Formulae & Notes
Joint PDF of \((X,Y)\) integrates to 1; probability over a region \(R\) is \(\iint_R f_{XY}\). Marginals: \(f_X(x)=\int f_{XY}(x,y)\,dy\), \(f_Y(y)=\int f_{XY}(x,y)\,dx\). Conditional density: \(f_{Y|X}(y|x)=\frac{f_{XY}(x,y)}{f_X(x)}\).
Mathematics used
\[ f_{XY}(x,y)=\frac{1}{2\pi\sigma_x\sigma_y\sqrt{1-\rho^2}} \exp\!\left(-\frac{1}{2(1-\rho^2)}\Big[(\tfrac{x-\mu_x}{\sigma_x})^2 -2\rho(\tfrac{x-\mu_x}{\sigma_x})(\tfrac{y-\mu_y}{\sigma_y}) +(\tfrac{y-\mu_y}{\sigma_y})^2\Big]\right). \] \(X\sim\mathcal N(\mu_x,\sigma_x^2)\), \(Y\sim\mathcal N(\mu_y,\sigma_y^2)\).
FYI Math — Worked Example (study vs. sleep)
Setup. Let \(X\) = study hours and \(Y\) = sleep hours. Use \(\mu_x=3,\ \sigma_x=1,\ \mu_y=7,\ \sigma_y=1.2,\ \rho=-0.4\). Rectangle: \([2,4]\times[6,8]\).
Goal. \[ p=\Pr(2\lt X\lt 4,\ 6\lt Y\lt 8) =\iint_{[2,4]\times[6,8]} f_{XY}(x,y)\,dx\,dy. \]
- Independence baseline (\(\rho=0\)). \[ \Pr(2\lt X\lt 4)=\Phi(1)-\Phi(-1)=0.682689\ldots \] \[ \Pr(6\lt Y\lt 8)=\Phi\!\Big(\tfrac{8-7}{1.2}\Big)-\Phi\!\Big(\tfrac{6-7}{1.2}\Big) =\Phi(0.8333\ldots)-\Phi(-0.8333\ldots)=0.595343\ldots \] \[ p_0=\Pr(2\lt X\lt 4)\Pr(6\lt Y\lt 8)\approx \boxed{0.4064}. \]
- With correlation via conditioning (\(\rho=-0.4\)). For the bivariate normal, \[ Y\mid X=x \sim \mathcal N\!\big(m(x),\, s^2\big),\quad m(x)=\mu_y+\rho\frac{\sigma_y}{\sigma_x}(x-\mu_x),\quad s=\sigma_y\sqrt{1-\rho^2}. \] Numerically, \[ m(x)=7+(-0.4)\cdot1.2\,(x-3)=\boxed{8.44-0.48x},\qquad s=\boxed{1.0998\ldots}. \] Hence \[ p=\int_{2}^{4} f_X(x)\left[ \Phi\!\Big(\tfrac{8-m(x)}{s}\Big)-\Phi\!\Big(\tfrac{6-m(x)}{s}\Big) \right]\,dx \ \approx\ \boxed{0.4258}. \] (Monte Carlo with \(n=60{,}000\): \(\hat p\approx 0.4275\).)
Interpretation. The negative correlation tilts contours; here it slightly increases mass in the rectangle compared with independence (\(0.4258\) vs \(0.4064\)).