# Find $P(X>1|Y=y)$ if $(X,Y)$ has bivatiate normal distribution

Let $(X,Y)$ bivariate normal distributed with $\mu_X=\mu_Y=0$ and $\sigma_X=\sigma_Y=1$ and correlation $\rho$. I want to express $P(X>1 | Y=y)$ in terms of $\rho$, $y$ and $Q$-function.

I am able to express $P(X>1)$ in terms of $Q$-function and it is equal to $Q(1)$. How should I incorporate the condition here?

• Bivariate normal distributed and the parameters are....? – Shashi Jan 29 '18 at 18:21
• mx=my=0 sigma(x)=sigma(y)=1 – Mona Jalal Jan 29 '18 at 19:01
• I hope you are okay with my edits. And do you mind to define the $Q$-function? Is it true that $Q(z)=1-\Phi(z)$ where $\Phi$ is the CDF of the standard normal distribution? – Shashi Jan 29 '18 at 19:22

## 1 Answer

Using the conditional density we get:

\begin{align} P(X>1|Y=y)=\int^\infty_1 f_{X|Y}(x|y)\,dx=\int^\infty_1 \frac{f_{X,Y}(x,y)}{f_Y(y)}\,dx \end{align} Now the problem is to find the integrand. After some simplifying the terms we get: \begin{align} \frac{f_{X,Y}(x,y)}{f_Y(y)}= \frac{1}{\sqrt[]{2\pi (1-\rho^2)}}\exp\left(-\frac{(x-\rho y)^2}{2(1-\rho^2)}\right) \end{align} So now we get: \begin{align} P(X>1|Y=y)=\int^\infty_1 \frac{1}{\sqrt[]{2\pi (1-\rho^2)}}\exp\left(-\frac{(x-\rho y)^2}{2(1-\rho^2)}\right)\,dx \end{align} Substitute $u=(x-\rho y)/\sqrt[]{1-\rho^2}$ so that $du=dx/\sqrt[]{1-\rho^2}$: \begin{align} P(X>1|Y=y)=\int^\infty_{\frac{1-\rho y}{\sqrt[]{1-\rho^2}}}\ \ \frac{1}{\sqrt[]{2\pi}}e^{-x^2/2}\,dx=1-\Phi\left(\frac{1-\rho y}{\sqrt[]{1-\rho^2}} \right) \end{align} where $\Phi(\cdot)$ is the CDF of the standard normal distribution. I think there is a relation between $\Phi$ and the $Q$-function you are talking about?