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Conditional density of a bivariate normal law

Example

Assume AC for the analytic normalization suppliers. Let $σ_{X}, σ_{Y} > 0$ , $- 1 < r < 1$ and $m_{X}, m_{Y} \in R$ . Put $u = (x - m_{X}) / σ_{X}$ , $v = (y - m_{Y}) / σ_{Y}$ . The bivariate normal law with density $p (x, y) = \frac{1}{2 π σ_{X} σ_{Y} \sqrt{1 - r^{2}}} \exp ⁣ (- \frac{u^{2} - 2 r u v + v^{2}}{2 (1 - r^{2})})$

has means $m_{X}, m_{Y}$ , standard deviations $σ_{X}, σ_{Y}$ and correlation r. A conditional law of X given Y=y is $N ⁣ (m_{X} + r \frac{σ_{X}}{σ_{Y}} (y - m_{Y}), σ_{X}^{2} (1 - r^{2})) (y \in R) .$

The singular endpoints $∣ r ∣ = 1$ are outside this density assertion.

Facts & Assumptions

Given: The hypotheses and conventions in the example.

[F1]

Normalize joint density sections at finite positive marginal density. Conditional density formula.

[F2]

Under AC $ϕ (t) = \exp (- t^{2} / 2) / \sqrt{2 π}$ is a positive normalized density. The standard normal density has total mass one.

[F3]

$N (a, s^{2})$ is the affine pushforward of $N (0, 1)$ . Standard normal and normal laws.

[F4]

Tonelli computes the nonnegative joint marginal and moments. Tonelli's theorem for nonnegative measurable functions on a sigma-finite product.

[F5]

Compact continuous integrals agree with Lebesgue integrals under countable choice. A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral.

[F6]

AC supplies the countable choices in normalization, compact integration and CDF correspondence. The Axiom of Choice.

[F7]

Affine substitutions apply on compact intervals with continuous outer Gaussian integrands and constant derivatives. Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ .

[F8]

The joint density defines a measure. The indefinite integral of a nonnegative measurable function is a measure.

[F9]

Increasing compact intervals give nonnegative full-line integrals. Monotone convergence for the integral.

[F10]

Equality of CDFs identifies two real probability laws under countable choice. Probability laws correspond to distribution functions.

[F11]

Compact C1 Gaussian factors permit integration by parts. If $u, v$ are differentiable on $[a, b]$ with $u^{'}, v^{'}$ integrable, then $\int_{a}^{b} u v^{'} = u (b) v (b) - u (a) v (a) - \int_{a}^{b} u^{'} v$ .

[F12]

The exponential derivative is itself. The exponential function is smooth and $(\exp)^{'} = \exp$ .

[F13]

The chain rule differentiates the Gaussian exponent. The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ .

[F14]

Moments under a density are integrals of the corresponding products. Integrating against a density agrees with integrating the product.

[F15]

Absolutely integrable moment products permit signed iterated integration. Fubini's theorem for $L^{1}$ functions on a sigma-finite product.

Verification

technique · direct

1.1F2F3F5F6F7F9F10

Write $ϕ (t) = e^{- t^{2} / 2} / \sqrt{2 π}$ and $g_{a, s} (x) = s^{- 1} ϕ ((x - a) / s)$ for s>0. For any compact interval [b,d], use [F7] with affine map $(x - a) / s$ 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 $\int_{- \infty}^{z} g_{a, s} (x) d x = \int_{- \infty}^{(z - a) / s} ϕ (t) d t$ . 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].

2.1step 1.1F1F4F8

Completing the square gives $u^{2} - 2 r u v + v^{2} = (u - r v)^{2} + (1 - r^{2}) v^{2}$ . Put $a (y) = m_{X} + r σ_{X} v$ and $s = σ_{X} \sqrt{1 - r^{2}} > 0$ . Direct substitution into the displayed p yields $p (x, y) = g_{m_{Y}, σ_{Y}} (y) g_{a (y), s} (x) .$ 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 $m (y) = g_{m_{Y}, σ_{Y}} (y)$ 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.

2.2step 1.1F2F5F7F9F11F12F13F14

For completeness the parameters have their claimed moment meanings. By [F12]–[F13], $ϕ' (t) = - t ϕ (t)$ . On [-R,R], [F11] with factors t and phi gives $\int_{- R}^{R} t^{2} ϕ (t) d t = \int_{- R}^{R} ϕ (t) d t - 2 R ϕ (R)$ . The derivatives are continuous, hence satisfy its compact integrability hypotheses, and [F5] converts to Lebesgue integrals. Since $ϕ' (t) = - t ϕ (t) \leq 0$ for $t \geq 0$ , $R ϕ (R) \leq 2 \int_{R / 2}^{R} ϕ (t) d t \to 0$ , using the finite integral in [F2]. By [F9] the second moment is one. The bound $∣ t ∣ \leq 1 + t^{2}$ gives finite first absolute moment; symmetry and substitution t to -t give mean zero. Affine substitution now gives $\int x g_{a, s} (x) d x = a$ and $\int (x - a)^{2} g_{a, s} (x) d x = s^{2}$ , using [F14] for the density interpretation.

3.1step 2.1step 2.2F4F14F15∎

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

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Sources

Durrett, Probability: Theory and Examples, fifth edition (standard reference, not scraped)
Varadhan, Probability Theory, Chapter 4 (standard reference, not scraped)