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Exam P · Bivariate Normal Basics · Tutorial 870 of 915

Conditioning on an Interval of the Other Variable

Use the conditional normal distribution and the marginal density to calculate probabilities given that the other variable lies in an interval.

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What You'll Learn

  • Set up conditional probabilities given that a bivariate normal variable lies in an interval.
  • Weight conditional normal probabilities by the marginal density of the conditioning variable.
  • Normalize the integral by the probability of the conditioning interval.
  • Approximate the required integrals with Simpson’s rule.
  • Recognize why conditioning at the interval midpoint is generally incorrect.

Why an Interval Condition Requires Averaging

Knowing that \(Y\) lies between two values does not tell us its exact value. In a bivariate normal model, the conditional distribution of \(X\) changes with \(Y\), so the probability of an event involving \(X\) can vary across the interval. The desired conditional probability must account for all possible values of \(Y\) in that interval, giving more weight to values that are more likely under the marginal distribution of \(Y\).

This is a continuous version of conditioning over several possible cases: calculate the conditional probability at each possible value of \(Y\), weight it by the density of \(Y\), and integrate. Finally, divide by the probability that \(Y\) is in the specified interval. This distinction is important in exam problems: conditioning on \(Y=y\) at one fixed value and conditioning on \(a<Y<b\) are different calculations.

Definition: Suppose \((X,Y)\) has a nondegenerate bivariate normal distribution, with \(|\rho|<1\), and \(a<b\). For an event \(A\) determined by \(X\), the conditional probability given that \(Y\) lies in \((a,b)\) is $$ P(A\mid a<Y<b) = \frac{\displaystyle\int_a^b P(A\mid Y=y) f_Y(y)\,dy} {\displaystyle\int_a^b f_Y(y)\,dy}, $$ provided \(P(a<Y<b)>0\). Here \(f_Y\) is the marginal density of \(Y\). Including or excluding the interval endpoints does not change the result for a continuous \(Y\).

The Bivariate Normal Calculation

Let \(X\) and \(Y\) have means \(\mu_X,\mu_Y\), standard deviations \(\sigma_X,\sigma_Y>0\), and correlation \(\rho\), where \(|\rho|<1\). The conditional distribution is

$$ X\mid Y=y\sim N\left(\mu_X+\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y),\ \sigma_X^2(1-\rho^2)\right). $$

The second parameter here is a variance. Thus the conditional standard deviation is \(\sigma_X\sqrt{1-\rho^2}\), which does not depend on \(y\). For an event \(A=\{X>c\}\), for example, the conditional probability in the numerator is

$$ P(X>c\mid Y=y) = 1-\Phi\left( \frac{c-\mu_X-\rho\frac{\sigma_X}{\sigma_Y}(y-\mu_Y)} {\sigma_X\sqrt{1-\rho^2}} \right), $$

where \(\Phi\) is the standard normal cumulative distribution function. The density weighting is \(f_Y(y)\), the marginal normal density with mean \(\mu_Y\) and standard deviation \(\sigma_Y\). The denominator is \(P(a<Y<b)\), which can be obtained from that marginal distribution.

Why the Density-Weighted Integral Holds

The joint density factors into the marginal density of \(Y\) and the conditional density of \(X\) given \(Y=y\). Integrating the joint density over \(X\) values in the event \(A\) and over \(a<y<b\) therefore gives the joint probability of \(A\) and the interval condition. Dividing by the probability of the interval gives the conditional probability.

$$ \begin{aligned} P(A,\ a<Y<b) &=\int_a^b\int_{\{x:\,x\in A\}} f_{X,Y}(x,y)\,dx\,dy\\ &=\int_a^b P(A\mid Y=y)f_Y(y)\,dy,\\[4pt] P(A\mid a<Y<b) &=\frac{P(A,\ a<Y<b)}{P(a<Y<b)}. \end{aligned} $$

The factor \(f_Y(y)\) is essential: values of \(y\) with greater density contribute more to the average. If the conditional probability is constant throughout the interval—for instance, when \(X\) and \(Y\) are independent—then the integral simplifies to that same constant. Otherwise, replacing the integral with a conditional probability at one selected value is generally not exact.

Worked Examples

Worked Example: A Positive Correlation and a Lower-Tail Event

Suppose \((X,Y)\) is standard bivariate normal with correlation \(\rho=0.6\). Calculate \(P(X>0\mid 0<Y<1)\). Use composite Simpson’s rule with four equal subintervals, and report the probability to four decimal places.

Method. Since both marginal variables are standard normal, \(f_Y(y)=\phi(y)\). The conditional distribution is \(X\mid Y=y\sim N(0.6y,0.64)\), with conditional standard deviation \(\sqrt{0.64}=0.8\). Hence

$$ P(X>0\mid Y=y)=\Phi\left(\frac{0.6y}{0.8}\right)=\Phi(0.75y). $$

The numerator is the integral of \(g(y)=\phi(y)\Phi(0.75y)\) over \([0,1]\). The denominator is \(P(0<Y<1)=\Phi(1)-\Phi(0)=0.8413447-0.5000000=0.3413447\). The Simpson spacing is \(h=(1-0)/4=0.25\).

Evaluating the integrand at the five equally spaced points gives, to six decimal places,

$$ \begin{array}{c|ccccc} y&0&0.25&0.50&0.75&1\\ \hline g(y)&0.199471&0.222089&0.227494&0.214745&0.187134 \end{array} $$

Thus the Simpson approximation to the numerator is

$$ \begin{aligned} \int_0^1 g(y)\,dy &\approx\frac{0.25}{3}\left[ 0.199471+0.187134 +4(0.222089+0.214745)+2(0.227494) \right]\\ &=\frac{0.25}{3}(2.588929)\\ &\approx 0.215744. \end{aligned} $$

The conditional probability is therefore \(0.215744/0.3413447\approx 0.6320\), to four decimal places. As a check, the result lies between zero and one, and the positive correlation makes \(X>0\) more likely when \(Y\) is positive than its unconditional probability of \(0.5\).

Common mistake avoided: Using \(P(X>0\mid Y=0.5)=\Phi(0.375)\) as the answer would condition on one fixed value, not on the whole interval. It does not account for how the conditional probability and density vary over \(0<Y<1\).

Worked Example: A Negative Conditional Mean

Let \((X,Y)\) be standard bivariate normal with correlation \(\rho=0.5\). Find \(P(X<0\mid 0<Y<1)\), using composite Simpson’s rule with four equal subintervals. Report the probability to four decimal places.

Method. Here \(X\mid Y=y\sim N(0.5y,0.75)\), so its conditional standard deviation is \(\sqrt{0.75}=\sqrt{3}/2\). Consequently,

$$ P(X<0\mid Y=y) =\Phi\left(\frac{-0.5y}{\sqrt{3}/2}\right) =\Phi\left(-\frac{y}{\sqrt{3}}\right). $$

The numerator integrand is \(g(y)=\phi(y)\Phi(-y/\sqrt{3})\). The denominator is again \(\Phi(1)-\Phi(0)=0.3413447\), and \(h=0.25\). Evaluating the displayed integrand gives the following values, rounded to six decimal places:

$$ \begin{array}{c|ccccc} y&0&0.25&0.50&0.75&1\\ \hline g(y)&0.199471&0.171146&0.136043&0.100129&0.068200 \end{array} $$

Applying Simpson’s rule to these values,

$$ \begin{aligned} \int_0^1 g(y)\,dy &\approx\frac{0.25}{3}\left[ 0.199471+0.068200 +4(0.171146+0.100129)+2(0.136043) \right]\\ &=\frac{0.25}{3}(1.624857)\\ &\approx 0.135405. \end{aligned} $$

Dividing by the interval probability gives \(0.135405/0.3413447\approx 0.3967\), to four decimal places. The answer is below \(0.5\), as expected: throughout the conditioning interval, \(Y\) is positive and the conditional mean of \(X\) is positive.

Common mistake avoided: At the midpoint \(y=0.5\), the conditional probability is approximately \(\Phi(-0.288675)=0.3864\). Substituting that single value for the interval-conditioned probability ignores the density-weighted contributions from all other \(y\) values; the interval answer is approximately \(0.3967\), not \(0.3864\).

Worked Example: Comparing \(X\) with the Conditioning Variable

Suppose \((X,Y)\) is standard bivariate normal with correlation \(\rho=0.6\). Calculate \(P(X>Y\mid 0<Y<1)\) by Simpson’s rule with four equal subintervals, reporting the result to four decimal places.

Method. Given \(Y=y\), the event \(X>Y\) is the event \(X>y\). The conditional distribution remains \(X\mid Y=y\sim N(0.6y,0.64)\), so

$$ P(X>Y\mid Y=y) =1-\Phi\left(\frac{y-0.6y}{0.8}\right) =\Phi(-0.5y). $$

The numerator integrand is now \(g(y)=\phi(y)\Phi(-0.5y)\), not the integrand for \(P(X>0\mid 0<Y<1)\). The denominator remains \(0.3413447\). At the five Simpson points, the integrand values rounded to six decimals are \(0.199471,\ 0.174102,\ 0.141281,\ 0.106551,\ 0.074657\). Therefore,

$$ \begin{aligned} \int_0^1g(y)\,dy &\approx\frac{0.25}{3}\left[ 0.199471+0.074657 +4(0.174102+0.106551)+2(0.141281) \right]\\ &=\frac{0.25}{3}(1.679302)\\ &\approx 0.139942. \end{aligned} $$

The required probability is approximately \(0.139942/0.3413447=0.4100\), to four decimal places. The event threshold changes with \(y\), so it must be substituted into the conditional normal probability before integrating. A useful check is that the computed conditional probability is below \(0.5\): on this interval, the conditional mean \(0.6y\) is less than the threshold \(y\).

Common mistake avoided: Treating \(X>Y\) as \(X>0\) would use the wrong event except at \(y=0\). The threshold in the conditional probability is \(y\), not zero.

A Reliable Calculation Process

For exam problems, separate the conditioning step from the integration step. The following process keeps the event, the density weight, and the normalization distinct.

1
Write the interval condition.
Identify the bounds \(a,b\) and calculate \(P(a<Y<b)\) from the marginal distribution of \(Y\).
2
Find the conditional distribution.
Use the bivariate normal conditional mean and variance, checking that the second parameter is a variance.
3
Express the event probability at \(Y=y\).
Standardize the event using the conditional mean and standard deviation. If the event threshold depends on \(Y\), retain \(y\) in that threshold.
4
Weight and integrate.
Multiply the conditional event probability by \(f_Y(y)\), integrate from \(a\) to \(b\), and divide by \(P(a<Y<b)\).

For a numerical integral, state the method and its grid. Composite Simpson’s rule with an even number \(n\) of subintervals uses spacing \(h=(b-a)/n\), weights \(1,4,2,4,\ldots,2,4,1\), and the multiplier \(h/3\). Keep enough digits in the integrand calculations to avoid accumulating rounding error, and round the final probability only at the end.

Common Exam Traps

  • Using a single conditional value. \(P(A\mid Y=y)\) is a function of \(y\); it is not generally equal to \(P(A\mid a<Y<b)\) at the midpoint.
  • Forgetting the density weight. Integrating \(P(A\mid Y=y)\) alone treats all values of \(Y\) in the interval as equally weighted. The correct integrand includes \(f_Y(y)\).
  • Omitting the denominator. The integral in the numerator is \(P(A,\ a<Y<b)\), a joint probability. To obtain a probability conditional on the interval, divide by \(P(a<Y<b)\).
  • Using the wrong conditional standard deviation. The conditional variance is \(\sigma_X^2(1-\rho^2)\); standardization uses \(\sigma_X\sqrt{1-\rho^2}\).
  • Changing a moving event into a fixed threshold. For \(X>Y\), the threshold in \(P(X>Y\mid Y=y)\) is \(y\). It is not the threshold for \(X>0\).

This method applies when the bivariate normal model is nondegenerate and the conditioning interval has positive probability. At perfect correlation, the conditional distribution is degenerate and the density formula above does not apply; the variables must instead be treated through their exact linear relationship. For ordinary nondegenerate bivariate normal models, interval conditioning is a direct application of the conditional normal distribution and the marginal density, aligned with the SOA Exam P (Probability) syllabus.

Check Your Understanding

For each problem, set up the density-weighted numerator and normalize by the probability of the conditioning interval.

  1. Let \((X,Y)\) be standard bivariate normal with correlation \(0.4\). Find \(P(X>0\mid -1<Y<1)\). Use symmetry of the interval and conditional probabilities at \(y\) and \(-y\). (Answer: \(0.5000\))
  2. Let \((X,Y)\) be standard bivariate normal with correlation \(0.6\). Estimate \(P(X>0\mid 0<Y<1)\) using the four-subinterval Simpson calculation in the tutorial. (Answer: approximately \(0.6320\))
  3. Let \((X,Y)\) be standard bivariate normal with correlation \(0.5\). Set up, but do not numerically evaluate, \(P(X<0\mid 0<Y<1)\) as a ratio of integrals using \(\phi\) and \(\Phi\). (Answer: \(\displaystyle\frac{\int_0^1\phi(y)\Phi(-y/\sqrt{3})\,dy}{\Phi(1)-\Phi(0)}\))
  4. Let \(X,Y\) be standard bivariate normal with correlation \(0.5\). Find \(P(X>0\mid Y>0)\). You may use the standard quadrant probability \(P(X>0,Y>0)=\frac14+\frac{\arcsin(\rho)}{2\pi}\). (Answer: \(\frac23\), approximately \(0.6667\))
  5. Let \((X,Y)\) be standard bivariate normal with correlation \(0.6\). For \(P(X>Y\mid 0<Y<1)\), write the conditional event probability as a function of \(y\), and identify the correct numerator integrand. (Answer: \(P(X>Y\mid Y=y)=\Phi(-0.5y)\); integrand \(=\phi(y)\Phi(-0.5y)\))

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