\sigma_X^2 = \sigma_Y^2 = \sigma^2$ and $Cov(X,Y) = \rho\sigma^2$. Using the sums of squares and cross-products ($S_X^2, S_Y^2, S_{XY}$) and relating them to the variances of $X_i-Y_i$ and $X_i+Y_i$, it can be shown that these formulas for $\hat{\sigma}^2$ and $\hat{\rho}$ are the correct MLEs.
- Specifically, $\hat{\sigma}^2 \approx \frac{1}{4n} (\hat{Var(X_i-Y_i)} + \hat{Var(X_i+Y_i)})$ and $\hat{\rho} \approx \frac{(\hat{Var(X_i+Y_i)} - \hat{Var(X_i-Y_i)})/4}{\hat{\sigma}^2}$, where the sample variances are calculated appropriately. This leads to the given formulas.
- Conclusion: Option B is correct.
Option C: Relation involving $\hat{\sigma}^2$ and $\hat{\rho}$ (Sum)
We verify the identity $\hat{\sigma}^2(1 + \hat{\rho}) = \frac{1}{2n}(S_X^2 + S_Y^2 + 2S_{XY})$ using the MLEs from Option B.
- Substitute $\hat{\sigma}^2$ and $\hat{\rho}$: $LHS = \left(\frac{S_X^2 + S_Y^2}{2n}\right) \left(1 + \frac{2S_{XY}}{S_X^2 + S_Y^2}\right)$
- Simplify the expression: $LHS = \left(\frac{S_X^2 + S_Y^2}{2n}\right) \left(\frac{S_X^2 + S_Y^2 + 2S_{XY}}{S_X^2 + S_Y^2}\right)$ $LHS = \frac{S_X^2 + S_Y^2 + 2S_{XY}}{2n}$
- Conclusion: The simplified LHS matches the RHS. Option C is correct.
Option D: Relation involving $\hat{\sigma}^2$ and $\hat{\rho}$ (Difference)
We verify the identity $\hat{\sigma}^2(1 - \hat{\rho}) = \frac{1}{2n}(S_X^2 + S_Y^2 - 2S_{XY})$ using the MLEs from Option B.
- Substitute $\hat{\sigma}^2$ and $\hat{\rho}$: $LHS = \left(\frac{S_X^2 + S_Y^2}{2n}\right) \left(1 - \frac{2S_{XY}}{S_X^2 + S_Y^2}\right)$
- Simplify the expression: $LHS = \left(\frac{S_X^2 + S_Y^2}{2n}\right) \left(\frac{S_X^2 + S_Y^2 - 2S_{XY}}{S_X^2 + S_Y^2}\right)$ $LHS = \frac{S_X^2 + S_Y^2 - 2S_{XY}}{2n}$
- Conclusion: The simplified LHS matches the RHS. Option D is correct.
Final Conclusion
Based on the analysis, statements A, B, C, and D are all correct properties and estimations related to the specified bivariate normal distribution.
Important Questions from Multivariate Normal Distribution
- Suppose $(X_1, X_2)$ follows a bivariate normal distribution with $E(X_1) = E(X_2) = 0$, $V(X_1) = V(X_2) = 2$ and $Cov(X_1, X_2) = -1$. If $\Phi(x) = \frac{1}{\sqrt{2\pi}} \int_{-\infty}^x e^{-y^2/2} dy$, then $P[X_1 - X_2 > 6]$ is equal to
Let $Y = (Y_1, \dots, Y_n)'$ have the multivariate normal distribution $N_n(0, I)$. Which of the following is the covariance matrix of the conditional distribution of $Y$ given $$\sum_{i=1}^n Y_i ?$$ ($1$ denotes the $n \times 1$ vector with all elements 1.)
Let $\Sigma = ((\sigma_{ij}))$ be an $n \times n$ symmetric and positive definite matrix such that $\sigma_{ij} \neq 0$ for all $i, j$. Which of the following matrices will always be the covariance matrix of a multivariate normal random vector?
- Suppose $X_1 \sim N_{p_1}(0, \Sigma_1), \ X_2 \sim N_{p_2}(0, \Sigma_2)$, where $X_1$ and $X_2$ are independently distributed. If $p_1 > p_2$ and $\Sigma_1, \Sigma_2$ are positive definite then which of the following statements is necessarily true?
- Let $X_1 \sim N(0, 1)$ and let
$$X_2 = \begin{cases} -X_1, & -2 \le X_1 \le 2 \\ X_1, & \text{otherwise}. \end{cases}$$
Then identify the correct statement.
