How to solve right-tail probability problems?

Question

I have been trying to learn probability by solving problems. I came across this question. I have no idea how to even start solving this one. Can someone help me? Where can I learn about solving these kind of problems.

Let $Q(b)$ denote the right-tail probability of a standard Gaussian random variable. Suppose that X and Y are jointly-distributed Gaussian random variables with $X \sim N( 3 , 8 )$ and $Y \sim N ( 10 , 5)$. If the correlation coefficient between X and Y is 0.8, then $P[Y>0|X = 3] = Q(b)$, where b is:

1. -7.453
2. -4.431
3. 2.639
4. -4.671

If I have to guess it then I would guess the positive value, which is $2.639$, because it is the right-tail value. Since I do not know how to solve it, I can't check. I would really love learn how to solve it as well as learn a way to make sense of it intuitively.

Any help is highly appreciated.

Thanks!

Answer 1

Let $T\sim N(0,1)$ independent from $X$, and take $Z=a + bX + cT$. The expected value of $Z$ is $a+3b$, the variance of $Z$ is $8b^2+c^2$, and the correlation coefficient between $X$ and $Z$ is $$\frac{E(XZ)-EXEZ}{\sigma_x \sigma z} = \frac{3a+ 17b - 3\cdot (a+3b)}{\sqrt{8(8b^2+c^2)}}=\frac{8b}{\sqrt{64b^2+8c^2}}$$

For $Y$ and $Z$ to have the same distribution, you need $a+3b=10$, $8b^2+c^2=5$ and $8b / \sqrt{64b^2+8c^2} = 0.8$. Now you just need to solve for $a$, $b$ and $c$. One solution is $a\approx 8.10263$, $b \approx 0.632456$ and $c\approx 1.34164$.

Now back to the question: $$P(Y > 0 | X=3) = P(a+bX+cT > 0 | X=3) = P(a+3b+cT > 0) = P(T > (-a-3b)/c)$$ So the answer is $(-a-3b)/c \approx -7.454$.

Answer 2

I solved it this way:

We need to calculate: $P(Y>0|X=3)$ and we are given: $\mu_x= 3$, $\mu_y= 10$, $\sigma_x = \sqrt{8}$, $\sigma_y = \sqrt{5}$, and $\rho= 0.8$

We have:

$E[Y|X=x] =\mu_y + \rho*\sigma_y \frac{x-\mu_x}{\sigma_x}$

$Var[Y|X=x]= (1-\rho^2)*\sigma_y^2$

We get :

$E[Y|X=3] =10$

$\sigma^2[Y|X=3]$=$Var[Y|X=3]=1.8$

Now using,

$P(Y>y|X=x)= 1- \phi(\frac{y-E[Y|X=x]}{\sigma[Y|X=x]})$

We get: $P(Y>y|X=x)= 1- \phi(\frac{-10}{\sqrt(1.8)})$ -------(1)

and since $Q(b) = 1- \phi(\frac{b}{1})$ ------(2)

Equating (1) and (2), we get $b = -7.454$

The problem with solving this way is that you need to memorize bunch of formula.