Conditional density of a bivariate normal law

Write $\varphi (t)=e^{-t^2\mathrm{/}2}\mathrm{/}\sqrt{2\pi }$ and $g_{a,s}(x)=s^{-1}\varphi ((x-a)\mathrm{/}s)$ for s>0. For any compact interval [b,d], use [F7] with affine map $(x-a)\mathrm{/}s$ and continuous outer function phi; its derivative is the integrable constant 1/s. By [F5] this proves the same substitution for Lebesgue integrals. Let b decrease to minus infinity and d increase to infinity, using [F9], to get $\int g_{a,s}=1$ from [F2]. For a fixed upper endpoint z, the identical limiting argument gives ${\int }_{-\mathrm{\infty }}^z{g}_{a,s}(x)dx={\int }_{-\mathrm{\infty }}^{(z-a)\mathrm{/}s}\varphi (t)dt$. This is the CDF of the affine law [F3], since s>0. Thus [F10] identifies the density law with $N(a,s^2)$. All countable-choice hypotheses are supplied by [F6].

Completing the square gives $u^2-2ruv+v^2=(u-rv{)}^2+(1-r^2)v^2$. Put $a(y)=m_X+r{\sigma }_{X}v$ and $s={\sigma }_X\sqrt{1-r^2}>0$. Direct substitution into the displayed p yields $p(x,y)=g_{m_Y,{\sigma }_Y}(y)g_{a(y),s}(x)\mathrm{.}$ It is positive and product-measurable: it is obtained from measurable coordinate projections by continuous arithmetic and exponential operations with fixed nonzero denominators. Step 1.1 and [F4] give marginal $m(y)=g_{m_Y,{\sigma }_Y}(y)$ and total mass $\int m=1$. Hence [F8] constructs the joint probability. Its marginal is finite positive at every y. Apply [F1]: the normalized section is exactly $g_{a(y),s}$, which step 1.1 identifies as the asserted normal law. No exceptional filler is needed here. When r=0 this conditional density is independent of y and has the original X parameters.

For completeness the parameters have their claimed moment meanings. By [F12]–[F13], $\varphi \mathrm{\prime }(t)=-t\varphi (t)$. On [-R,R], [F11] with factors t and phi gives ${\int }_{-R}^R{t}^2\varphi (t)dt={\int }_{-R}^R\varphi (t)dt-2R\varphi (R)$. The derivatives are continuous, hence satisfy its compact integrability hypotheses, and [F5] converts to Lebesgue integrals. Since $\varphi \mathrm{\prime }(t)=-t\varphi (t)\le 0$ for $t\ge 0$, $R\varphi (R)\le 2{\int }_{R\mathrm{/}2}^R\varphi (t)dt\to 0$, using the finite integral in [F2]. By [F9] the second moment is one. The bound $\mathrm{\mid }t\mathrm{\mid }\le 1+t^2$ gives finite first absolute moment; symmetry and substitution t to -t give mean zero. Affine substitution now gives $\int x{g}_{a,s}(x)dx=a$ and $\int (x-a{)}^2{g}_{a,s}(x)dx=s^2$, using [F14] for the density interpretation.

Using the factorization of step 2.1 and step 2.2, nonnegative Tonelli gives $E(Y-m_Y{)}^2={\sigma }_Y^2$ and $E(X-m_X{)}^2=\int (s^2+(a(y)-m_X{)}^2)m(y)dy={\sigma }_X^2(1-r^2)+r^2{\sigma }_X^2={\sigma }_X^2\mathrm{.}$ This also gives finite absolute first moments, so [F15] permits signed integration and yields $EX=m_X$, $EY=m_Y$. The product $(X-m_X)(Y-m_Y)$ is absolutely integrable because $2\mathrm{\mid }ab\mathrm{\mid }\le a^2+b^2$. Thus [F15] again gives $E[(X-m_X)(Y-m_Y)]=\int r\frac{{\sigma }_X}{{\sigma }_Y}(y-m_Y{)}^{2}m(y)dy=r{\sigma }_X{\sigma }_Y\mathrm{.}$ Division by the positive standard deviations gives correlation r. At $\mathrm{\mid }r\mathrm{\mid }=1$ the displayed density denominator and conditional scale cease to be positive, so neither the normalized density nor this density argument asserts that singular case.