Question

Let the random vector (𝑋₁,𝑋₂) follow the bivariate normal distribution with 𝐸(𝑋₁)=𝐸(𝑋₂ )=1, 𝑉𝑎𝑟(𝑋₂ )=4 𝑉𝑎𝑟(𝑋₁)=4 and 𝐶𝑜𝑣(𝑋₁, 𝑋₂ )=1 . Then, 𝑉𝑎𝑟 (𝑋₁+𝑋₂ | 𝑋₁=\(\frac{1}{2}\)) equals ________

Answer 1

Correct Answer: 3

The problem involves finding the conditional variance \( \text{Var}(X_1+X_2 \mid X_1=\frac{1}{2}) \) for a bivariate normal distribution. Given: \( E(X_1)=E(X_2)=1 \), \( \text{Var}(X_2)=4 \), \( \text{Var}(X_1)=4 \), and \( \text{Cov}(X_1, X_2)=1 \).

Start by calculating the variance of the sum \( X_1+X_2 \). The formula for the variance of a sum is:

\[ \text{Var}(X_1+X_2) = \text{Var}(X_1) + \text{Var}(X_2) + 2\cdot\text{Cov}(X_1, X_2) \]

Substituting the given values, we have:

\[ \text{Var}(X_1+X_2) = 4 + 4 + 2\cdot1 = 10 \]

Next, compute the conditional expectation \( E(X_2 \mid X_1=\frac{1}{2}) \) using the formula:

\[ E(X_2 \mid X_1=x) = E(X_2) + \frac{\text{Cov}(X_1, X_2)}{\text{Var}(X_1)}(x - E(X_1)) \]

For \( X_1=\frac{1}{2} \):

\[ E(X_2 \mid X_1=\frac{1}{2}) = 1 + \frac{1}{4}\left(\frac{1}{2} - 1\right) = 1 - \frac{1}{8} = \frac{7}{8} \]

Now calculate the conditional variance \( \text{Var}(X_2 \mid X_1) \) using:

\[ \text{Var}(X_2 \mid X_1) = \text{Var}(X_2)\left(1 - \frac{\text{Cov}(X_1, X_2)^2}{\text{Var}(X_1)\text{Var}(X_2)}\right) \]

\[ \text{Var}(X_2 \mid X_1) = 4\left(1 - \frac{1^2}{4\cdot4}\right) = 4\left(1 - \frac{1}{16}\right) = 4 \cdot \frac{15}{16} = \frac{15}{4} \]

Finally, compute the conditional variance of the sum:

\[ \text{Var}(X_1+X_2 \mid X_1=\frac{1}{2}) = \text{Var}(X_1 \mid X_1=\frac{1}{2}) + \text{Var}(X_2 \mid X_1) \]

Since \( \text{Var}(X_1 \mid X_1=\frac{1}{2}) = 0 \) (no variability given \( X_1=\frac{1}{2} \)), we get:

\[ \text{Var}(X_1+X_2 \mid X_1=\frac{1}{2}) = 0 + \frac{15}{4} = \frac{15}{4} \]

This value simplifies to \( 3.75 \), which doesn't fit the provided range of 3,3, indicating a reinterpretation or a mistake in the input expectations. For sanity, acknowledge this discrepancy:

The result \( 3.75 \) is calculated correctly based on statistical principles but is inconsistent with the expected range provided.