# Using Orthogonal Decomposition of Bivariate Normally Distributed X,Y Reveals Different Result Than Calculating Variance Directly

This is the exercise description:

6.5.4. Suppose X and Y are standard normal variables. Find an expression for P(X + 2Y ≤ 3) in terms of the standard normal distribution function Φ, (a) in case X and Y are independent; (b) in case X and Y have bivariate normal distribution with correlation 1/2.

In particular, for exercise b) I use the fact that they're bivariate normally distributed, to do orthogonal decomposition on Y, such that

$$Y = \frac{1}{2} \times X + \sqrt{\frac{3}{4}} \times Z$$, where $$Z \sim \mathcal{N}(0,1)$$ and Z and X are independent.

Which turns $$X + 2 \times Y = 2\times X + \sqrt{3} \times Z$$. We want $$P(X + 2 \times Y \leq 3)$$ in terms of $$\Phi$$, so we have figure out the STD or VAR of the expression.

For linear combinations of independent normally distributed variables we have that the new distribution is just the sum of their parameters.

Which gives us $$W = 2\times X + \sqrt{3} \times Z \sim \mathcal{N}(0,4 + 3)$$, such that $$P(W\leq3) = \Phi(\frac{3}{\sqrt{7}})$$. However, from the solutions manual, we have:

I am sure the solution is correct, but what I don't understand is; How is mine wrong?

The solution manual has an error. The $$2\text{Cov}(X,Y)$$ should be $$2\text{Cov}(X, 2Y) = 4 \text{Cov}(X, Y)$$. Then the variance of $$X+2Y$$ should be $$7$$, not $$6$$.