Bivariate Normal Distribution Probability
This solution calculates $P[X_1 - X_2 > 6]$ for a bivariate normal distribution $(X_1, X_2)$ using its statistical properties.
Distribution Parameters
The given parameters are:
- Means: $E(X_1) = 0$, $E(X_2) = 0$
- Variances: $V(X_1) = 2$, $V(X_2) = 2$
- Covariance: $Cov(X_1, X_2) = -1$
Linear Combination $X_1 - X_2$
Let $Y = X_1 - X_2$. A linear combination of variables from a bivariate normal distribution is itself normally distributed.
Mean and Variance of Y
Calculate the mean and variance of $Y$:
- Mean: $E(Y) = E(X_1) - E(X_2) = 0 - 0 = 0$
- Variance: $V(Y) = V(X_1) + V(X_2) - 2 Cov(X_1, X_2) = 2 + 2 - 2(-1) = 4 + 2 = 6$
Therefore, $Y$ follows a normal distribution $N(0, 6)$.
Standardizing Y for Probability Calculation
To calculate $P[Y > 6]$, we standardize $Y$. The standard deviation is $\sigma_Y = \sqrt{V(Y)} = \sqrt{6}$. The standard normal variable is $Z = \frac{Y - E(Y)}{\sigma_Y} = \frac{Y - 0}{\sqrt{6}}$.
The probability transforms as:
$ P(Y > 6) = P\left( Z > \frac{6}{\sqrt{6}} \right) = P(Z > \sqrt{6}) $Probability using Standard Normal CDF $\Phi$
The standard normal cumulative distribution function is $\Phi(x)$. We use the property $P(Z > z) = \Phi(-z)$.
Applying this property:
$ P(Z > \sqrt{6}) = \Phi(-\sqrt{6}) $Conclusion
The probability $P[X_1 - X_2 > 6]$ equals $\Phi(-\sqrt{6})$.
Important Questions from Multivariate Normal Distribution
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. - Let $X \sim N_3 (\underline{\mu}, \Sigma)$ where $\underline{\mu} = (1, 1, 1)$ and
$\Sigma = \begin{pmatrix} 1 & 1 & 1 \\ 1 & 3 & c \\ 1 & c & 2 \end{pmatrix}$. The value of $c$ such that
$X_2$ and $-X_1 + X_2 - X_3$ are independent is
