Question

Let \((X,Y)\) be a random vector having the bivariate normal distribution with \(E(X)=2\), \(E(Y)=10\), \(Var(X)=9\), \(Var(Y)=25\), and the correlation coefficient between \(X\) and \(Y\) equals \(\rho>0\). If

Hint:

For bivariate normal distributions, the conditional variance is reduced by the factor \((1-\rho^2)\).

Answer 1

Correct Answer: 0.6

Step 1: Write the conditional distribution of \(Y|X=x\).
For a bivariate normal distribution,
\[ Y|X=x \sim N\left( \mu_Y+\rho\frac{\sigma_Y}{\sigma_X}(x-\mu_X), \; \sigma_Y^2(1-\rho^2) \right) \] Given
\[ \mu_X=2,\qquad \mu_Y=10 \] \[ \sigma_X=\sqrt{9}=3,\qquad \sigma_Y=\sqrt{25}=5 \] Now, conditioning on
\[ X=2 \] gives
\[ E(Y|X=2) = 10+\rho\frac{5}{3}(2-2) = 10 \] and
\[ Var(Y|X=2) = 25(1-\rho^2) \] Therefore,
\[ Y|X=2 \sim N\left(10,\;25(1-\rho^2)\right) \]

Step 2: Standardize the probability.
We are given
\[ P(4<Y<16\mid X=2)=0.8664 \] Now,
\[ P\left( \frac{4-10}{5\sqrt{1-\rho^2}} < Z < \frac{16-10}{5\sqrt{1-\rho^2}} \right) = 0.8664 \] So,
\[ P\left( -\frac{6}{5\sqrt{1-\rho^2}} < Z < \frac{6}{5\sqrt{1-\rho^2}} \right) = 0.8664 \] Using symmetry of the standard normal distribution,
\[ 2\Phi\left( \frac{6}{5\sqrt{1-\rho^2}} \right)-1 = 0.8664 \] Thus,
\[ \Phi\left( \frac{6}{5\sqrt{1-\rho^2}} \right) = 0.9332 \] From the given table value,
\[ \Phi(1.5)=0.9332 \] Hence,
\[ \frac{6}{5\sqrt{1-\rho^2}}=1.5 \]

Step 3: Solve for \(\rho\).
\[ 6 = 7.5\sqrt{1-\rho^2} \] \[ \sqrt{1-\rho^2} = \frac{6}{7.5} = 0.8 \] Squaring both sides,
\[ 1-\rho^2=0.64 \] \[ \rho^2=0.36 \] Since \(\rho>0\),
\[ \rho=0.6 \] Rounded off to two decimal places,
\[ 0.60 \]

Step 4: Final conclusion.
Hence,
\[ \boxed{0.60} \]